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Maths exam-style questions with worked solutions on MathsGenie.\",\"learningResourceType\":\"Practice questions\",\"educationalLevel\":\"A Level\",\"about\":{\"@type\":\"Thing\",\"name\":\"1.5.5 Trigonometry\"},\"url\":\"https://mathsgenie.co.uk/a-level/maths/wjec/trigonometry-6/questions/q37\",\"inLanguage\":\"en-GB\",\"provider\":{\"@id\":\"https://mathsgenie.co.uk/#organization\"}}"}}],["$","div",null,{"className":"min-h-0 w-full flex-1 overflow-hidden bg-background","children":["$","$L3b",null,{"backHref":"/a-level/maths/wjec/questions","basePath":"/a-level/maths/wjec/trigonometry-6/questions","formulaUrl":"https://www.wjec.co.uk/media/4jfdytcz/ss-w27-18-e.pdf","initialIndex":36,"initialQuestionWindow":[{"id":"99044bf8-6ee5-436d-aa4a-345d00112f02","revisionId":"58836003-b5f0-4da8-ac37-5e2be6272215","questionNumber":null,"questionType":"LA","specification":"A robotic arm rotates such that its displacement angle \$\\alpha\$ in radians satisfies the drive torque equation:\n\\[ 3 \\sin \\alpha (4 \\cot^2 2\\alpha - 5) = 8 \\sec \\alpha \\]","difficulty":7,"difficultyBand":"hard","tier":null,"totalMarks":9,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"a626451f-3830-4222-932b-3e2c8704efc8","label":"a","content":"Show that this equation can be expressed in the form\n\\[ 12 \\csc^2 2\\alpha - 16 \\csc 2\\alpha - 27 = 0 \\]","markscheme":"- Re-writes the equation as \$ 4 \\cot^2 2\\alpha - 5 = \\frac{8 \\sec \\alpha}{3 \\sin \\alpha} \$ or \$ 3 \\sin \\alpha \\cos \\alpha (4 \\cot^2 2\\alpha - 5) = 8 \$ M1\n- Recognises that \$ \\frac{8}{3 \\sin \\alpha \\cos \\alpha} = \\frac{16}{3 \\sin 2\\alpha} \$ or uses the double angle identity \$ \\sin 2\\alpha = 2 \\sin \\alpha \\cos \\alpha \$ to obtain \$ \\frac{16}{3} \\csc 2\\alpha \$ M1\n- Substitutes the identity \$ \\cot^2 2\\alpha = \\csc^2 2\\alpha - 1 \$ into their expression M1\n- Rearranges to reaching the target quadratic form: \$ 12 (\\csc^2 2\\alpha - 1) - 15 = 16 \\csc 2\\alpha \\rightarrow 12 \\csc^2 2\\alpha - 16 \\csc 2\\alpha - 27 = 0 \$ A1","marks":4,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"rhs_first","label":"Simplifying Right-Hand Side First","steps":[{"type":"text","order":0,"title":"Isolate the bracketed term","content":"To begin, we want to isolate the term containing the double angle, $4 \\cot^2 2\\alpha - 5$. We can do this by dividing both sides of the equation by $3 \\sin \\alpha$:\n\n$$4 \\cot^2 2\\alpha - 5 = \\frac{8 \\sec \\alpha}{3 \\sin \\alpha}$$\n\nThis separates the functions of the single angle $\\alpha$ on the right-hand side so we can simplify them first."},{"type":"text","order":1,"title":"Simplify the right-hand side","content":"Next, let's rewrite the right-hand side by expressing $\\sec \\alpha$ in terms of $\\cos \\alpha$ using the reciprocal identity $\\sec \\alpha = \\frac{1}{\\cos \\alpha}$:\n\n$$\\frac{8 \\sec \\alpha}{3 \\sin \\alpha} = \\frac{8}{3 \\sin \\alpha \\cos \\alpha}$$"},{"type":"text","order":2,"title":"Apply the double-angle formula","content":"Recall the double-angle formula for sine:\n\n$$\\sin 2\\alpha = 2 \\sin \\alpha \\cos \\alpha \\implies \\sin \\alpha \\cos \\alpha = \\frac{1}{2} \\sin 2\\alpha$$\n\nSubstituting this into our expression for the right-hand side gives:\n\n$$\\frac{8}{3 \\sin \\alpha \\cos \\alpha} = \\frac{8}{3 \\left(\\frac{1}{2} \\sin 2\\alpha\\right)} = \\frac{16}{3 \\sin 2\\alpha}$$\n\nUsing the reciprocal identity $\\csc 2\\alpha = \\frac{1}{\\sin 2\\alpha}$, we get:\n\n$$4 \\cot^2 2\\alpha - 5 = \\frac{16}{3} \\csc 2\\alpha$$"},{"type":"text","order":3,"title":"Use the Pythagorean identity","content":"To form a quadratic equation solely in terms of $\\csc 2\\alpha$, we substitute the Pythagorean identity:\n\n$$\\cot^2 2\\alpha = \\csc^2 2\\alpha - 1$$\n\nThis gives:\n\n$$4(\\csc^2 2\\alpha - 1) - 5 = \\frac{16}{3} \\csc 2\\alpha$$\n\nMultiply the entire equation by $3$ to eliminate the fraction:\n\n$$12(\\csc^2 2\\alpha - 1) - 15 = 16 \\csc 2\\alpha$$\n\n$$12 \\csc^2 2\\alpha - 12 - 15 = 16 \\csc 2\\alpha$$"},{"type":"text","order":4,"title":"Rearrange to the final form","content":"Finally, group all terms on the left-hand side to obtain the target quadratic form:\n\n$$12 \\csc^2 2\\alpha - 16 \\csc 2\\alpha - 27 = 0$$\n\nThis successfully shows the required expression."