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bg-background","children":["$","$L3b",null,{"backHref":"/a-level/maths/ccea/questions","basePath":"/a-level/maths/ccea/trigonometry-8/questions","formulaUrl":"https://ccea.org.uk/downloads/docs/Support/General/2019/Mathematical%20Formulae%20and%20Tables.pdf","initialIndex":77,"initialQuestionWindow":[{"id":"162bd4d0-a91c-4e9f-a578-46fb86c6548b","revisionId":"0511afa3-d030-498d-9a44-c2d9062b5997","questionNumber":null,"questionType":"LA","specification":"Let \$ f(\\theta) = 3\\cos\\theta - 6\\sin\\theta \$ for \$ \\theta \\in \\mathbb{R} \$.","difficulty":5,"difficultyBand":"medium","tier":null,"totalMarks":10,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"4d7c0a63-9f56-41ea-b63b-defe1e6fa859","label":"a","content":"Express \$ f(\\theta) \$ in the form \$ R\\cos(\\theta + \\alpha) \$, where \$ R > 0 \$ and \$ 0 < \\alpha < \\frac{\\pi}{2} \$. Give the exact value of \$ R \$ and the value of \$ \\alpha \$, in radians, to 3 decimal places.","markscheme":"- Calculates $R = \\sqrt{3^2 + 6^2} = \\sqrt{45} = 3\\sqrt{5}$ B1\n- Uses $\\tan \\alpha = \\frac{6}{3}$ (or $2$) M1\n- Finds $\\alpha = 1.107$ (to 3 d.p.) A1","marks":3,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"expansion_and_comparison","label":"Expansion and Equating Coefficients","steps":[{"type":"text","order":0,"title":"State the identity","content":"To express $3\\cos\\theta - 6\\sin\\theta$ in the form $R\\cos(\\theta + \\alpha)$, we start by using the compound angle identity for cosine:\n\n$$\\cos(A + B) = \\cos A \\cos B - \\sin A \\sin B$$\n\nApplying this to our expression gives:\n\n$$R\\cos(\\theta + \\alpha) = R(\\cos\\theta\\cos\\alpha - \\sin\\theta\\sin\\alpha)$$\n\nExpanding the brackets:\n\n$$R\\cos(\\theta + \\alpha) = (R\\cos\\alpha)\\cos\\theta - (R\\sin\\alpha)\\sin\\theta$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=orig}{\\color{@sky}{3}}\\cos\\theta - \\htmlData{anchor=orig-sin}{\\color{@amber}{6}}\\sin\\theta = \\htmlData{anchor=exp-cos}{\\color{@sky}{R\\cos\\alpha}}\\cos\\theta - \\htmlData{anchor=exp-sin}{\\color{@amber}{R\\sin\\alpha}}\\sin\\theta$"},{"latex":"$$\\htmlData{anchor=eq1}{\\color{@sky}{R\\cos\\alpha = 3}}$"},{"latex":"$$\\htmlData{anchor=eq2}{\\color{@amber}{R\\sin\\alpha = 6}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"orig"},{"color":"amber","style":"text-color","anchor":"orig-sin"},{"color":"sky","style":"box","anchor":"eq1"},{"color":"amber","style":"box","anchor":"eq2"}],"connections":[{"to":"eq1","from":"orig","color":"sky","label":"equate cos coefficient","style":"solid"},{"to":"eq2","from":"orig-sin","color":"amber","label":"equate sin coefficient","style":"solid"}]},"order":1,"title":"Equate the coefficients","description":"We now compare our expanded expression with the given function $f(\\theta) = 3\\cos\\theta - 6\\sin\\theta$ to equate the coefficients of $\\cos\\theta$ and $\\sin\\theta$."},{"type":"text","order":2,"title":"Solve for R","content":"To find the value of $R$, we square and add our two equations:\n\n$$(R\\sin\\alpha)^2 + (R\\cos\\alpha)^2 = 6^2 + 3^2$$\n\nFactor out $R^2$ using the Pythagorean identity $\\sin^2\\alpha + \\cos^2\\alpha = 1$:\n\n$$R^2(\\sin^2\\alpha + \\cos^2\\alpha) = 36 + 9$$\n\n$$R^2 = 45$$\n\nSince $R > 0$, we find the exact value of $R$ by simplifying the square root:\n\n$$R = \\sqrt{45} = \\sqrt{9 \\times 5} = 3\\sqrt{5}$$\n\n(This earns 1 mark)"},{"type":"text","order":3,"title":"Solve for alpha","content":"To find $\\alpha$, we divide the equation for $\\sin\\alpha$ by the equation for $\\cos\\alpha$ to use the identity $\\tan\\alpha = \\frac{\\sin\\alpha}{\\cos\\alpha}$:\n\n$$\\frac{R\\sin\\alpha}{R\\cos\\alpha} = \\frac{6}{3}$$\n\n$$\\tan\\alpha = 2$$\n\n(This earns 1 mark)\n\nNow, solve for $\\alpha$ using the inverse tangent function in radian mode:\n\n$$\\alpha = \\arctan(2) \\approx 1.107\\text{ radians}$$\n\n(This earns 1 mark)"},{"type":"text","order":4,"title":"State the final expression","content":"Substitute the exact value of $R$ and the rounded value of $\\alpha$ back into our target form:\n\n$$f(\\theta) = 3\\sqrt{5}\\cos(\\theta + 1.107)$$"}]},{"id":"direct_formula","label":"Direct R-Formula Application","steps":[{"type":"text","order":0,"title":"Identify the standard form","content":"We want to express $f(\\theta) = 3\\cos\\theta - 6\\sin\\theta$ in the form $R\\cos(\\theta + \\alpha)$.\n\nUsing the direct $R$-formula for a cosine expansion:\n\n$$a\\cos\\theta - b\\sin\\theta = R\\cos(\\theta + \\alpha)$$\n\nComparing both expressions, we can directly identify the coefficients:\n- $a = 3$\n- $b = 6$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$R = \\sqrt{\\htmlData{anchor=a-val}{\\color{@sky}{a}}^2 + \\htmlData{anchor=b-val}{\\color{@amber}{b}}^2}$"},{"latex":"$$R = \\sqrt{\\htmlData{anchor=a-sub}{\\color{@sky}{3}}^2 + \\htmlData{anchor=b-sub}{\\color{@amber}{6}}^2}$"},{"latex":"$$R = \\sqrt{9 + 36} = \\sqrt{45} = 3\\sqrt{5}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-val"},{"color":"amber","style":"text-color","anchor":"b-val"},{"color":"sky","style":"text-color","anchor":"a-sub"},{"color":"amber","style":"text-color","anchor":"b-sub"}],"connections":[{"to":"a-sub","from":"a-val","color":"sky","label":"a = 3","style":"solid"},{"to":"b-sub","from":"b-val","color":"amber","label":"b = 6","style":"solid"}]},"order":1,"title":"Calculate R","description":"We find the exact value of $R$ using the formula $R = \\sqrt{a^2 + b^2}$. This earns 1 mark."},{"type":"text","order":2,"title":"Calculate alpha","content":"Next, we find the phase angle $\\alpha$ using the relationship:\n\n$$\\tan\\alpha = \\frac{b}{a}$$\n\nSubstituting $a = 3$ and $b = 6$:\n\n$$\\tan\\alpha = \\frac{6}{3} = 2$$\n\nTo find $\\alpha$, we compute the inverse tangent in radians:\n\n$$\\alpha = \\arctan(2) \\approx 1.107\\text{ radians}$$\n\nThis calculation of $\\alpha$ to 3 decimal places completes the final marks."},{"type":"text","order":3,"title":"Write the final expression","content":"Now we substitute our values of $R$ and $\\alpha$ back into the requested format:\n\n$$f(\\theta) = 3\\sqrt{5}\\cos(\\theta + 1.107)$$\n\nWhere the exact value of $R$ is $3\\sqrt{5}$ and the value of $\\alpha$ is $1.107$ radians (to 3 decimal places)."}]}],"generatedAt":"2026-06-09T03:38:54.171Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"177190d5-687b-4c1d-b7c0-5ea024340504","label":"b","content":"The curve with equation \$ y = \\cos\\theta \$ is transformed onto the curve with equation \$ y = f(\\theta) \$ by a sequence of two transformations.\n\nGiven that the first transformation is a stretch and the second is a translation:\n\n(i) Describe fully the transformation that is a stretch.\n(ii) Describe fully the transformation that is a translation.","markscheme":"- (i) Identifies a stretch parallel to the $y$-axis (or vertical stretch) B1\n- States the scale factor is $3\\sqrt{5}$ (or $\\sqrt{45}$) B1\n- (ii) Identifies a horizontal translation M1\n- Describes the translation by the vector $\\begin{pmatrix} -1.107 \\\\ 0 \\end{pmatrix}$ or \"left by $1.107$\" A1","marks":4,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"analytical","label":"Analytical Harmonic Form","steps":[{"type":"text","order":0,"title":"Convert f(θ) to R cos(θ + α)","content":"To understand how the curve $y = \\cos\\theta$ is transformed into $y = f(\\theta)$, we first need to express $f(\\theta) = 3\\cos\\theta - 6\\sin\\theta$ in the harmonic form $R\\cos(\\theta + \\alpha)$.\n\nUsing the cosine addition formula:\n$$R\\cos(\\theta + \\alpha) = R(\\cos\\theta\\cos\\alpha - \\sin\\theta\\sin\\alpha) = (R\\cos\\alpha)\\cos\\theta - (R\\sin\\alpha)\\sin\\theta$$\n\nBy comparing coefficients with $3\\cos\\theta - 6\\sin\\theta$, we get:\n$$\\begin{aligned}\nR\\cos\\alpha &= 3 \\\\\nR\\sin\\alpha &= 6\n\\end{aligned}$$\n\nNow, we find $R$ (the amplitude) and $\\alpha$ (the phase angle):\n* **Find $R$**:\n $$R^2 = 3^2 + 6^2 = 9 + 36 = 45 \\implies R = \\sqrt{45} = 3\\sqrt{5}$$\n* **Find $\\alpha$**:\n $$\\tan\\alpha = \\frac{R\\sin\\alpha}{R\\cos\\alpha} = \\frac{6}{3} = 2 \\implies \\alpha = \\arctan(2) \\approx 1.107 \\text{ radians}$$"},{"type":"text","order":1,"title":"Analyze the transformations","content":"Now we have the simplified equation:\n$$f(\\theta) = 3\\sqrt{5}\\cos(\\theta + 1.107)$$\n\nWe want to find the sequence of two transformations that maps the base curve $y = \\cos\\theta$ to $y = f(\\theta)$:\n1. A **stretch** transformation.\n2. A **translation** transformation.\n\nLet's apply these step by step to $y = \\cos\\theta$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\cos\\theta \\xrightarrow{\\text{Stretch}} y = \\htmlData{anchor=stretch-val}{\\color{@sky}{3\\sqrt{5}}}\\cos\\theta$"}],"highlights":[{"color":"sky","style":"box","anchor":"stretch-val"}],"connections":[]},"order":2,"title":"Describe the stretch","description":"The first transformation is a stretch. Multiplying the entire function by $3\\sqrt{5}$ represents a vertical stretch. This is a stretch parallel to the $y$-axis with a scale factor of $3\\sqrt{5}$ (or $\\sqrt{45}$)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = 3\\sqrt{5}\\cos\\theta \\xrightarrow{\\text{Translation}} y = 3\\sqrt{5}\\cos(\\theta + \\htmlData{anchor=trans-val}{\\color{@amber}{1.107}})$"}],"highlights":[{"color":"amber","style":"box","anchor":"trans-val"}],"connections":[]},"order":3,"title":"Describe the translation","description":"The second transformation is a translation. Replacing $\\theta$ with $\\theta + 1.107$ shifts the graph horizontally to the left by $1.107$ units. In vector form, this is a translation by the vector $\\begin{pmatrix} -1.107 \\\\ 0 \\end{pmatrix}$."},{"type":"text","order":4,"title":"Summarize final answers","content":"Here are the final complete descriptions for the transformations:\n\n* **(i) Stretch**:\n A stretch parallel to the $y$-axis (vertical stretch) with scale factor $3\\sqrt{5}$ (or $\\sqrt{45}$).\n\n* **(ii) Translation**:\n A horizontal translation by the vector $\\begin{pmatrix} -1.107 \\\\ 0 \\end{pmatrix}$ (or translation left by $1.107$ units)."}]},{"id":"graphical","label":"Graphical Calculator Approach","steps":[{"type":"text","order":0,"title":"Analyze the problem","content":"We need to find a sequence of two transformations that maps the parent curve $y = \\cos\\theta$ onto the curve $y = 3\\cos\\theta - 6\\sin\\theta$.\n\nThe question specifies that the transformations are:\n1. A **stretch**\n2. A **translation**\n\nBy graphing the function on a graphing calculator, we can easily find the key features of the transformed graph (such as the peak's height and position) and describe these transformations fully.","highlights":[{"text":"first transformation is a stretch and the second is a translation","location":"specification"}]},{"type":"text","order":1,"title":"Graph the function","content":"Let's input the function $y = 3\\cos x - 6\\sin x$ (using $x$ as the variable instead of $\\theta$) into our graphing calculator in radian mode.","calculator":[{"gdc":{"expression":"3\\cos(x) - 6\\sin(x)","instructions":"1. Set the calculator angle mode to Radians.\n2. In the Graph menu, enter the equation \$ y_1 = 3\\cos(x) - 6\\sin(x) \$.\n3. Set a suitable window (e.g., \$ x \\in [-2\\pi, 2\\pi] \$, \$ y \\in [-8, 8] \$) and plot the graph."},"ti84":{"expression":"3\\cos(X) - 6\\sin(X)","instructions":"1. Press [MODE], select [RADIAN], then press [ENTER].\n2. Press [Y=] and enter \$ Y_1 = 3\\cos(X) - 6\\sin(X) \$.\n3. Press [WINDOW] and set: \$ X_{\\min} = -2\\pi \$, \$ X_{\\max} = 2\\pi \$, \$ Y_{\\min} = -8 \$, \$ Y_{\\max} = 8 \$.\n4. Press [GRAPH] to view the curve."},"desmos":{"expression":"y = 3\\cos(x) - 6\\sin(x)","instructions":"1. In the settings wrench menu, make sure the angle setting is set to Radians.\n2. In the expression box, type \$ y = 3\\cos(x) - 6\\sin(x) \$.\n3. Observe the wave and click on the highest peak to reveal its coordinates."},"description":"Graph the function $y = 3\\cos(x) - 6\\sin(x)$ in radian mode to find its maximum value and the $x$-coordinate of the peak closest to the origin."}]},{"type":"text","order":2,"title":"Find the stretch factor","content":"Using the maximum-finder feature of your calculator, find the maximum value (the peak) of the wave.\n\nThe calculator shows that the maximum value is $y \\approx 6.708$.\n\nSince the maximum value of $y = \\cos\\theta$ is normally $1$, a maximum value of $6.708$ means the graph has been stretched vertically.\n\nLet's write this value in its exact form:\n$$R = \\sqrt{3^2 + (-6)^2} = \\sqrt{9 + 36} = \\sqrt{45} = 3\\sqrt{5} \\approx 6.708$$\n\nThus, the first transformation is a **stretch parallel to the $y$-axis with a scale factor of $3\\sqrt{5}$** (or $\\sqrt{45}$). This earns the first two marks.","calculator":[{"gdc":{"instructions":"1. Use the G-Solve / Analyze Graph tool.\n2. Select Maximum.\n3. The calculator displays the coordinates of the peak near the origin: \$ (-1.107, 6.708) \$."},"ti84":{"instructions":"1. Press [2nd] [TRACE] (CALC) and select option 4: maximum.\n2. Move the cursor to the left of the peak and press [ENTER] (Left Bound).\n3. Move the cursor to the right of the peak and press [ENTER] (Right Bound).\n4. Press [ENTER] again (Guess). The calculator displays the maximum value: \$ Y \\approx 6.7082 \$ (which is \$ 3\\sqrt{5} \$)."},"desmos":{"instructions":"1. Click on the local maximum point on the graph near the y-axis.\n2. The coordinates displayed are approximately \$ (-1.107, 6.708) \$."},"description":"Find the maximum point of the curve."}]},{"type":"text","order":3,"title":"Find the horizontal translation","content":"Now look at the horizontal position of the first peak.\n\nFor the parent curve $y = \\cos\\theta$, the peak is located at $\\theta = 0$.\n\nFrom the calculator's analysis in the previous step, the peak of our transformed function $y = 3\\cos\\theta - 6\\sin\\theta$ is located at:\n$$\\theta \\approx -1.11 \\text{ (or more precisely, } -1.107\\text{)}$$\n\nThis shift to the left by $1.107$ represents a horizontal translation.\n\nWe can describe this transformation as a **translation by the vector $\\begin{pmatrix} -1.107 \\\\ 0 \\end{pmatrix}$** (or **translation of $1.107$ units to the left**). This earns the remaining two marks."},{"type":"text","order":4,"title":"State the final answer","content":"The full descriptions of the transformations are:\n\n**(i) Stretch:**\nA stretch parallel to the $y$-axis (vertical stretch) with a scale factor of $3\\sqrt{5}$ (or $\\sqrt{45}$).\n\n**(ii) Translation:**\nA translation by the vector $\\begin{pmatrix} -1.107 \\\\ 0 \\end{pmatrix}$ (or a horizontal translation of $1.107$ units in the negative $\\theta$-direction/to the left)."}]}],"generatedAt":"2026-06-09T11:05:59.703Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"dafa3855-832d-45f9-918d-aed58a09109b","label":"c","content":"Given \$ g(\\theta) = \\frac{75}{5 + (f(\\theta))^2} \$, find the range of \$ g \$.","markscheme":"- Recognises that the maximum value of $f^2$ is $(3\\sqrt{5})^2 = 45$ and the minimum is $0$ M1\n- Evaluates $g$ at these boundaries: $g = \\frac{75}{5+45}$ and $g = \\frac{75}{5+0}$ M1\n- States range as $1.5 \\le g(\\theta) \\le 15$ (or $\\frac{3}{2} \\le g(\\theta) \\le 15$) A1","marks":3,"order":2,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"analytical_r_formula","label":"Analytical using R-formula","steps":[{"type":"text","order":0,"title":"Convert to R-formula form","content":"To find the range of $f(\\theta) = 3\\cos\\theta - 6\\sin\\theta$, we first express it in the form $R\\cos(\\theta + \\alpha)$.\n\nUsing the R-formula:\n$$R = \\sqrt{a^2 + b^2}$$\n\nSubstituting our coefficients $a = 3$ and $b = 6$:\n$$R = \\sqrt{3^2 + (-6)^2} = \\sqrt{9 + 36} = \\sqrt{45} = 3\\sqrt{5}$$\n\nThis gives:\n$$f(\\theta) = 3\\sqrt{5}\\cos(\\theta + \\alpha)$$\n\nSince $-1 \\le \\cos(\\theta + \\alpha) \\le 1$, the range of $f(\\theta)$ is:\n$$-3\\sqrt{5} \\le f(\\theta) \\le 3\\sqrt{5}$$"},{"type":"text","order":1,"title":"Determine the range of f^2","content":"Now we need to find the range of the squared term $(f(\\theta))^2$.\n\nSince $f(\\theta)$ can take any value between $-3\\sqrt{5}$ and $3\\sqrt{5}$ (including $0$), squaring these values gives:\n- The minimum value: $0$\n- The maximum value: $(3\\sqrt{5})^2 = 45$\n\nTherefore, the range of $(f(\\theta))^2$ is:\n$$0 \\le (f(\\theta))^2 \\le 45$$\n\nThis key step earns the first mark M1."},{"type":"text","order":2,"title":"Find the range of g","content":"Next, we substitute the minimum and maximum boundaries of $(f(\\theta))^2$ into the expression for $g(\\theta)$:\n$$g(\\theta) = \\frac{75}{5 + (f(\\theta))^2}$$\n\n- At the minimum boundary $(f(\\theta))^2 = 0$:\n$$g_{\\text{max}} = \\frac{75}{5 + 0} = 15$$\n\n- At the maximum boundary $(f(\\theta))^2 = 45$:\n$$g_{\\text{min}} = \\frac{75}{5 + 45} = \\frac{75}{50} = 1.5$$\n\nEvaluating at these boundaries earns the second mark M1.\n\nThus, the range of $g$ is:\n$$1.5 \\le g(\\theta) \\le 15$$ (or $\\frac{3}{2} \\le g(\\theta) \\le 15$)\n\nThis final range earns the final accuracy mark A1."}]},{"id":"graphical_calculator","label":"Using a Graphing Calculator","steps":[{"type":"text","order":0,"title":"Understand the function","content":"We want to find the range of the function $g(\\theta) = \\frac{75}{5 + (f(\\theta))^2}$, where $f(\\theta) = 3\\cos\\theta - 6\\sin\\theta$.