},{"type":"text","order":5,"title":"Solve the quadratic equation","content":"For Part 2, let $x = \\csc 2\\alpha$. The quadratic equation is:\n\n$$12x^2 - 16x - 27 = 0$$\n\nApplying the quadratic formula:\n\n$$x = \\frac{16 \\pm \\sqrt{(-16)^2 - 4(12)(-27)}}{2(12)}$$\n\n$$x = \\frac{16 \\pm \\sqrt{256 + 1296}}{24} = \\frac{16 \\pm \\sqrt{1552}}{24} = \\frac{4 \\pm \\sqrt{97}}{6}$$\n\nCalculating the two values:\n- $x \\approx 2.308$\n- $x \\approx -0.975$\n\nSince $0 < \\alpha < \\frac{\\pi}{2}$, we have $0 < 2\\alpha < \\pi$, which means $\\sin 2\\alpha > 0$. Thus, we reject the negative root, leaving:\n\n$$\\csc 2\\alpha \\approx 2.308$$"},{"type":"text","order":6,"title":"Determine the displacement angles","content":"Now, we convert back to $\\sin 2\\alpha$:\n\n$$\\sin 2\\alpha = \\frac{1}{2.308...} \\approx 0.4333$$\n\nWe find the principal value in the interval $0 < 2\\alpha < \\pi$:\n\n$$2\\alpha = \\arcsin(0.4333...) \\approx 0.4481\\text{ rad}$$\n\nThe second possible value in this interval is:\n\n$$2\\alpha = \\pi - 0.4481 \\approx 2.6935\\text{ rad}$$\n\nDividing both solutions by 2, we get:\n\n$$\\alpha \\approx 0.224\\text{ rad} \\quad \\text{and} \\quad \\alpha \\approx 1.35\\text{ rad}$$"}]},{"id":"pythagorean_first","label":"Substituting Pythagorean Identity First","steps":[{"type":"text","order":0,"title":"Substitute the Pythagorean Identity","content":"To start with our second method, we substitute the Pythagorean identity relating cotangent and cosecant:\n\n$$\\cot^2 2\\alpha = \\csc^2 2\\alpha - 1$$\n\nSubstituting this into the left-hand side of our original equation:\n\n$$3 \\sin \\alpha \\left( 4(\\csc^2 2\\alpha - 1) - 5 \\right) = 8 \\sec \\alpha$$\n\nNow, let's simplify the terms inside the parentheses:\n\n$$3 \\sin \\alpha (4 \\csc^2 2\\alpha - 4 - 5) = 8 \\sec \\alpha$$\n\n$$3 \\sin \\alpha (4 \\csc^2 2\\alpha - 9) = 8 \\sec \\alpha$$"},{"type":"text","order":1,"title":"Rearrange to group single angles","content":"Next, let's rearrange the equation to gather all the single-angle trigonometric terms ($\\sin \\alpha$ and $\\sec \\alpha$) on one side.\n\nSince $\\sec \\alpha = \\frac{1}{\\cos \\alpha}$, we can divide both sides of the equation by $3 \\sin \\alpha$:\n\n$$4 \\csc^2 2\\alpha - 9 = \\frac{8 \\sec \\alpha}{3 \\sin \\alpha}$$\n\nSubstitute $\\sec \\alpha = \\frac{1}{\\cos \\alpha}$ into the right-hand side:\n\n$$4 \\csc^2 2\\alpha - 9 = \\frac{8}{3 \\sin \\alpha \\cos \\alpha}$$\n\nThis step sets us up to convert the single-angle terms into double-angle terms."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$2 \\sin \\alpha \\cos \\alpha = \\sin 2\\alpha$"},{"latex":"$$\\frac{8}{3 \\sin \\alpha \\cos \\alpha} = \n\\frac{8 \\times 2}{3(2 \\sin \\alpha \\cos \\alpha)} = \\frac{16}{3 \\sin 2\\alpha}$"},{"latex":"$$\\htmlData{anchor=double-angle}{\\frac{16}{3 \\sin 2\\alpha}} = \\htmlData{anchor=csc-term}{\\frac{16}{3} \\csc 2\\alpha}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"double-angle"},{"color":"sky","style":"text-color","anchor":"csc-term"}],"connections":[{"to":"csc-term","from":"double-angle","color":"sky","label":"using csc(x) = 1/sin(x)","style":"solid"}]},"order":2,"title":"Apply double-angle identity"},{"type":"text","order":3,"title":"Form the quadratic equation","content":"Now we substitute our simplified right-hand side back into the main equation:\n\n$$4 \\csc^2 2\\alpha - 9 = \\frac{16}{3} \\csc 2\\alpha$$\n\nTo eliminate the fraction, multiply the entire equation by 3:\n\n$$3(4 \\csc^2 2\\alpha - 9) = 16 \\csc 2\\alpha$$\n\n$$12 \\csc^2 2\\alpha - 27 = 16 \\csc 2\\alpha$$\n\nFinally, subtract $16 \\csc 2\\alpha$ from both sides to form the required quadratic equation:\n\n$$12 \\csc^2 2\\alpha - 16 \\csc 2\\alpha - 27 = 0$$"},{"type":"text","order":4,"title":"Solve for the angle values","content":"Using the quadratic formula for $12 \\csc^2 2\\alpha - 16 \\csc 2\\alpha - 27 = 0$, we find:\n\n$$\\csc 2\\alpha = \\frac{16 \\pm \\sqrt{(-16)^2 - 4(12)(-27)}}{24} = \\frac{4 \\pm \\sqrt{97}}{6}$$\n\nSince $0 < \\alpha < \\frac{\\pi}{2}$ implies $0 < 2\\alpha < \\pi$, we must have $\\csc 2\\alpha > 0$. We therefore select the positive solution:\n\n$$\\csc 2\\alpha = \\frac{4 + \\sqrt{97}}{6} \\approx 2.308$$\n\nThis gives:\n\n$$\\sin 2\\alpha = \\frac{1}{2.308...} \\approx 0.4333$$\n\nSolving for $2\\alpha$ in the range $0 < 2\\alpha < \\pi$:\n\n$$2\\alpha = \\arcsin(0.4333...) \\approx 0.4481\\text{ rad}$$\n\n$$2\\alpha = \\pi - 0.4481 \\approx 2.6935\\text{ rad}$$\n\nDividing each by 2 yields:\n\n$$\\alpha \\approx 0.224\\text{ rad} \\quad \\text{and} \\quad \\alpha \\approx 1.35\\text{ rad}$$"}]}],"generatedAt":"2026-06-09T02:00:19.377Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"9114b3ef-64e5-4942-979d-56603f6d0b35","label":"b","content":"Hence, for \$0 < \\alpha < \\frac{\\pi}{2}\$, determine the possible values of the displacement angle \$\\alpha\$, giving your answers to 3 significant figures.","markscheme":"- Attempts to solve the quadratic \$ 12x^2 - 16x - 27 = 0 \$ using the quadratic formula: \$ \\csc 2\\alpha = \\frac{16 \\pm \\sqrt{256 - 4(12)(-27)}}{24} \$ M1\n- Obtains \$ \\csc 2\\alpha = \\frac{4 + \\sqrt{97}}{6} \\approx 2.308 \$ and identifies that the negative root is rejected for the given interval A1\n- Calculates the first primary value for \$ 2\\alpha = \\arcsin \\left( \\frac{1}{2.308...