\n\nSubstituting $f(\\theta)$ into our expression for $g(\\theta)$ gives:\n\n$$g(\\theta) = \\frac{75}{5 + (3\\cos\\theta - 6\\sin\\theta)^2}$$\n\nTo find the range, we need to find the maximum and minimum values of $g(\\theta)$ over all real numbers $\\theta$. Using a graphing calculator is an efficient way to visualize this function and read off these extreme values directly."},{"type":"text","order":1,"title":"Graph the function","content":"Let's enter the function $y = \\frac{75}{5 + (3\\cos(x) - 6\\sin(x))^2}$ into our graphing tool. \n\nBecause the trigonometric functions are periodic, the graph will repeat. A standard viewing window such as $x \\in [-2\\pi, 2\\pi]$ and $y \\in [0, 20]$ will allow us to see several complete cycles clearly.","calculator":[{"gdc":{"expression":"75 / (5 + (3*cos(x) - 6*sin(x))^2)","instructions":"1. In the Graph menu, define \$ f1(x) = \\frac{75}{5 + (3\\cos(x) - 6\\sin(x))^2} \$.\n2. Adjust the window settings to show \$ x \$ from \$ -2\\pi \$ to \$ 2\\pi \$ and \$ y \$ from \$ 0 \$ to \$ 20 \$.\n3. Plot the graph."},"ti84":{"expression":"75 / (5 + (3*cos(X) - 6*sin(X))^2)","instructions":"1. Press [Y=] and enter \$ Y_1 = 75 / (5 + (3*\\cos(X) - 6*\\sin(X))^2) \$.\n2. Press [WINDOW] and set: \$ X_{\\text{min}} = -2\\pi \$, \$ X_{\\text{max}} = 2\\pi \$, \$ Y_{\\text{min}} = 0 \$, and \$ Y_{\\text{max}} = 20 \$.\n3. Press [GRAPH]."},"desmos":{"expression":"y = 75 / (5 + (3\\cos(x) - 6\\sin(x))^2)","instructions":"1. In the input bar, type \$ y = 75 / (5 + (3\\cos(x) - 6\\sin(x))^2) \$.\n2. Adjust the axes zoom so that you can see the peaks and troughs clearly."},"description":"Plot the function and analyze its peaks (maxima) and valleys (minima)."}]},{"type":"text","order":2,"title":"Find the maximum","content":"Using the analyzer tools on your calculator, find the coordinates of any local maximum on the graph.\n\nYou will find that the peaks of the curve reach a vertical height of exactly $15$.\n\n$$\\text{Maximum value of } g(\\theta) = 15$$","calculator":[{"gdc":{"instructions":"1. Go to Menu > Analyze Graph > Maximum.\n2. Click on the left and right sides of any peak.\n3. The maximum point is displayed with a y-coordinate of \$ 15 \$."},"ti84":{"instructions":"1. Press [2nd] [CALC] and select '4: maximum'.\n2. Set a Left Bound to the left of a peak, a Right Bound to the right, and press [ENTER] for a Guess.\n3. The calculator displays the maximum y-coordinate as \$ 15 \$."},"desmos":{"instructions":"1. Click on one of the grey dots at the top of a peak.\n2. The coordinates of the vertex will appear, showing a y-value of \$ 15 \$."},"description":"Find the maximum value of the function on your calculator screen."}]},{"type":"text","order":3,"title":"Find the minimum","content":"Now, locate a local minimum (trough) on the graph.\n\nYour calculator will show that the lowest points of the curve reach a vertical value of exactly $1.5$ (or $\\frac{3}{2}$).\n\n$$\\text{Minimum value of } g(\\theta) = 1.5$$","calculator":[{"gdc":{"instructions":"1. Go to Menu > Analyze Graph > Minimum.\n2. Click on the left and right sides of any trough.\n3. The minimum point is displayed with a y-coordinate of \$ 1.5 \$."},"ti84":{"instructions":"1. Press [2nd] [CALC] and select '3: minimum'.\n2. Set a Left Bound to the left of a trough, a Right Bound to the right, and press [ENTER] for a Guess.\n3. The calculator displays the minimum y-coordinate as \$ 1.5 \$."},"desmos":{"instructions":"1. Click on one of the grey dots at the lowest point of a curve.\n2. The coordinates will show a y-value of \$ 1.5 \$."},"description":"Find the minimum value of the function on your calculator screen."}]},{"type":"text","order":4,"title":"State the range","content":"Since the function $g(\\theta)$ is continuous and oscillates smoothly between these two values, the range is everything between the minimum and maximum boundaries.\n\nThus, the range of $g$ is:\n\n$$1.5 \\le g(\\theta) \\le 15 \\quad \\text{or} \\quad \\frac{3}{2} \\le g(\\theta) \\le 15$$"}]}],"generatedAt":"2026-06-09T11:08:02.400Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"4d7c0a63-9f56-41ea-b63b-defe1e6fa859:expansion_and_comparison","label":"(a) Expansion and Equating Coefficients","steps":[{"type":"text","order":0,"title":"State the identity","content":"To express $3\\cos\\theta - 6\\sin\\theta$ in the form $R\\cos(\\theta + \\alpha)$, we start by using the compound angle identity for cosine:\n\n$$\\cos(A + B) = \\cos A \\cos B - \\sin A \\sin B$$\n\nApplying this to our expression gives:\n\n$$R\\cos(\\theta + \\alpha) = R(\\cos\\theta\\cos\\alpha - \\sin\\theta\\sin\\alpha)$$\n\nExpanding the brackets:\n\n$$R\\cos(\\theta + \\alpha) = (R\\cos\\alpha)\\cos\\theta - (R\\sin\\alpha)\\sin\\theta$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=orig}{\\color{@sky}{3}}\\cos\\theta - \\htmlData{anchor=orig-sin}{\\color{@amber}{6}}\\sin\\theta = \\htmlData{anchor=exp-cos}{\\color{@sky}{R\\cos\\alpha}}\\cos\\theta - \\htmlData{anchor=exp-sin}{\\color{@amber}{R\\sin\\alpha}}\\sin\\theta$"},{"latex":"$$\\htmlData{anchor=eq1}{\\color{@sky}{R\\cos\\alpha = 3}}$"},{"latex":"$$\\htmlData{anchor=eq2}{\\color{@amber}{R\\sin\\alpha = 6}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"orig"},{"color":"amber","style":"text-color","anchor":"orig-sin"},{"color":"sky","style":"box","anchor":"eq1"},{"color":"amber","style":"box","anchor":"eq2"}],"connections":[{"to":"eq1","from":"orig","color":"sky","label":"equate cos coefficient","style":"solid"},{"to":"eq2","from":"orig-sin","color":"amber","label":"equate sin coefficient","style":"solid"}]},"order":1,"title":"Equate the coefficients","description":"We now compare our expanded expression with the given function $f(\\theta) = 3\\cos\\theta - 6\\sin\\theta$ to equate the coefficients of $\\cos\\theta$ and $\\sin\\theta$."},{"type":"text","order":2,"title":"Solve for R","content":"To find the value of $R$, we square and add our two equations:\n\n$$(R\\sin\\alpha)^2 + (R\\cos\\alpha)^2 = 6^2 + 3^2$$\n\nFactor out $R^2$ using the Pythagorean identity $\\sin^2\\alpha + \\cos^2\\alpha = 1$:\n\n$$R^2(\\sin^2\\alpha + \\cos^2\\alpha) = 36 + 9$$\n\n$$R^2 = 45$$\n\nSince $R > 0$, we find the exact value of $R$ by simplifying the square root:\n\n$$R = \\sqrt{45} = \\sqrt{9 \\times 5} = 3\\sqrt{5}$$\n\n(This earns 1 mark)"},{"type":"text","order":3,"title":"Solve for alpha","content":"To find $\\alpha$, we divide the equation for $\\sin\\alpha$ by the equation for $\\cos\\alpha$ to use the identity $\\tan\\alpha = \\frac{\\sin\\alpha}{\\cos\\alpha}$:\n\n$$\\frac{R\\sin\\alpha}{R\\cos\\alpha} = \\frac{6}{3}$$\n\n$$\\tan\\alpha = 2$$\n\n(This earns 1 mark)\n\nNow, solve for $\\alpha$ using the inverse tangent function in radian mode:\n\n$$\\alpha = \\arctan(2) \\approx 1.107\\text{ radians}$$\n\n(This earns 1 mark)"},{"type":"text","order":4,"title":"State the final expression","content":"Substitute the exact value of $R$ and the rounded value of $\\alpha$ back into our target form:\n\n$$f(\\theta) = 3\\sqrt{5}\\cos(\\theta + 1.107)$$"}]},{"id":"4d7c0a63-9f56-41ea-b63b-defe1e6fa859:direct_formula","label":"(a) Direct R-Formula Application","steps":[{"type":"text","order":0,"title":"Identify the standard form","content":"We want to express $f(\\theta) = 3\\cos\\theta - 6\\sin\\theta$ in the form $R\\cos(\\theta + \\alpha)$.\n\nUsing the direct $R$-formula for a cosine expansion:\n\n$$a\\cos\\theta - b\\sin\\theta = R\\cos(\\theta + \\alpha)$$\n\nComparing both expressions, we can directly identify the coefficients:\n- $a = 3$\n- $b = 6$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$R = \\sqrt{\\htmlData{anchor=a-val}{\\color{@sky}{a}}^2 + \\htmlData{anchor=b-val}{\\color{@amber}{b}}^2}$"},{"latex":"$$R = \\sqrt{\\htmlData{anchor=a-sub}{\\color{@sky}{3}}^2 + \\htmlData{anchor=b-sub}{\\color{@amber}{6}}^2}$"},{"latex":"$$R = \\sqrt{9 + 36} = \\sqrt{45} = 3\\sqrt{5}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-val"},{"color":"amber","style":"text-color","anchor":"b-val"},{"color":"sky","style":"text-color","anchor":"a-sub"},{"color":"amber","style":"text-color","anchor":"b-sub"}],"connections":[{"to":"a-sub","from":"a-val","color":"sky","label":"a = 3","style":"solid"},{"to":"b-sub","from":"b-val","color":"amber","label":"b = 6","style":"solid"}]},"order":1,"title":"Calculate R","description":"We find the exact value of $R$ using the formula $R = \\sqrt{a^2 + b^2}$. This earns 1 mark."