} \\right) \\approx 0.4481 \$ radians M1\n- Calculates the second value within the interval: \$ 2\\alpha = \\pi - 0.4481 \\approx 2.6935 \$ radians M1\n- Divides by 2 to find \$ \\alpha = 0.224 \$ and \$ \\alpha = 1.35 \$ A1","marks":5,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"algebraic","label":"Algebraic Reduction","steps":[{"type":"text","order":0,"title":"Set up the quadratic","content":"From the algebraic reduction of the initial equation, we can express the equation as a quadratic in terms of $\\csc 2\\alpha$:\n\n$$12\\csc^2 2\\alpha - 16\\csc 2\\alpha - 27 = 0$$\n\nTo make this easier to solve, we can let $x = \\csc 2\\alpha$. This gives us the standard quadratic equation:\n\n$$12x^2 - 16x - 27 = 0$$"},{"type":"text","order":1,"title":"Solve the quadratic equation","content":"We use the quadratic formula to solve for $x$:\n\n$$x = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$$\n\nSubstituting $a = 12$, $b = -16$, and $c = -27$:\n\n$$x = \\frac{16 \\pm \\sqrt{(-16)^2 - 4(12)(-27)}}{2(12)}$$\n\n$$x = \\frac{16 \\pm \\sqrt{256 + 1296}}{24} = \\frac{16 \\pm \\sqrt{1552}}{24}$$\n\nSimplifying the surd $\\sqrt{1552} = 4\\sqrt{97}$:\n\n$$x = \\frac{16 \\pm 4\\sqrt{97}}{24} = \\frac{4 \\pm \\sqrt{97}}{6}$$"},{"type":"text","order":2,"title":"Find valid values for csc","content":"Let's calculate the two numerical values of $x = \\csc 2\\alpha$:\n\n1. $$x_1 = \\frac{4 + \\sqrt{97}}{6} \\approx 2.308$$\n2. $$x_2 = \\frac{4 - \\sqrt{97}}{6} \\approx -0.975$$\n\nSince the domain for $\\alpha$ is $0 < \\alpha < \\frac{\\pi}{2}$, the double angle $2\\alpha$ must satisfy $0 < 2\\alpha < \\pi$. \n\nIn the first and second quadrants, the cosecant function (which is $\\frac{1}{\\sin}$) is always positive. Therefore, we must reject the negative root, leaving us with:\n\n$$\\csc 2\\alpha \\approx 2.308$$"},{"type":"text","order":3,"title":"Solve for the double angle","content":"Since $\\csc 2\\alpha = \\frac{1}{\\sin 2\\alpha}$, we can rewrite the equation in terms of sine:\n\n$$\\sin 2\\alpha = \\frac{1}{2.308...} \\approx 0.4333$$\n\nNow, we find the angles within the interval $0 < 2\\alpha < \\pi$:\n\n* **First value (principal angle):**\n $$2\\alpha = \\arcsin(0.4333) \\approx 0.4481\\text{ radians}$$\n\n* **Second value (symmetric angle in quadrant II):**\n $$2\\alpha = \\pi - 0.4481 \\approx 2.6935\\text{ radians}$$"},{"type":"text","order":4,"title":"Find displacement angle alpha","content":"To find the final values for the displacement angle $\\alpha$, we divide each of our solutions for $2\\alpha$ by $2$:\n\n* $$\\alpha = \\frac{0.4481}{2} \\approx 0.224\\text{ radians}$$\n* $$\\alpha = \\frac{2.6935}{2} \\approx 1.35\\text{ radians}$$\n\nBoth values lie within the required interval $0 < \\alpha < \\frac{\\pi}{2} \\approx 1.57$ radians.\n\nThus, to $3$ significant figures, the possible values of the displacement angle are:\n\n$$\\alpha = 0.224\\text{ and } \\alpha = 1.35$$"}]},{"id":"graphical","label":"Graphical Approach","steps":[{"type":"text","order":0,"title":"Formulate the graphical problem","content":"To solve the equation using a **graphing approach**, we can define two functions corresponding to the left-hand and right-hand sides of the equation. \n\nLet $x$ represent the angle $\\alpha$. We define:\n\n$$y_1 = 3 \\sin x \\left(4 \\cot^2 2x - 5\\right)$$\n\n$$y_2 = 8 \\sec x = \\frac{8}{\\cos x}$$\n\nWe want to find the intersection points of these two curves, $(x, y)$, or equivalently find the roots of the single function $f(x) = y_1 - y_2 = 0$ in the specified interval $0 < x < \\frac{\\pi}{2}$ (which is approximately $0 < x < 1.57$ radians)."},{"type":"text","order":1,"title":"Set up your calculator","content":"Before graphing, ensure your calculator is set to **Radian** mode, as the interval $0 < \\alpha < \\frac{\\pi}{2}$ is given in radians.\n\nSince standard calculators do not have direct keys for $\\cot$ and $\\sec$, we rewrite them using fundamental trigonometric reciprocal identities:\n\n$$\\cot 2x = \\frac{1}{\\tan 2x} \\quad \\text{and} \\quad \\sec x = \\frac{1}{\\cos x}$$\n\nSo we enter the equations as:\n\n$$y_1 = 3 \\sin x \\left(4 \\left(\\frac{1}{\\tan 2x}\\right)^2 - 5\\right)$$\n\n$$y_2 = \\frac{8}{\\cos x}$$","calculator":[{"gdc":{"instructions":"1. Go to the Graph menu.\n2. Define \$ Y_1 = 3 \\sin(x) (4 (\\tan(2x))^{-2} - 5) \$ and \$ Y_2 = 8 / \\cos(x) \$.\n3. Set your view window for \$ x \$ between \$ 0 \$ and \$ \\pi/2 \$.\n4. Use the **G-Solve** -> **Intersection (INTSECT)** tool to find the coordinate values of the intersection points."},"ti84":{"instructions":"1. Press [MODE] and ensure **RADIAN** is selected.\n2. Press [Y=] and enter:\n \$ Y_1 = 3 \\sin(X) \\cdot (4 / (\\tan(2X))^2 - 5) \$\n \$ Y_2 = 8 / \\cos(X) \$\n3. Press [WINDOW] and set:\n \$ X_{\\min} = 0 \$\n \$ X_{\\max} = 1.57 \$ (which is \$ \\pi/2 \$)\n \$ Y_{\\min} = 0 \$\n \$ Y_{\\max} = 30 \$\n4. Press [GRAPH].\n5. Press [2nd] [CALC] and select **5: intersect** to find where the curves cross."},"desmos":{"instructions":"1. Set the graph settings (wrench icon) to **Radians**.\n2. In the first input box, type: \$ y = 3 \\sin(x) (4 (\\tan(2x))^{-2} - 5) \$\n3. In the second input box, type: \$ y = 8 \\sec(x) \$\n4. Set the x-axis limits from \$ 0 \$ to \$ \\pi/2 \$.\n5. Click on the gray intersection points on the graph to display their coordinates."},"description":"Enter the equations in your graphing tool, adjust the window settings, and find the intersection points."}]},{"type":"text","order":2,"title":"Identify the intersection points","content":"By plotting these functions over the domain $0 < x < \\frac{\\pi}{2}$, you will observe that the two curves intersect at exactly two points within this interval.\n\nUsing your calculator's intersection tool, you will find the $x$-coordinates of these intersections:\n\n$$x_1 \\approx 0.22409...$$\n\n$$x_2 \\approx 1.3467...$$"},{"type":"text","order":3,"title":"Round to significant figures","content":"The question asks for the values to **3 significant figures**.\n\nRounding each value:\n- For $x_1 = 0.22409...$, the first three significant figures are $0.224$.\n- For $x_2 = 1.3467...$, rounding to three significant figures gives $1.35$.\n\nThus, the possible values for the displacement angle $\\alpha$ are:\n\n$$\\alpha = 0.224 \\quad \\text{and} \\quad \\alpha = 1.35 \\text{ radians}$$"}]}],"generatedAt":"2026-06-09T02:01:00.445Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"a626451f-3830-4222-932b-3e2c8704efc8:rhs_first","label":"(a) Simplifying Right-Hand Side First","steps":[{"type":"text","order":0,"title":"Isolate the bracketed term","content":"To begin, we want to isolate the term containing the double angle, $4 \\cot^2 2\\alpha - 5$. We can do this by dividing both sides of the equation by $3 \\sin \\alpha$:\n\n$$4 \\cot^2 2\\alpha - 5 = \\frac{8 \\sec \\alpha}{3 \\sin \\alpha}$$\n\nThis separates the functions of the single angle $\\alpha$ on the right-hand side so we can simplify them first."},{"type":"text","order":1,"title":"Simplify the right-hand side","content":"Next, let's rewrite the right-hand side by expressing $\\sec \\alpha$ in terms of $\\cos \\alpha$ using the reciprocal identity $\\sec \\alpha = \\frac{1}{\\cos \\alpha}$:\n\n$$\\frac{8 \\sec \\alpha}{3 \\sin \\alpha} = \\frac{8}{3 \\sin \\alpha \\cos \\alpha}$$"},{"type":"text","order":2,"title":"Apply the double-angle formula","content":"Recall the double-angle formula for sine:\n\n$$\\sin 2\\alpha = 2 \\sin \\alpha \\cos \\alpha \\implies \\sin \\alpha \\cos \\alpha = \\frac{1}{2} \\sin 2\\alpha$$\n\nSubstituting this into our expression for the right-hand side gives:\n\n$$\\frac{8}{3 \\sin \\alpha \\cos \\alpha} = \\frac{8}{3 \\left(\\frac{1}{2} \\sin 2\\alpha\\right)} = \\frac{16}{3 \\sin 2\\alpha}$$\n\nUsing the reciprocal identity $\\csc 2\\alpha = \\frac{1}{\\sin 2\\alpha}$, we get:\n\n$$4 \\cot^2 2\\alpha - 5 = \\frac{16}{3} \\csc 2\\alpha$$"},{"type":"text","order":3,"title":"Use the Pythagorean identity","content":"To form a quadratic equation solely in terms of $\\csc 2\\alpha$, we substitute the Pythagorean identity:\n\n$$\\cot^2 2\\alpha = \\csc^2 2\\alpha - 1$$\n\nThis gives:\n\n$$4(\\csc^2 2\\alpha - 1) - 5 = \\frac{16}{3} \\csc 2\\alpha$$\n\nMultiply the entire equation by $3$ to eliminate the fraction:\n\n$$12(\\csc^2 2\\alpha - 1) - 15 = 16 \\csc 2\\alpha$$\n\n$$12 \\csc^2 2\\alpha - 12 - 15 = 16 \\csc 2\\alpha$$"},{"type":"text","order":4,"title":"Rearrange to the final form","content":"Finally, group all terms on the left-hand side to obtain the target quadratic form:\n\n$$12 \\csc^2 2\\alpha - 16 \\csc 2\\alpha - 27 = 0$$\n\nThis successfully shows the required expression."