},{"type":"text","order":2,"title":"Calculate alpha","content":"Next, we find the phase angle $\\alpha$ using the relationship:\n\n$$\\tan\\alpha = \\frac{b}{a}$$\n\nSubstituting $a = 3$ and $b = 6$:\n\n$$\\tan\\alpha = \\frac{6}{3} = 2$$\n\nTo find $\\alpha$, we compute the inverse tangent in radians:\n\n$$\\alpha = \\arctan(2) \\approx 1.107\\text{ radians}$$\n\nThis calculation of $\\alpha$ to 3 decimal places completes the final marks."},{"type":"text","order":3,"title":"Write the final expression","content":"Now we substitute our values of $R$ and $\\alpha$ back into the requested format:\n\n$$f(\\theta) = 3\\sqrt{5}\\cos(\\theta + 1.107)$$\n\nWhere the exact value of $R$ is $3\\sqrt{5}$ and the value of $\\alpha$ is $1.107$ radians (to 3 decimal places)."}]},{"id":"177190d5-687b-4c1d-b7c0-5ea024340504:analytical","label":"(b) Analytical Harmonic Form","steps":[{"type":"text","order":0,"title":"Convert f(θ) to R cos(θ + α)","content":"To understand how the curve $y = \\cos\\theta$ is transformed into $y = f(\\theta)$, we first need to express $f(\\theta) = 3\\cos\\theta - 6\\sin\\theta$ in the harmonic form $R\\cos(\\theta + \\alpha)$.\n\nUsing the cosine addition formula:\n$$R\\cos(\\theta + \\alpha) = R(\\cos\\theta\\cos\\alpha - \\sin\\theta\\sin\\alpha) = (R\\cos\\alpha)\\cos\\theta - (R\\sin\\alpha)\\sin\\theta$$\n\nBy comparing coefficients with $3\\cos\\theta - 6\\sin\\theta$, we get:\n$$\\begin{aligned}\nR\\cos\\alpha &= 3 \\\\\nR\\sin\\alpha &= 6\n\\end{aligned}$$\n\nNow, we find $R$ (the amplitude) and $\\alpha$ (the phase angle):\n* **Find $R$**:\n $$R^2 = 3^2 + 6^2 = 9 + 36 = 45 \\implies R = \\sqrt{45} = 3\\sqrt{5}$$\n* **Find $\\alpha$**:\n $$\\tan\\alpha = \\frac{R\\sin\\alpha}{R\\cos\\alpha} = \\frac{6}{3} = 2 \\implies \\alpha = \\arctan(2) \\approx 1.107 \\text{ radians}$$"},{"type":"text","order":1,"title":"Analyze the transformations","content":"Now we have the simplified equation:\n$$f(\\theta) = 3\\sqrt{5}\\cos(\\theta + 1.107)$$\n\nWe want to find the sequence of two transformations that maps the base curve $y = \\cos\\theta$ to $y = f(\\theta)$:\n1. A **stretch** transformation.\n2. A **translation** transformation.\n\nLet's apply these step by step to $y = \\cos\\theta$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\cos\\theta \\xrightarrow{\\text{Stretch}} y = \\htmlData{anchor=stretch-val}{\\color{@sky}{3\\sqrt{5}}}\\cos\\theta$"}],"highlights":[{"color":"sky","style":"box","anchor":"stretch-val"}],"connections":[]},"order":2,"title":"Describe the stretch","description":"The first transformation is a stretch. Multiplying the entire function by $3\\sqrt{5}$ represents a vertical stretch. This is a stretch parallel to the $y$-axis with a scale factor of $3\\sqrt{5}$ (or $\\sqrt{45}$)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = 3\\sqrt{5}\\cos\\theta \\xrightarrow{\\text{Translation}} y = 3\\sqrt{5}\\cos(\\theta + \\htmlData{anchor=trans-val}{\\color{@amber}{1.107}})$"}],"highlights":[{"color":"amber","style":"box","anchor":"trans-val"}],"connections":[]},"order":3,"title":"Describe the translation","description":"The second transformation is a translation. Replacing $\\theta$ with $\\theta + 1.107$ shifts the graph horizontally to the left by $1.107$ units. In vector form, this is a translation by the vector $\\begin{pmatrix} -1.107 \\\\ 0 \\end{pmatrix}$."},{"type":"text","order":4,"title":"Summarize final answers","content":"Here are the final complete descriptions for the transformations:\n\n* **(i) Stretch**:\n A stretch parallel to the $y$-axis (vertical stretch) with scale factor $3\\sqrt{5}$ (or $\\sqrt{45}$).\n\n* **(ii) Translation**:\n A horizontal translation by the vector $\\begin{pmatrix} -1.107 \\\\ 0 \\end{pmatrix}$ (or translation left by $1.107$ units)."}]},{"id":"177190d5-687b-4c1d-b7c0-5ea024340504:graphical","label":"(b) Graphical Calculator Approach","steps":[{"type":"text","order":0,"title":"Analyze the problem","content":"We need to find a sequence of two transformations that maps the parent curve $y = \\cos\\theta$ onto the curve $y = 3\\cos\\theta - 6\\sin\\theta$.\n\nThe question specifies that the transformations are:\n1. A **stretch**\n2. A **translation**\n\nBy graphing the function on a graphing calculator, we can easily find the key features of the transformed graph (such as the peak's height and position) and describe these transformations fully.","highlights":[{"text":"first transformation is a stretch and the second is a translation","location":"specification"}]},{"type":"text","order":1,"title":"Graph the function","content":"Let's input the function $y = 3\\cos x - 6\\sin x$ (using $x$ as the variable instead of $\\theta$) into our graphing calculator in radian mode.","calculator":[{"gdc":{"expression":"3\\cos(x) - 6\\sin(x)","instructions":"1. Set the calculator angle mode to Radians.\n2. In the Graph menu, enter the equation \$ y_1 = 3\\cos(x) - 6\\sin(x) \$.\n3. Set a suitable window (e.g., \$ x \\in [-2\\pi, 2\\pi] \$, \$ y \\in [-8, 8] \$) and plot the graph."},"ti84":{"expression":"3\\cos(X) - 6\\sin(X)","instructions":"1. Press [MODE], select [RADIAN], then press [ENTER].\n2. Press [Y=] and enter \$ Y_1 = 3\\cos(X) - 6\\sin(X) \$.\n3. Press [WINDOW] and set: \$ X_{\\min} = -2\\pi \$, \$ X_{\\max} = 2\\pi \$, \$ Y_{\\min} = -8 \$, \$ Y_{\\max} = 8 \$.\n4. Press [GRAPH] to view the curve."},"desmos":{"expression":"y = 3\\cos(x) - 6\\sin(x)","instructions":"1. In the settings wrench menu, make sure the angle setting is set to Radians.\n2. In the expression box, type \$ y = 3\\cos(x) - 6\\sin(x) \$.\n3. Observe the wave and click on the highest peak to reveal its coordinates."},"description":"Graph the function $y = 3\\cos(x) - 6\\sin(x)$ in radian mode to find its maximum value and the $x$-coordinate of the peak closest to the origin."}]},{"type":"text","order":2,"title":"Find the stretch factor","content":"Using the maximum-finder feature of your calculator, find the maximum value (the peak) of the wave.\n\nThe calculator shows that the maximum value is $y \\approx 6.708$.\n\nSince the maximum value of $y = \\cos\\theta$ is normally $1$, a maximum value of $6.708$ means the graph has been stretched vertically.\n\nLet's write this value in its exact form:\n$$R = \\sqrt{3^2 + (-6)^2} = \\sqrt{9 + 36} = \\sqrt{45} = 3\\sqrt{5} \\approx 6.708$$\n\nThus, the first transformation is a **stretch parallel to the $y$-axis with a scale factor of $3\\sqrt{5}$** (or $\\sqrt{45}$). This earns the first two marks.","calculator":[{"gdc":{"instructions":"1. Use the G-Solve / Analyze Graph tool.\n2. Select Maximum.\n3. The calculator displays the coordinates of the peak near the origin: \$ (-1.107, 6.708) \$."},"ti84":{"instructions":"1. Press [2nd] [TRACE] (CALC) and select option 4: maximum.\n2. Move the cursor to the left of the peak and press [ENTER] (Left Bound).\n3. Move the cursor to the right of the peak and press [ENTER] (Right Bound).\n4. Press [ENTER] again (Guess). The calculator displays the maximum value: \$ Y \\approx 6.7082 \$ (which is \$ 3\\sqrt{5} \$)."},"desmos":{"instructions":"1. Click on the local maximum point on the graph near the y-axis.\n2. The coordinates displayed are approximately \$ (-1.107, 6.708) \$."},"description":"Find the maximum point of the curve."}]},{"type":"text","order":3,"title":"Find the horizontal translation","content":"Now look at the horizontal position of the first peak.\n\nFor the parent curve $y = \\cos\\theta$, the peak is located at $\\theta = 0$.\n\nFrom the calculator's analysis in the previous step, the peak of our transformed function $y = 3\\cos\\theta - 6\\sin\\theta$ is located at:\n$$\\theta \\approx -1.11 \\text{ (or more precisely, } -1.107\\text{)}$$\n\nThis shift to the left by $1.107$ represents a horizontal translation.\n\nWe can describe this transformation as a **translation by the vector $\\begin{pmatrix} -1.107 \\\\ 0 \\end{pmatrix}$** (or **translation of $1.107$ units to the left**). This earns the remaining two marks."},{"type":"text","order":4,"title":"State the final answer","content":"The full descriptions of the transformations are:\n\n**(i) Stretch:**\nA stretch parallel to the $y$-axis (vertical stretch) with a scale factor of $3\\sqrt{5}$ (or $\\sqrt{45}$).\n\n**(ii) Translation:**\nA translation by the vector $\\begin{pmatrix} -1.107 \\\\ 0 \\end{pmatrix}$ (or a horizontal translation of $1.107$ units in the negative $\\theta$-direction/to the left)."}]},{"id":"dafa3855-832d-45f9-918d-aed58a09109b:analytical_r_formula","label":"(c) Analytical using R-formula","steps":[{"type":"text","order":0,"title":"Convert to R-formula form","content":"To find the range of $f(\\theta) = 3\\cos\\theta - 6\\sin\\theta$, we first express it in the form $R\\cos(\\theta + \\alpha)$.