},{"type":"text","order":5,"title":"Solve the quadratic equation","content":"For Part 2, let $x = \\csc 2\\alpha$. The quadratic equation is:\n\n$$12x^2 - 16x - 27 = 0$$\n\nApplying the quadratic formula:\n\n$$x = \\frac{16 \\pm \\sqrt{(-16)^2 - 4(12)(-27)}}{2(12)}$$\n\n$$x = \\frac{16 \\pm \\sqrt{256 + 1296}}{24} = \\frac{16 \\pm \\sqrt{1552}}{24} = \\frac{4 \\pm \\sqrt{97}}{6}$$\n\nCalculating the two values:\n- $x \\approx 2.308$\n- $x \\approx -0.975$\n\nSince $0 < \\alpha < \\frac{\\pi}{2}$, we have $0 < 2\\alpha < \\pi$, which means $\\sin 2\\alpha > 0$. Thus, we reject the negative root, leaving:\n\n$$\\csc 2\\alpha \\approx 2.308$$"},{"type":"text","order":6,"title":"Determine the displacement angles","content":"Now, we convert back to $\\sin 2\\alpha$:\n\n$$\\sin 2\\alpha = \\frac{1}{2.308...} \\approx 0.4333$$\n\nWe find the principal value in the interval $0 < 2\\alpha < \\pi$:\n\n$$2\\alpha = \\arcsin(0.4333...) \\approx 0.4481\\text{ rad}$$\n\nThe second possible value in this interval is:\n\n$$2\\alpha = \\pi - 0.4481 \\approx 2.6935\\text{ rad}$$\n\nDividing both solutions by 2, we get:\n\n$$\\alpha \\approx 0.224\\text{ rad} \\quad \\text{and} \\quad \\alpha \\approx 1.35\\text{ rad}$$"}]},{"id":"a626451f-3830-4222-932b-3e2c8704efc8:pythagorean_first","label":"(a) Substituting Pythagorean Identity First","steps":[{"type":"text","order":0,"title":"Substitute the Pythagorean Identity","content":"To start with our second method, we substitute the Pythagorean identity relating cotangent and cosecant:\n\n$$\\cot^2 2\\alpha = \\csc^2 2\\alpha - 1$$\n\nSubstituting this into the left-hand side of our original equation:\n\n$$3 \\sin \\alpha \\left( 4(\\csc^2 2\\alpha - 1) - 5 \\right) = 8 \\sec \\alpha$$\n\nNow, let's simplify the terms inside the parentheses:\n\n$$3 \\sin \\alpha (4 \\csc^2 2\\alpha - 4 - 5) = 8 \\sec \\alpha$$\n\n$$3 \\sin \\alpha (4 \\csc^2 2\\alpha - 9) = 8 \\sec \\alpha$$"},{"type":"text","order":1,"title":"Rearrange to group single angles","content":"Next, let's rearrange the equation to gather all the single-angle trigonometric terms ($\\sin \\alpha$ and $\\sec \\alpha$) on one side.\n\nSince $\\sec \\alpha = \\frac{1}{\\cos \\alpha}$, we can divide both sides of the equation by $3 \\sin \\alpha$:\n\n$$4 \\csc^2 2\\alpha - 9 = \\frac{8 \\sec \\alpha}{3 \\sin \\alpha}$$\n\nSubstitute $\\sec \\alpha = \\frac{1}{\\cos \\alpha}$ into the right-hand side:\n\n$$4 \\csc^2 2\\alpha - 9 = \\frac{8}{3 \\sin \\alpha \\cos \\alpha}$$\n\nThis step sets us up to convert the single-angle terms into double-angle terms."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$2 \\sin \\alpha \\cos \\alpha = \\sin 2\\alpha$"},{"latex":"$$\\frac{8}{3 \\sin \\alpha \\cos \\alpha} = \n\\frac{8 \\times 2}{3(2 \\sin \\alpha \\cos \\alpha)} = \\frac{16}{3 \\sin 2\\alpha}$"},{"latex":"$$\\htmlData{anchor=double-angle}{\\frac{16}{3 \\sin 2\\alpha}} = \\htmlData{anchor=csc-term}{\\frac{16}{3} \\csc 2\\alpha}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"double-angle"},{"color":"sky","style":"text-color","anchor":"csc-term"}],"connections":[{"to":"csc-term","from":"double-angle","color":"sky","label":"using csc(x) = 1/sin(x)","style":"solid"}]},"order":2,"title":"Apply double-angle identity"},{"type":"text","order":3,"title":"Form the quadratic equation","content":"Now we substitute our simplified right-hand side back into the main equation:\n\n$$4 \\csc^2 2\\alpha - 9 = \\frac{16}{3} \\csc 2\\alpha$$\n\nTo eliminate the fraction, multiply the entire equation by 3:\n\n$$3(4 \\csc^2 2\\alpha - 9) = 16 \\csc 2\\alpha$$\n\n$$12 \\csc^2 2\\alpha - 27 = 16 \\csc 2\\alpha$$\n\nFinally, subtract $16 \\csc 2\\alpha$ from both sides to form the required quadratic equation:\n\n$$12 \\csc^2 2\\alpha - 16 \\csc 2\\alpha - 27 = 0$$"},{"type":"text","order":4,"title":"Solve for the angle values","content":"Using the quadratic formula for $12 \\csc^2 2\\alpha - 16 \\csc 2\\alpha - 27 = 0$, we find:\n\n$$\\csc 2\\alpha = \\frac{16 \\pm \\sqrt{(-16)^2 - 4(12)(-27)}}{24} = \\frac{4 \\pm \\sqrt{97}}{6}$$\n\nSince $0 < \\alpha < \\frac{\\pi}{2}$ implies $0 < 2\\alpha < \\pi$, we must have $\\csc 2\\alpha > 0$. We therefore select the positive solution:\n\n$$\\csc 2\\alpha = \\frac{4 + \\sqrt{97}}{6} \\approx 2.308$$\n\nThis gives:\n\n$$\\sin 2\\alpha = \\frac{1}{2.308...