\n\nUsing the R-formula:\n$$R = \\sqrt{a^2 + b^2}$$\n\nSubstituting our coefficients $a = 3$ and $b = 6$:\n$$R = \\sqrt{3^2 + (-6)^2} = \\sqrt{9 + 36} = \\sqrt{45} = 3\\sqrt{5}$$\n\nThis gives:\n$$f(\\theta) = 3\\sqrt{5}\\cos(\\theta + \\alpha)$$\n\nSince $-1 \\le \\cos(\\theta + \\alpha) \\le 1$, the range of $f(\\theta)$ is:\n$$-3\\sqrt{5} \\le f(\\theta) \\le 3\\sqrt{5}$$"},{"type":"text","order":1,"title":"Determine the range of f^2","content":"Now we need to find the range of the squared term $(f(\\theta))^2$.\n\nSince $f(\\theta)$ can take any value between $-3\\sqrt{5}$ and $3\\sqrt{5}$ (including $0$), squaring these values gives:\n- The minimum value: $0$\n- The maximum value: $(3\\sqrt{5})^2 = 45$\n\nTherefore, the range of $(f(\\theta))^2$ is:\n$$0 \\le (f(\\theta))^2 \\le 45$$\n\nThis key step earns the first mark M1."},{"type":"text","order":2,"title":"Find the range of g","content":"Next, we substitute the minimum and maximum boundaries of $(f(\\theta))^2$ into the expression for $g(\\theta)$:\n$$g(\\theta) = \\frac{75}{5 + (f(\\theta))^2}$$\n\n- At the minimum boundary $(f(\\theta))^2 = 0$:\n$$g_{\\text{max}} = \\frac{75}{5 + 0} = 15$$\n\n- At the maximum boundary $(f(\\theta))^2 = 45$:\n$$g_{\\text{min}} = \\frac{75}{5 + 45} = \\frac{75}{50} = 1.5$$\n\nEvaluating at these boundaries earns the second mark M1.\n\nThus, the range of $g$ is:\n$$1.5 \\le g(\\theta) \\le 15$$ (or $\\frac{3}{2} \\le g(\\theta) \\le 15$)\n\nThis final range earns the final accuracy mark A1."}]},{"id":"dafa3855-832d-45f9-918d-aed58a09109b:graphical_calculator","label":"(c) Using a Graphing Calculator","steps":[{"type":"text","order":0,"title":"Understand the function","content":"We want to find the range of the function $g(\\theta) = \\frac{75}{5 + (f(\\theta))^2}$, where $f(\\theta) = 3\\cos\\theta - 6\\sin\\theta$.\n\nSubstituting $f(\\theta)$ into our expression for $g(\\theta)$ gives:\n\n$$g(\\theta) = \\frac{75}{5 + (3\\cos\\theta - 6\\sin\\theta)^2}$$\n\nTo find the range, we need to find the maximum and minimum values of $g(\\theta)$ over all real numbers $\\theta$. Using a graphing calculator is an efficient way to visualize this function and read off these extreme values directly."},{"type":"text","order":1,"title":"Graph the function","content":"Let's enter the function $y = \\frac{75}{5 + (3\\cos(x) - 6\\sin(x))^2}$ into our graphing tool. \n\nBecause the trigonometric functions are periodic, the graph will repeat. A standard viewing window such as $x \\in [-2\\pi, 2\\pi]$ and $y \\in [0, 20]$ will allow us to see several complete cycles clearly.","calculator":[{"gdc":{"expression":"75 / (5 + (3*cos(x) - 6*sin(x))^2)","instructions":"1. In the Graph menu, define \$ f1(x) = \\frac{75}{5 + (3\\cos(x) - 6\\sin(x))^2} \$.\n2. Adjust the window settings to show \$ x \$ from \$ -2\\pi \$ to \$ 2\\pi \$ and \$ y \$ from \$ 0 \$ to \$ 20 \$.\n3. Plot the graph."},"ti84":{"expression":"75 / (5 + (3*cos(X) - 6*sin(X))^2)","instructions":"1. Press [Y=] and enter \$ Y_1 = 75 / (5 + (3*\\cos(X) - 6*\\sin(X))^2) \$.\n2. Press [WINDOW] and set: \$ X_{\\text{min}} = -2\\pi \$, \$ X_{\\text{max}} = 2\\pi \$, \$ Y_{\\text{min}} = 0 \$, and \$ Y_{\\text{max}} = 20 \$.\n3. Press [GRAPH]."},"desmos":{"expression":"y = 75 / (5 + (3\\cos(x) - 6\\sin(x))^2)","instructions":"1. In the input bar, type \$ y = 75 / (5 + (3\\cos(x) - 6\\sin(x))^2) \$.\n2. Adjust the axes zoom so that you can see the peaks and troughs clearly."},"description":"Plot the function and analyze its peaks (maxima) and valleys (minima)."}]},{"type":"text","order":2,"title":"Find the maximum","content":"Using the analyzer tools on your calculator, find the coordinates of any local maximum on the graph.\n\nYou will find that the peaks of the curve reach a vertical height of exactly $15$.\n\n$$\\text{Maximum value of } g(\\theta) = 15$$","calculator":[{"gdc":{"instructions":"1. Go to Menu > Analyze Graph > Maximum.\n2. Click on the left and right sides of any peak.\n3. The maximum point is displayed with a y-coordinate of \$ 15 \$."},"ti84":{"instructions":"1. Press [2nd] [CALC] and select '4: maximum'.\n2. Set a Left Bound to the left of a peak, a Right Bound to the right, and press [ENTER] for a Guess.\n3. The calculator displays the maximum y-coordinate as \$ 15 \$."},"desmos":{"instructions":"1. Click on one of the grey dots at the top of a peak.\n2. The coordinates of the vertex will appear, showing a y-value of \$ 15 \$."},"description":"Find the maximum value of the function on your calculator screen."}]},{"type":"text","order":3,"title":"Find the minimum","content":"Now, locate a local minimum (trough) on the graph.\n\nYour calculator will show that the lowest points of the curve reach a vertical value of exactly $1.5$ (or $\\frac{3}{2}$).\n\n$$\\text{Minimum value of } g(\\theta) = 1.5$$","calculator":[{"gdc":{"instructions":"1. Go to Menu > Analyze Graph > Minimum.\n2. Click on the left and right sides of any trough.\n3. The minimum point is displayed with a y-coordinate of \$ 1.5 \$."},"ti84":{"instructions":"1. Press [2nd] [CALC] and select '3: minimum'.\n2. Set a Left Bound to the left of a trough, a Right Bound to the right, and press [ENTER] for a Guess.\n3. The calculator displays the minimum y-coordinate as \$ 1.5 \$."},"desmos":{"instructions":"1. Click on one of the grey dots at the lowest point of a curve.\n2. The coordinates will show a y-value of \$ 1.5 \$."},"description":"Find the minimum value of the function on your calculator screen."}]},{"type":"text","order":4,"title":"State the range","content":"Since the function $g(\\theta)$ is continuous and oscillates smoothly between these two values, the range is everything between the minimum and maximum boundaries.\n\nThus, the range of $g$ is:\n\n$$1.5 \\le g(\\theta) \\le 15 \\quad \\text{or} \\quad \\frac{3}{2} \\le g(\\theta) \\le 15$$"}]}],"generatedAt":"2026-06-09T11:08:02.515Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"1940ae10-9142-4756-ab7a-e58291094be1","revisionId":"fc54f358-caa1-4d28-944b-665711b75466","questionNumber":null,"questionType":"LA","specification":"A chord of a circle of radius $9$ cm subtends an angle of $1.4$ radians at the centre.","difficulty":null,"difficultyBand":"medium","tier":null,"totalMarks":5,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"67c83821-b400-459c-ac17-540caaa78275","label":"a","content":"Find the length of the chord.","markscheme":"- Chord $=2(9)\\sin(1.4/2)$ M1\n- $11.6$ cm A1","marks":2,"order":0,"markingRubric":null,"tableImages":[],"explanation":null,"options":[]},{"id":"abe41b23-a7c2-4f8b-9c5d-7320ed6986d1","label":"b","content":"Find the area of the minor segment.\n\nGive both answers to three significant figures.","markscheme":"- Uses segment area $=\\frac12r^2(\\theta-\\sin\\theta)$ M1\n- $=\\frac{81}{2}(1.4-\\sin1.4)$ M1\n- $16.8$ cm$^2$ A1","marks":3,"order":1,"markingRubric":null,"tableImages":[],"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"1fde01cc-33ca-4271-a99a-2994b71cce0f","revisionId":"4df773d7-1c54-4fa5-9ddc-204edc330d23","questionNumber":null,"questionType":"LA","specification":"","difficulty":5,"difficultyBand":"medium","tier":null,"totalMarks":8,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"b5c616b6-a7b9-4936-b911-3a31f43f62d7","label":"i","content":"Show that\n\\[ \\frac{\\cot \\theta - \\tan \\theta}{\\cot \\theta + \\tan \\theta} \\equiv \\cos 2\\theta \\]\nfor \$ \\theta \\neq \\frac{n\\pi}{2} \$ where \$ n \\in \\mathbb{Z} \$.","markscheme":"- Substitutes $\\tan \\theta = \\frac{\\sin \\theta}{\\cos \\theta}$ and $\\cot \\theta = \\frac{\\cos \\theta}{\\sin \\theta}$ into the expression M1\n- Simplifies the numerator and denominator to single fractions: $\\frac{\\frac{\\cos^2 \\theta - \\sin^2 \\theta}{\\sin \\theta \\cos \\theta}}{\\frac{\\cos^2 \\theta + \\sin^2 \\theta}{\\sin \\theta \\cos \\theta}}$ M1\n- Recognises that $\\cos^2 \\theta + \\sin^2 \\theta = 1$ and $\\cos^2 \\theta - \\sin^2 \\theta = \\cos 2\\theta$ to reach the RHS A1","marks":3,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"sincos","label":"Converting to Sine and Cosine","steps":[{"type":"text","order":0,"title":"Convert to sine and cosine","content":"To prove the identity, let's start with the Left-Hand Side (LHS):\n\n$$\\text{LHS} = \\frac{\\cot \\theta - \\tan \\theta}{\\cot \\theta + \\tan \\theta}$$\n\nUsing the fundamental trigonometric definitions, we substitute:\n\n$$\\cot \\theta = \\frac{\\cos \\theta}{\\sin \\theta} \\quad \\text{and} \\quad \\tan \\theta = \\frac{\\sin \\theta}{\\cos \\theta}$$\n\nSubstituting these into the LHS gives:\n\n$$\\text{LHS} = \\frac{\\frac{\\cos \\theta}{\\sin \\theta} - \\frac{\\sin \\theta}{\\cos \\theta}}{\\frac{\\cos \\theta}{\\sin \\theta} + \\frac{\\sin \\theta}{\\cos \\theta}}$$\n\n*This first step of substituting definitions earns you your first mark (M1).*","highlights":[{"text":"\\frac{\\cot \\theta - \\tan \\theta}{\\cot \\theta + \\tan \\theta} \\equiv \\cos 2\\theta","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\text{LHS} = \\frac{\\frac{\\htmlData{anchor=num1}{\\cos^2 \\theta} - \\htmlData{anchor=num2}{\\sin^2 \\theta}}{\\htmlData{anchor=denom-common1}{\\sin \\theta \\cos \\theta}}}{\\frac{\\htmlData{anchor=denom1}{\\cos^2 \\theta} + \\htmlData{anchor=denom2}{\\sin^2 \\theta}}{\\htmlData{anchor=denom-common2}{\\sin \\theta \\cos \\theta}}}$$"},{"latex":"$$$\\text{LHS} = \\frac{\\htmlData{anchor=num-diff}{\\cos^2 \\theta - \\sin^2 \\theta}}{\\htmlData{anchor=denom-sum}{\\cos^2 \\theta + \\sin^2 \\theta}}$$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"num-diff"},{"color":"teal","style":"text-color","anchor":"denom-sum"}],"connections":[{"to":"denom-common2","from":"denom-common1","color":"sky","label":"cancel common denominator","style":"dashed"}]},"order":1,"title":"Simplify the compound fraction","description":"Next, find a common denominator for the numerator and the denominator separately, and simplify. The common denominator for both is $\\sin\\theta\\cos\\theta$."},{"type":"text","order":2,"title":"Apply key identities","content":"Now, we recall two standard trigonometric identities from the formula booklet:\n\n1. The Pythagorean identity:\n $$\\cos^2 \\theta + \\sin^2 \\theta = 1$$\n\n2. The double-angle identity for cosine:\n $$\\cos^2 \\theta - \\sin^2 \\theta = \\cos 2\\theta$$\n\nSubstituting these two identities back into our simplified fraction:\n\n$$\\text{LHS} = \\frac{\\cos 2\\theta}{1} = \\cos 2\\theta$$\n\nSince this is equal to the Right-Hand Side (RHS), the identity is proven:\n\n$$\\text{LHS} \\equiv \\text{RHS}$$\n\n*Using these identities to obtain the final simplified expression completes the proof and earns the remaining marks (M1, A1).*"}]},{"id":"tangent","label":"Converting to Tangent","steps":[{"type":"text","order":0,"title":"Substitute cotangent","content":"We start with the Left-Hand Side (LHS) of the identity:\n\n$$\\text{LHS} = \\frac{\\cot\\theta - \\tan\\theta}{\\cot\\theta + \\tan\\theta}$$\n\nTo use the **Converting to Tangent** approach, we substitute the fundamental identity for cotangent, $\\cot\\theta = \\frac{1}{\\tan\\theta}$, into the expression.","highlights":[{"text":"\\frac{\\cot \\theta - \\tan \\theta}{\\cot \\theta + \\tan \\theta} \\equiv \\cos 2\\theta","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{\\htmlData{anchor=cot1}{\\color{@sky}{\\cot\\theta}} - \\tan\\theta}{\\htmlData{anchor=cot2}{\\color{@sky}{\\cot\\theta}} + \\tan\\theta}$"},{"latex":"$$\\frac{\\htmlData{anchor=tan1}{\\color{@sky}{\\frac{1}{\\tan\\theta}}} - \\tan\\theta}{\\htmlData{anchor=tan2}{\\color{@sky}{\\frac{1}{\\tan\\theta}}} + \\tan\\theta}$"},{"latex":"$$\\frac{\\left(\\frac{1}{\\tan\\theta} - \\tan\\theta\\right) \\times \\color{@amber}{\\tan\\theta}}{\\left(\\frac{1}{\\tan\\theta} + \\tan\\theta\\right) \\times \\color{@amber}{\\tan\\theta}} = \\frac{1 - \\tan^2\\theta}{1 + \\tan^2\\theta}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"cot1"},{"color":"sky","style":"text-color","anchor":"cot2"},{"color":"sky","style":"text-color","anchor":"tan1"},{"color":"sky","style":"text-color","anchor":"tan2"}],"connections":[{"to":"tan1","from":"cot1","color":"sky","label":"substitute","style":"solid"},{"to":"tan2","from":"cot2","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Simplify the fraction","description":"Substitute $\\cot\\theta$ with $\\frac{1}{\\tan\\theta}$ and simplify by multiplying both the numerator and denominator by $\\tan\\theta$."},{"type":"text","order":2,"title":"Convert tangent to sine and cosine","content":"Next, let's rewrite the tangent terms using $\\tan\\theta = \\frac{\\sin\\theta}{\\cos\\theta}$:\n\n$$1 - \\tan^2\\theta = 1 - \\frac{\\sin^2\\theta}{\\cos^2\\theta} = \\frac{\\cos^2\\theta - \\sin^2\\theta}{\\cos^2\\theta}$$\n\nAnd for the denominator, using the Pythagorean identity:\n\n$$1 + \\tan^2\\theta = \\sec^2\\theta = \\frac{1}{\\cos^2\\theta}$$\n\nSubstituting these expressions into our fraction:\n\n$$\\frac{1 - \\tan^2\\theta}{1 + \\tan^2\\theta} = \\frac{\\frac{\\cos^2\\theta - \\sin^2\\theta}{\\cos^2\\theta}}{\\frac{1}{\\cos^2\\theta}}$$"},{"type":"text","order":3,"title":"Simplify and conclude","content":"The $\\cos^2\\theta$ terms in the denominators cancel out:\n\n$$\\frac{\\cos^2\\theta - \\sin^2\\theta}{\\cos^2\\theta} \\times \\frac{\\cos^2\\theta}{1} = \\cos^2\\theta - \\sin^2\\theta$$\n\nUsing the double-angle identity for cosine:\n\n$$\\cos 2\\theta \\equiv \\cos^2\\theta - \\sin^2\\theta$$\n\nThus, we have successfully shown that:\n\n$$\\frac{\\cot\\theta - \\tan\\theta}{\\cot\\theta + \\tan\\theta} \\equiv \\cos 2\\theta$$\n\nThis completes the proof!"}]}],"generatedAt":"2026-06-11T03:28:45.690Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"0fb80fc6-d968-4264-8bab-e2be1fb03811","label":"ii","content":"The power output \$ P \$ (in Watts) of a specific AC circuit component is modelled by the function \$ P(t) = 11 \\cos^2(2t - 40^\\circ) \$, where \$ t \$ is a phase angle measured in degrees and \$ 0^\\circ \\le t < 90^\\circ \$.\n\nDetermine the values of \$ t \$ for which the power output is exactly \$ 3 \$ Watts, giving your answers to one decimal place. (Solutions based entirely on graphical or numerical methods are not acceptable.)","markscheme":"- Sets up the equation $11 \\cos^2(2t - 40^\\circ) = 3$ and deduces $\\cos^2(2t - 40^\\circ) = \\frac{3}{11}$ B1\n- Takes the square root to find $\\cos(2t - 40^\\circ) = \\pm \\sqrt{\\frac{3}{11}}$ M1\n- Finds at least one correct principal value for $(2t - 40^\\circ)$, e.g., $58.51...^\\circ$ or $121.48...^\\circ$ A1\n- Solves for $t$ by adding $40^\\circ$ and dividing by $2$ for both cases within the range $-40^\\circ \\le 2t - 40^\\circ < 140^\\circ$ M1\n- Obtains $t = 49.3^\\circ$ and $t = 80.7^\\circ$ A1","marks":5,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"direct_algebraic","label":"Direct Square Root Method","steps":[{"type":"text","order":0,"title":"Set up the equation","content":"To find when the power output $P(t)$ is exactly $3$ Watts, we set the model equation equal to $3$:\n\n$$11 \\cos^2(2t - 40^\\circ) = 3$$\n\nNext, isolate the squared term by dividing both sides of the equation by $11$:\n\n$$\\cos^2(2t - 40^\\circ) = \\frac{3}{11}$$\n\nThis gives us our starting point for the direct square root method.","highlights":[{"text":"P(t) = 11 \\cos^2(2t - 40^\\circ)","location":"specification"},{"text":"3","location":"specification"}]},{"type":"text","order":1,"title":"Apply direct square root","content":"Taking the square root of both sides gives two possible cases (positive and negative):\n\n$$\\cos(2t - 40^\\circ) = \\pm\\sqrt{\\frac{3}{11}}$$\n\nThis splits our problem into two simpler first-order trigonometric equations to solve:\n\n1. $$\\cos(2t - 40^\\circ) = \\sqrt{\\frac{3}{11}}$$\n2. $$\\cos(2t - 40^\\circ) = -\\sqrt{\\frac{3}{11}}$$"},{"type":"text","order":2,"title":"Determine the angle interval","content":"Before finding the angles, let's find the valid interval for the compound angle $2t - 40^\\circ$.\n\nWe are given the domain for $t$:\n$$0^\\circ \\le t < 90^\\circ$$\n\nMultiply the entire inequality by $2$:\n$$0^\\circ \\le 2t < 180^\\circ$$\n\nSubtract $40^\\circ$:\n$$-40^\\circ \\le 2t - 40^\\circ < 140^\\circ$$\n\nAny solutions for $2t - 40^\\circ$ must lie within this interval.","highlights":[{"text":"0^\\circ \\le t < 90^\\circ","location":"specification"}]},{"type":"text","order":3,"title":"Find principal values","content":"Now, let's calculate the principal value for each case using the inverse cosine function.\n\nFor the positive case:\n$$2t - 40^\\circ = \\arccos\\left(\\sqrt{\\frac{3}{11}}\\right) \\approx 58.515^\\circ$$\n\nSince $58.515^\\circ$ is within our interval $[-40^\\circ, 140^\\circ)$, this is a valid solution.\n\nFor the negative case:\n$$2t - 40^\\circ = \\arccos\\left(-\\sqrt{\\frac{3}{11}}\\right) \\approx 121.485^\\circ$$\n\nSince $121.485^\\circ$ is also within our interval $[-40^\\circ, 140^\\circ)$, this is another valid solution.\n\n*(Note: Other symmetric solutions like $-58.515^\\circ$ or $-121.485^\\circ$ fall outside our interval $[-40^\\circ, 140^\\circ)$.)*"},{"type":"text","order":4,"title":"Solve for t","content":"We now solve for $t$ in each of our two cases by adding $40^\\circ$ and dividing by $2$.\n\n**Case 1:**\n$$\\begin{aligned}\n2t - 40^\\circ &= 58.515^\\circ \\\\\n2t &= 98.515^\\circ \\\\\nt &= 49.257...^\\circ \\approx 49.3^\\circ\n\\end{aligned}$$\n\n**Case 2:**\n$$\\begin{aligned}\n2t - 40^\\circ &= 121.485^\\circ \\\\\n2t &= 161.485^\\circ \\\\\nt &= 80.742...