} \\approx 0.4333$$\n\nSolving for $2\\alpha$ in the range $0 < 2\\alpha < \\pi$:\n\n$$2\\alpha = \\arcsin(0.4333...) \\approx 0.4481\\text{ rad}$$\n\n$$2\\alpha = \\pi - 0.4481 \\approx 2.6935\\text{ rad}$$\n\nDividing each by 2 yields:\n\n$$\\alpha \\approx 0.224\\text{ rad} \\quad \\text{and} \\quad \\alpha \\approx 1.35\\text{ rad}$$"}]},{"id":"9114b3ef-64e5-4942-979d-56603f6d0b35:algebraic","label":"(b) Algebraic Reduction","steps":[{"type":"text","order":0,"title":"Set up the quadratic","content":"From the algebraic reduction of the initial equation, we can express the equation as a quadratic in terms of $\\csc 2\\alpha$:\n\n$$12\\csc^2 2\\alpha - 16\\csc 2\\alpha - 27 = 0$$\n\nTo make this easier to solve, we can let $x = \\csc 2\\alpha$. This gives us the standard quadratic equation:\n\n$$12x^2 - 16x - 27 = 0$$"},{"type":"text","order":1,"title":"Solve the quadratic equation","content":"We use the quadratic formula to solve for $x$:\n\n$$x = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$$\n\nSubstituting $a = 12$, $b = -16$, and $c = -27$:\n\n$$x = \\frac{16 \\pm \\sqrt{(-16)^2 - 4(12)(-27)}}{2(12)}$$\n\n$$x = \\frac{16 \\pm \\sqrt{256 + 1296}}{24} = \\frac{16 \\pm \\sqrt{1552}}{24}$$\n\nSimplifying the surd $\\sqrt{1552} = 4\\sqrt{97}$:\n\n$$x = \\frac{16 \\pm 4\\sqrt{97}}{24} = \\frac{4 \\pm \\sqrt{97}}{6}$$"},{"type":"text","order":2,"title":"Find valid values for csc","content":"Let's calculate the two numerical values of $x = \\csc 2\\alpha$:\n\n1. $$x_1 = \\frac{4 + \\sqrt{97}}{6} \\approx 2.308$$\n2. $$x_2 = \\frac{4 - \\sqrt{97}}{6} \\approx -0.975$$\n\nSince the domain for $\\alpha$ is $0 < \\alpha < \\frac{\\pi}{2}$, the double angle $2\\alpha$ must satisfy $0 < 2\\alpha < \\pi$. \n\nIn the first and second quadrants, the cosecant function (which is $\\frac{1}{\\sin}$) is always positive. Therefore, we must reject the negative root, leaving us with:\n\n$$\\csc 2\\alpha \\approx 2.308$$"},{"type":"text","order":3,"title":"Solve for the double angle","content":"Since $\\csc 2\\alpha = \\frac{1}{\\sin 2\\alpha}$, we can rewrite the equation in terms of sine:\n\n$$\\sin 2\\alpha = \\frac{1}{2.308...} \\approx 0.4333$$\n\nNow, we find the angles within the interval $0 < 2\\alpha < \\pi$:\n\n* **First value (principal angle):**\n $$2\\alpha = \\arcsin(0.4333) \\approx 0.4481\\text{ radians}$$\n\n* **Second value (symmetric angle in quadrant II):**\n $$2\\alpha = \\pi - 0.4481 \\approx 2.6935\\text{ radians}$$"},{"type":"text","order":4,"title":"Find displacement angle alpha","content":"To find the final values for the displacement angle $\\alpha$, we divide each of our solutions for $2\\alpha$ by $2$:\n\n* $$\\alpha = \\frac{0.4481}{2} \\approx 0.224\\text{ radians}$$\n* $$\\alpha = \\frac{2.6935}{2} \\approx 1.35\\text{ radians}$$\n\nBoth values lie within the required interval $0 < \\alpha < \\frac{\\pi}{2} \\approx 1.57$ radians.\n\nThus, to $3$ significant figures, the possible values of the displacement angle are:\n\n$$\\alpha = 0.224\\text{ and } \\alpha = 1.35$$"}]},{"id":"9114b3ef-64e5-4942-979d-56603f6d0b35:graphical","label":"(b) Graphical Approach","steps":[{"type":"text","order":0,"title":"Formulate the graphical problem","content":"To solve the equation using a **graphing approach**, we can define two functions corresponding to the left-hand and right-hand sides of the equation. \n\nLet $x$ represent the angle $\\alpha$. We define:\n\n$$y_1 = 3 \\sin x \\left(4 \\cot^2 2x - 5\\right)$$\n\n$$y_2 = 8 \\sec x = \\frac{8}{\\cos x}$$\n\nWe want to find the intersection points of these two curves, $(x, y)$, or equivalently find the roots of the single function $f(x) = y_1 - y_2 = 0$ in the specified interval $0 < x < \\frac{\\pi}{2}$ (which is approximately $0 < x < 1.57$ radians)."},{"type":"text","order":1,"title":"Set up your calculator","content":"Before graphing, ensure your calculator is set to **Radian** mode, as the interval $0 < \\alpha < \\frac{\\pi}{2}$ is given in radians.\n\nSince standard calculators do not have direct keys for $\\cot$ and $\\sec$, we rewrite them using fundamental trigonometric reciprocal identities:\n\n$$\\cot 2x = \\frac{1}{\\tan 2x} \\quad \\text{and} \\quad \\sec x = \\frac{1}{\\cos x}$$\n\nSo we enter the equations as:\n\n$$y_1 = 3 \\sin x \\left(4 \\left(\\frac{1}{\\tan 2x}\\right)^2 - 5\\right)$$\n\n$$y_2 = \\frac{8}{\\cos x}$$","calculator":[{"gdc":{"instructions":"1. Go to the Graph menu.\n2. Define \$ Y_1 = 3 \\sin(x) (4 (\\tan(2x))^{-2} - 5) \$ and \$ Y_2 = 8 / \\cos(x) \$.\n3. Set your view window for \$ x \$ between \$ 0 \$ and \$ \\pi/2 \$.\n4. Use the **G-Solve** -> **Intersection (INTSECT)** tool to find the coordinate values of the intersection points."},"ti84":{"instructions":"1. Press [MODE] and ensure **RADIAN** is selected.\n2. Press [Y=] and enter:\n \$ Y_1 = 3 \\sin(X) \\cdot (4 / (\\tan(2X))^2 - 5) \$\n \$ Y_2 = 8 / \\cos(X) \$\n3. Press [WINDOW] and set:\n \$ X_{\\min} = 0 \$\n \$ X_{\\max} = 1.57 \$ (which is \$ \\pi/2 \$)\n \$ Y_{\\min} = 0 \$\n \$ Y_{\\max} = 30 \$\n4. Press [GRAPH].