^\\circ \\approx 80.7^\\circ\n\\end{aligned}$$\n\nBoth values lie within $0^\\circ \\le t < 90^\\circ$.\n\nThus, to one decimal place, the values of $t$ are **$49.3^\\circ$** and **$80.7^\\circ$**.","highlights":[{"text":"one decimal place","location":"specification"}]}]},{"id":"double_angle","label":"Double-Angle Identity","steps":[{"type":"text","order":0,"title":"Set up the equation","content":"We want to find the values of $t$ for which the power output $P(t)$ is exactly $3$ Watts. We start by setting the function expression equal to $3$:\n\n$$11 \text{cos}^2(2t - 40^\\circ) = 3$$","highlights":[{"text":"power output P (in Watts) of a specific AC circuit component is modelled by the function P(t) = 11 \\cos^2(2t - 40^\\circ)","location":"specification"},{"text":"exactly 3 Watts","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"text":"We can rewrite the squared cosine term using the double-angle identity:"},{"latex":"$$$\\htmlData{anchor=double-angle-id}{\\cos^2\\theta = \\frac{1 + \\cos(2\\theta)}{2}}$$"},{"text":"Substituting $\\theta = 2t - 40^\\circ$ gives:"},{"latex":"$$$\\htmlData{anchor=cos-sq-sub}{\\cos^2(2t - 40^\\circ)} = \\frac{1 + \\cos(4t - 80^\\circ)}{2}$$"}],"highlights":[{"color":"sky","style":"box","anchor":"double-angle-id"},{"color":"sky","style":"box","anchor":"cos-sq-sub"}],"connections":[{"to":"cos-sq-sub","from":"double-angle-id","color":"sky","label":"Substitute \\theta = 2t - 40^\\circ","style":"solid"}]},"order":1,"title":"Apply double-angle identity","description":"Using the double-angle identity allows us to linearise the quadratic trigonometric expression."},{"type":"text","order":2,"title":"Substitute and simplify","content":"Now substitute this identity back into our original equation:\n\n$$11 \\left( \\frac{1 + \\cos(4t - 80^\\circ)}{2} \\right) = 3$$\n\nMultiply both sides by $2$ and divide by $11$:\n\n$$1 + \\cos(4t - 80^\\circ) = \\frac{6}{11}$$\n\nSubtract $1$ from both sides to isolate the cosine term:\n\n$$\\cos(4t - 80^\\circ) = \\frac{6}{11} - 1 = -\\frac{5}{11}$$"},{"type":"text","order":3,"title":"Determine the interval","content":"Before solving for $t$, we need to find the interval of the angle argument $(4t - 80^\\circ)$ based on the given domain $0^\\circ \\le t < 90^\\circ$:\n\n$$\\begin{aligned}\n0^\\circ &\\le t < 90^\\circ \\\\\n0^\\circ &\\le 4t < 360^\\circ \\\\\n-80^\\circ &\\le 4t - 80^\\circ < 280^\\circ\n\\end{aligned}$$\n\nSo we are looking for values of $\\theta = 4t - 80^\\circ$ in the range $[-80^\\circ, 280^\\circ)$ that satisfy:\n\n$$\\cos(\\theta) = -\\frac{5}{11}$$","highlights":[{"text":"0^\\circ \\le t < 90^\\circ","location":"specification"}]},{"type":"text","order":4,"title":"Find principal values","content":"First, find the principal value using the inverse cosine function:\n\n$$\\theta_1 = \\arccos\\left(-\\frac{5}{11}\\right) \\approx 117.035^\\circ$$\n\nSince cosine is symmetric about the y-axis, the other solution in the standard interval is:\n\n$$\\theta_2 = 360^\\circ - 117.035^\\circ = 242.965^\\circ$$\n\nAnother possible value is the negative reflection of the principal value:\n\n$$\\theta_3 = -117.035^\\circ$$\n\nHowever, we must check which of these lies in our interval $[-80^\\circ, 280^\\circ)$:\n- $\\theta_3 = -117.035^\\circ$ is less than $-80^\\circ$ (out of range).\n- $\\theta_1 = 117.035^\\circ$ is within $[-80^\\circ, 280^\\circ)$ (valid).\n- $\\theta_2 = 242.965^\\circ$ is within $[-80^\\circ, 280^\\circ)$ (valid)."},{"type":"text","order":5,"title":"Solve for t","content":"We now set $4t - 80^\\circ$ equal to our two valid angles and solve for $t$:\n\n**Case 1:**\n$$\\begin{aligned}\n4t - 80^\\circ &= 117.035^\\circ \\\\\n4t &= 197.035^\\circ \\\\\nt &= 49.258...^\\circ \\approx 49.3^\\circ\n\\end{aligned}$$\n\n**Case 2:**\n$$\\begin{aligned}\n4t - 80^\\circ &= 242.965^\\circ \\\\\n4t &= 322.965^\\circ \\\\\nt &= 80.741...^\\circ \\approx 80.7^\\circ\n\\end{aligned}$$\n\nBoth values lie within our original domain of $0^\\circ \\le t < 90^\\circ$.\n\nThus, the values of $t$ to one decimal place are $t = 49.3^\\circ$ and $t = 80.7^\\circ$.","highlights":[{"text":"one decimal place","location":"specification"}]}]}],"generatedAt":"2026-06-11T03:29:07.325Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"b5c616b6-a7b9-4936-b911-3a31f43f62d7:sincos","label":"(i) Converting to Sine and Cosine","steps":[{"type":"text","order":0,"title":"Convert to sine and cosine","content":"To prove the identity, let's start with the Left-Hand Side (LHS):\n\n$$\\text{LHS} = \\frac{\\cot \\theta - \\tan \\theta}{\\cot \\theta + \\tan \\theta}$$\n\nUsing the fundamental trigonometric definitions, we substitute:\n\n$$\\cot \\theta = \\frac{\\cos \\theta}{\\sin \\theta} \\quad \\text{and} \\quad \\tan \\theta = \\frac{\\sin \\theta}{\\cos \\theta}$$\n\nSubstituting these into the LHS gives:\n\n$$\\text{LHS} = \\frac{\\frac{\\cos \\theta}{\\sin \\theta} - \\frac{\\sin \\theta}{\\cos \\theta}}{\\frac{\\cos \\theta}{\\sin \\theta} + \\frac{\\sin \\theta}{\\cos \\theta}}$$\n\n*This first step of substituting definitions earns you your first mark (M1).*","highlights":[{"text":"\\frac{\\cot \\theta - \\tan \\theta}{\\cot \\theta + \\tan \\theta} \\equiv \\cos 2\\theta","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\text{LHS} = \\frac{\\frac{\\htmlData{anchor=num1}{\\cos^2 \\theta} - \\htmlData{anchor=num2}{\\sin^2 \\theta}}{\\htmlData{anchor=denom-common1}{\\sin \\theta \\cos \\theta}}}{\\frac{\\htmlData{anchor=denom1}{\\cos^2 \\theta} + \\htmlData{anchor=denom2}{\\sin^2 \\theta}}{\\htmlData{anchor=denom-common2}{\\sin \\theta \\cos \\theta}}}$$"},{"latex":"$$$\\text{LHS} = \\frac{\\htmlData{anchor=num-diff}{\\cos^2 \\theta - \\sin^2 \\theta}}{\\htmlData{anchor=denom-sum}{\\cos^2 \\theta + \\sin^2 \\theta}}$$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"num-diff"},{"color":"teal","style":"text-color","anchor":"denom-sum"}],"connections":[{"to":"denom-common2","from":"denom-common1","color":"sky","label":"cancel common denominator","style":"dashed"}]},"order":1,"title":"Simplify the compound fraction","description":"Next, find a common denominator for the numerator and the denominator separately, and simplify. The common denominator for both is $\\sin\\theta\\cos\\theta$."},{"type":"text","order":2,"title":"Apply key identities","content":"Now, we recall two standard trigonometric identities from the formula booklet:\n\n1. The Pythagorean identity:\n $$\\cos^2 \\theta + \\sin^2 \\theta = 1$$\n\n2. The double-angle identity for cosine:\n $$\\cos^2 \\theta - \\sin^2 \\theta = \\cos 2\\theta$$\n\nSubstituting these two identities back into our simplified fraction:\n\n$$\\text{LHS} = \\frac{\\cos 2\\theta}{1} = \\cos 2\\theta$$\n\nSince this is equal to the Right-Hand Side (RHS), the identity is proven:\n\n$$\\text{LHS} \\equiv \\text{RHS}$$\n\n*Using these identities to obtain the final simplified expression completes the proof and earns the remaining marks (M1, A1).*"}]},{"id":"b5c616b6-a7b9-4936-b911-3a31f43f62d7:tangent","label":"(i) Converting to Tangent","steps":[{"type":"text","order":0,"title":"Substitute cotangent","content":"We start with the Left-Hand Side (LHS) of the identity:\n\n$$\\text{LHS} = \\frac{\\cot\\theta - \\tan\\theta}{\\cot\\theta + \\tan\\theta}$$\n\nTo use the **Converting to Tangent** approach, we substitute the fundamental identity for cotangent, $\\cot\\theta = \\frac{1}{\\tan\\theta}$, into the expression.","highlights":[{"text":"\\frac{\\cot \\theta - \\tan \\theta}{\\cot \\theta + \\tan \\theta} \\equiv \\cos 2\\theta","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{\\htmlData{anchor=cot1}{\\color{@sky}{\\cot\\theta}} - \\tan\\theta}{\\htmlData{anchor=cot2}{\\color{@sky}{\\cot\\theta}} + \\tan\\theta}$"},{"latex":"$$\\frac{\\htmlData{anchor=tan1}{\\color{@sky}{\\frac{1}{\\tan\\theta}}} - \\tan\\theta}{\\htmlData{anchor=tan2}{\\color{@sky}{\\frac{1}{\\tan\\theta}}} + \\tan\\theta}$"},{"latex":"$$\\frac{\\left(\\frac{1}{\\tan\\theta} - \\tan\\theta\\right) \\times \\color{@amber}{\\tan\\theta}}{\\left(\\frac{1}{\\tan\\theta} + \\tan\\theta\\right) \\times \\color{@amber}{\\tan\\theta}} = \\frac{1 - \\tan^2\\theta}{1 + \\tan^2\\theta}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"cot1"},{"color":"sky","style":"text-color","anchor":"cot2"},{"color":"sky","style":"text-color","anchor":"tan1"},{"color":"sky","style":"text-color","anchor":"tan2"}],"connections":[{"to":"tan1","from":"cot1","color":"sky","label":"substitute","style":"solid"},{"to":"tan2","from":"cot2","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Simplify the fraction","description":"Substitute $\\cot\\theta$ with $\\frac{1}{\\tan\\theta}$ and simplify by multiplying both the numerator and denominator by $\\tan\\theta$."},{"type":"text","order":2,"title":"Convert tangent to sine and cosine","content":"Next, let's rewrite the tangent terms using $\\tan\\theta = \\frac{\\sin\\theta}{\\cos\\theta}$:\n\n$$1 - \\tan^2\\theta = 1 - \\frac{\\sin^2\\theta}{\\cos^2\\theta} = \\frac{\\cos^2\\theta - \\sin^2\\theta}{\\cos^2\\theta}$$\n\nAnd for the denominator, using the Pythagorean identity:\n\n$$1 + \\tan^2\\theta = \\sec^2\\theta = \\frac{1}{\\cos^2\\theta}$$\n\nSubstituting these expressions into our fraction:\n\n$$\\frac{1 - \\tan^2\\theta}{1 + \\tan^2\\theta} = \\frac{\\frac{\\cos^2\\theta - \\sin^2\\theta}{\\cos^2\\theta}}{\\frac{1}{\\cos^2\\theta}}$$"},{"type":"text","order":3,"title":"Simplify and conclude","content":"The $\\cos^2\\theta$ terms in the denominators cancel out:\n\n$$\\frac{\\cos^2\\theta - \\sin^2\\theta}{\\cos^2\\theta} \\times \\frac{\\cos^2\\theta}{1} = \\cos^2\\theta - \\sin^2\\theta$$\n\nUsing the double-angle identity for cosine:\n\n$$\\cos 2\\theta \\equiv \\cos^2\\theta - \\sin^2\\theta$$\n\nThus, we have successfully shown that:\n\n$$\\frac{\\cot\\theta - \\tan\\theta}{\\cot\\theta + \\tan\\theta} \\equiv \\cos 2\\theta$$\n\nThis completes the proof!"}]},{"id":"0fb80fc6-d968-4264-8bab-e2be1fb03811:direct_algebraic","label":"(ii) Direct Square Root Method","steps":[{"type":"text","order":0,"title":"Set up the equation","content":"To find when the power output $P(t)$ is exactly $3$ Watts, we set the model equation equal to $3$:\n\n$$11 \\cos^2(2t - 40^\\circ) = 3$$\n\nNext, isolate the squared term by dividing both sides of the equation by $11$:\n\n$$\\cos^2(2t - 40^\\circ) = \\frac{3}{11}$$\n\nThis gives us our starting point for the direct square root method.","highlights":[{"text":"P(t) = 11 \\cos^2(2t - 40^\\circ)","location":"specification"},{"text":"3","location":"specification"}]},{"type":"text","order":1,"title":"Apply direct square root","content":"Taking the square root of both sides gives two possible cases (positive and negative):\n\n$$\\cos(2t - 40^\\circ) = \\pm\\sqrt{\\frac{3}{11}}$$\n\nThis splits our problem into two simpler first-order trigonometric equations to solve:\n\n1. $$\\cos(2t - 40^\\circ) = \\sqrt{\\frac{3}{11}}$$\n2. $$\\cos(2t - 40^\\circ) = -\\sqrt{\\frac{3}{11}}$$"},{"type":"text","order":2,"title":"Determine the angle interval","content":"Before finding the angles, let's find the valid interval for the compound angle $2t - 40^\\circ$.\n\nWe are given the domain for $t$:\n$$0^\\circ \\le t < 90^\\circ$$\n\nMultiply the entire inequality by $2$:\n$$0^\\circ \\le 2t < 180^\\circ$$\n\nSubtract $40^\\circ$:\n$$-40^\\circ \\le 2t - 40^\\circ < 140^\\circ$$\n\nAny solutions for $2t - 40^\\circ$ must lie within this interval.","highlights":[{"text":"0^\\circ \\le t < 90^\\circ","location":"specification"}]},{"type":"text","order":3,"title":"Find principal values","content":"Now, let's calculate the principal value for each case using the inverse cosine function.\n\nFor the positive case:\n$$2t - 40^\\circ = \\arccos\\left(\\sqrt{\\frac{3}{11}}\\right) \\approx 58.515^\\circ$$\n\nSince $58.515^\\circ$ is within our interval $[-40^\\circ, 140^\\circ)$, this is a valid solution.\n\nFor the negative case:\n$$2t - 40^\\circ = \\arccos\\left(-\\sqrt{\\frac{3}{11}}\\right) \\approx 121.485^\\circ$$\n\nSince $121.485^\\circ$ is also within our interval $[-40^\\circ, 140^\\circ)$, this is another valid solution.\n\n*(Note: Other symmetric solutions like $-58.515^\\circ$ or $-121.485^\\circ$ fall outside our interval $[-40^\\circ, 140^\\circ)$.)*"},{"type":"text","order":4,"title":"Solve for t","content":"We now solve for $t$ in each of our two cases by adding $40^\\circ$ and dividing by $2$.\n\n**Case 1:**\n$$\\begin{aligned}\n2t - 40^\\circ &= 58.515^\\circ \\\\\n2t &= 98.515^\\circ \\\\\nt &= 49.257...^\\circ \\approx 49.3^\\circ\n\\end{aligned}$$\n\n**Case 2:**\n$$\\begin{aligned}\n2t - 40^\\circ &= 121.485^\\circ \\\\\n2t &= 161.485^\\circ \\\\\nt &= 80.742...^\\circ \\approx 80.7^\\circ\n\\end{aligned}$$\n\nBoth values lie within $0^\\circ \\le t < 90^\\circ$.\n\nThus, to one decimal place, the values of $t$ are **$49.3^\\circ$** and **$80.7^\\circ$**.","highlights":[{"text":"one decimal place","location":"specification"}]}]},{"id":"0fb80fc6-d968-4264-8bab-e2be1fb03811:double_angle","label":"(ii) Double-Angle Identity","steps":[{"type":"text","order":0,"title":"Set up the equation","content":"We want to find the values of $t$ for which the power output $P(t)$ is exactly $3$ Watts. We start by setting the function expression equal to $3$:\n\n$$11 \\text{cos}^2(2t - 40^\\circ) = 3$$","highlights":[{"text":"power output P (in Watts) of a specific AC circuit component is modelled by the function P(t) = 11 \\cos^2(2t - 40^\\circ)","location":"specification"},{"text":"exactly 3 Watts","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"text":"We can rewrite the squared cosine term using the double-angle identity:"},{"latex":"$$$\\htmlData{anchor=double-angle-id}{\\cos^2\\theta = \\frac{1 + \\cos(2\\theta)}{2}}$$"},{"text":"Substituting $\\theta = 2t - 40^\\circ$ gives:"},{"latex":"$$$\\htmlData{anchor=cos-sq-sub}{\\cos^2(2t - 40^\\circ)} = \\frac{1 + \\cos(4t - 80^\\circ)}{2}$$"}],"highlights":[{"color":"sky","style":"box","anchor":"double-angle-id"},{"color":"sky","style":"box","anchor":"cos-sq-sub"}],"connections":[{"to":"cos-sq-sub","from":"double-angle-id","color":"sky","label":"Substitute \\theta = 2t - 40^\\circ","style":"solid"}]},"order":1,"title":"Apply double-angle identity","description":"Using the double-angle identity allows us to linearise the quadratic trigonometric expression."},{"type":"text","order":2,"title":"Substitute and simplify","content":"Now substitute this identity back into our original equation:\n\n$$11 \\left( \\frac{1 + \\cos(4t - 80^\\circ)}{2} \\right) = 3$$\n\nMultiply both sides by $2$ and divide by $11$:\n\n$$1 + \\cos(4t - 80^\\circ) = \\frac{6}{11}$$\n\nSubtract $1$ from both sides to isolate the cosine term:\n\n$$\\cos(4t - 80^\\circ) = \\frac{6}{11} - 1 = -\\frac{5}{11}$$"},{"type":"text","order":3,"title":"Determine the interval","content":"Before solving for $t$, we need to find the interval of the angle argument $(4t - 80^\\circ)$ based on the given domain $0^\\circ \\le t < 90^\\circ$:\n\n$$\\begin{aligned}\n0^\\circ &\\le t < 90^\\circ \\\\\n0^\\circ &\\le 4t < 360^\\circ \\\\\n-80^\\circ &\\le 4t - 80^\\circ < 280^\\circ\n\\end{aligned}$$\n\nSo we are looking for values of $\\theta = 4t - 80^\\circ$ in the range $[-80^\\circ, 280^\\circ)$ that satisfy:\n\n$$\\cos(\\theta) = -\\frac{5}{11}$$","highlights":[{"text":"0^\\circ \\le t < 90^\\circ","location":"specification"}]},{"type":"text","order":4,"title":"Find principal values","content":"First, find the principal value using the inverse cosine function:\n\n$$\\theta_1 = \\arccos\\left(-\\frac{5}{11}\\right) \\approx 117.035^\\circ$$\n\nSince cosine is symmetric about the y-axis, the other solution in the standard interval is:\n\n$$\\theta_2 = 360^\\circ - 117.035^\\circ = 242.965^\\circ$$\n\nAnother possible value is the negative reflection of the principal value:\n\n$$\\theta_3 = -117.035^\\circ$$\n\nHowever, we must check which of these lies in our interval $[-80^\\circ, 280^\\circ)$:\n- $\\theta_3 = -117.035^\\circ$ is less than $-80^\\circ$ (out of range).\n- $\\theta_1 = 117.035^\\circ$ is within $[-80^\\circ, 280^\\circ)$ (valid).\n- $\\theta_2 = 242.965^\\circ$ is within $[-80^\\circ, 280^\\circ)$ (valid)."},{"type":"text","order":5,"title":"Solve for t","content":"We now set $4t - 80^\\circ$ equal to our two valid angles and solve for $t$:\n\n**Case 1:**\n$$\\begin{aligned}\n4t - 80^\\circ &= 117.035^\\circ \\\\\n4t &= 197.035^\\circ \\\\\nt &= 49.258...^\\circ \\approx 49.3^\\circ\n\\end{aligned}$$\n\n**Case 2:**\n$$\\begin{aligned}\n4t - 80^\\circ &= 242.965^\\circ \\\\\n4t &= 322.965^\\circ \\\\\nt &= 80.741...^\\circ \\approx 80.7^\\circ\n\\end{aligned}$$\n\nBoth values lie within our original domain of $0^\\circ \\le t < 90^\\circ$.\n\nThus, the values of $t$ to one decimal place are $t = 49.3^\\circ$ and $t = 80.7^\\circ$.","highlights":[{"text":"one decimal 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Trigonometry (A-level only)","topicId":"97cb7c2e-f32f-45ff-ada1-651a91e4cf22","topicSidebar":"$L3c","topicSlug":"trigonometry-8","topicTitle":"3.4 Trigonometry (A-level only)"}]}],"$L3d","$L3e"]}]

3.4 Trigonometry (A-level only)

Question 63

3.4 Trigonometry (A-level only) Questions

3.4 Trigonometry (A-level only)

Question 63

3.4 Trigonometry (A-level only) Questions