\n5. Press [2nd] [CALC] and select **5: intersect** to find where the curves cross."},"desmos":{"instructions":"1. Set the graph settings (wrench icon) to **Radians**.\n2. In the first input box, type: \$ y = 3 \\sin(x) (4 (\\tan(2x))^{-2} - 5) \$\n3. In the second input box, type: \$ y = 8 \\sec(x) \$\n4. Set the x-axis limits from \$ 0 \$ to \$ \\pi/2 \$.\n5. Click on the gray intersection points on the graph to display their coordinates."},"description":"Enter the equations in your graphing tool, adjust the window settings, and find the intersection points."}]},{"type":"text","order":2,"title":"Identify the intersection points","content":"By plotting these functions over the domain $0 < x < \\frac{\\pi}{2}$, you will observe that the two curves intersect at exactly two points within this interval.\n\nUsing your calculator's intersection tool, you will find the $x$-coordinates of these intersections:\n\n$$x_1 \\approx 0.22409...$$\n\n$$x_2 \\approx 1.3467...$$"},{"type":"text","order":3,"title":"Round to significant figures","content":"The question asks for the values to **3 significant figures**.\n\nRounding each value:\n- For $x_1 = 0.22409...$, the first three significant figures are $0.224$.\n- For $x_2 = 1.3467...$, rounding to three significant figures gives $1.35$.\n\nThus, the possible values for the displacement angle $\\alpha$ are:\n\n$$\\alpha = 0.224 \\quad \\text{and} \\quad \\alpha = 1.35 \\text{ radians}$$"}]}],"generatedAt":"2026-06-09T02:01:15.022Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"b68578c7-96e4-40ae-9f93-3abd5ab27699","revisionId":"cd2fa995-8c83-4eb3-9b07-64ef2d3aa899","questionNumber":null,"questionType":"LA","specification":"Find all the values of $x$ in the interval $0° \\leqslant x < 360°$ for which\n$2\\cos^2 x - 3\\sin^2 x = 14\\cos x$","difficulty":null,"difficultyBand":"hard","tier":null,"totalMarks":8,"sourceTitle":null,"calculatorUse":null,"isCalculation":false,"parts":[{"id":"a0873e6b-321f-429d-8cb7-3343bbb320b9","label":null,"content":"Give your answers to $1$ decimal place.","markscheme":"- Substitutes $\\sin^2 x = 1 - \\cos^2 x$ M1\n- $2\\cos^2 x - 3\\left(1 - \\cos^2 x\\right) = 14\\cos x$ A1\n- Expands and rearranges to $5\\cos^2 x - 14\\cos x - 3 = 0$ M1\n- Factorises: $(5\\cos x + 1)(\\cos x - 3) = 0$ A1\n- $\\cos x = -\\dfrac{1}{5}$ or $\\cos x = 3$ A1\n- Rejects $\\cos x = 3$, as it lies outside the range of cosine B1\n- Solves $\\cos x = -\\dfrac{1}{5}$ in the given interval M1\n- $x = 101.5°$ and $x = 258.5°$ A1\n\n$101.537\\ldots°$ and $258.463\\ldots°$. Both solutions required.","marks":8,"order":0,"markingRubric":null,"tableImages":[],"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"ed516dfd-c130-4f70-adc0-5af08df92d4b","revisionId":"3b571222-125c-4948-9824-36b0bdfd7d4d","questionNumber":null,"questionType":"LA","specification":"","difficulty":null,"difficultyBand":"hard","tier":null,"totalMarks":10,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"b2ae6bc4-e5b4-415a-b6e0-af8ea58605aa","label":"a","content":"Show that the equation\n$3\\sin 2x \\tan 2x = \\cos 2x + 2$\ncan be written in the form\n$4\\cos^2 2x + 2\\cos 2x - 3 = 0$","markscheme":"- Uses $\\tan 2x = \\dfrac{\\sin 2x}{\\cos 2x}$ to obtain $\\dfrac{3\\sin^2 2x}{\\cos 2x} = \\cos 2x + 2$ M1\n- Multiplies through by $\\cos 2x$: $3\\sin^2 2x = \\cos^2 2x + 2\\cos 2x$ M1\n- Substitutes $\\sin^2 2x = 1 - \\cos^2 2x$ M1\n- Rearranges to $4\\cos^2 2x + 2\\cos 2x - 3 = 0$, as required A1\n\nAnswer given, so fully correct working is required. The multiplication by $\\cos 2x$ assumes $\\cos 2x \\neq 0$; neither $x = 45°$ nor $x = 135°$ turns out to be a solution.","marks":4,"order":0,"markingRubric":null,"tableImages":[],"explanation":null,"options":[]},{"id":"1ef512d2-5605-4e31-b9a5-44f384914622","label":"b","content":"Find all the values of $x$ in the interval $0° \\leqslant x < 180°$ for which\n$3\\sin 2x \\tan 2x = \\cos 2x + 2$\nGive your answers to $2$ decimal places.","markscheme":"- Applies the quadratic formula to $4\\cos^2 2x + 2\\cos 2x - 3 = 0$ M1\n- $\\cos 2x = \\dfrac{-1 \\pm \\sqrt{13}}{4}$, i.e. $0.6514$ or $-1.1514$ A1\n- Rejects $\\cos 2x = -1.1514$, as it lies outside the range of cosine B1\n- $2x = \\cos^{-1}(0.6514) = 49.36°$ M1\n- Second angle for $2x$ in $0° \\leqslant 2x < 360°$: $2x = 310.64°$ A1\n- $x = 24.68°$ and $x = 155.32°$ A1\n\nTransform the interval first: $0° \\leqslant x < 180°$ gives $0° \\leqslant 2x < 360°$, so exactly two values of $2x$ are needed. $24.679\\ldots°$ and $155.320\\ldots°$.","marks":6,"order":1,"markingRubric":null,"tableImages":[],"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths__wjec","summaries":[{"id":"0aec2d07-eeb2-4854-99e5-78c5846bc99a","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"4ed7094e-cf03-4e1e-8691-1d3d0175a970","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"2cb328b9-19c8-4983-a09d-31ba9ff0ece8","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"02abd9f1-3793-4eae-9173-31c3de431a06","difficultyBand":"hard","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"0bd36201-bbb9-416e-a36c-c1c055178ab5","difficultyBand":"easy","calculatorUse":null,"totalMarks":1,"questionType":"MC","isCalculation":true},{"id":"16625c96-dbb8-4a12-aaca-d77403e7dbac","difficultyBand":"hard","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"1eeb206a-a348-4e3a-a08f-bbe35d6be702","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"20b983b4-4d93-4d18-a2b3-f77479714025","difficultyBand":"hard","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"245935ae-b393-4618-b200-7ceb22ea40be","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"2867863b-b9e7-4f84-ae17-8331b1c6c9e5","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"2a2d5a62-4963-473f-a4bb-a7e8dd56da56","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"2dd44c3f-bdf0-42d9-bd30-910ce4b32418","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"38c35132-2384-4b6c-8845-8f63f74ad764","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"3ac9e0d1-6515-488d-b25f-aca54b9664e0","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"3cc96763-3087-4987-be28-0948c2f5f671","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"4e6f71db-2e58-4456-91f4-83c231964608","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"513e051a-a107-415f-a26f-5a645b94caae","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"515db5da-764b-47d7-8bd9-8ce0ec7e690f","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"5326fecf-c197-46ff-a3b1-eb44cf54fd0e","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"5cda2d5f-6094-411a-8acd-84f4b8050608","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"5dbd6a16-7dfb-44ce-aece-fe9cee1738bd","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"6241cfd3-b352-4d84-ba16-859861bec3b8","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"6e256c92-2764-44c4-bdb0-431f4684f0c5","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"700f480d-116e-44f7-a3d5-8da252854a2e","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"70499ec1-c56f-420f-adde-ce7a7491e875","difficultyBand":"easy","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"7e4cbf9c-3042-48ed-bf3f-45b563ab2a42","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"85a5a693-25a7-4a40-8b84-3e96aa4a8a7b","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"867e1c4c-5bdb-4ccf-ac8a-f128bc99febe","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"8a1fa494-fdf6-48ec-a9d8-b6866980397d","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"8d4206fd-76f3-46f8-98e1-08940aaef0f4","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"8d82aca5-86df-43e2-873e-af06a2934430","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"94a4665e-3a10-44ef-9810-326acd4c49d0","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"99044bf8-6ee5-436d-aa4a-345d00112f02","difficultyBand":"hard","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"9e134eed-4e06-4005-804b-e37007bf9af1","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"9ea04f6f-ce65-4283-8cff-0c256fc188da","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"b658e46a-2d28-4b3d-8c6d-677b07f5d6db","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"b68578c7-96e4-40ae-9f93-3abd5ab27699","difficultyBand":"hard","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":false},{"id":"b88231c0-c4e7-4bf9-9b40-8dfa97f3e170","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"d5905b02-7f0f-48a0-9d61-39baaede11c7","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"db08208d-c428-4a8d-a4fb-84fbf9f58ef6","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"e9fd3e58-1fc3-415c-8bfb-44bd919805a6","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"ed516dfd-c130-4f70-adc0-5af08df92d4b","difficultyBand":"hard","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true}],"title":"1.5.5 Trigonometry","topicId":"b2b12074-99bf-4028-9d5b-1b072fedf61f","topicSidebar":"$L3c","topicSlug":"trigonometry-6","topicTitle":"1.5.5 Trigonometry"}]}],"$L3d","$L3e"]}]

1.5.5 Trigonometry

Question 5

1.5.5 Trigonometry Questions

1.5.5 Trigonometry

Question 5

1.5.5 Trigonometry Questions