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Questions\",\"description\":\"Practise 1.5.15 Trigonometric equations A Level Maths exam-style questions with worked solutions on MathsGenie.\",\"learningResourceType\":\"Practice questions\",\"educationalLevel\":\"A Level\",\"about\":{\"@type\":\"Thing\",\"name\":\"1.5.15 Trigonometric equations\"},\"url\":\"https://mathsgenie.co.uk/a-level/maths/ocr/trigonometric-equations/questions/q14\",\"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/ocr/questions","basePath":"/a-level/maths/ocr/trigonometric-equations/questions","formulaUrl":null,"initialIndex":13,"initialQuestionWindow":[{"id":"21b4f24d-1066-4db1-87e6-25a1d3e10f8c","revisionId":"cd0c97e1-e537-4381-b941-34f568ce64a7","questionNumber":null,"questionType":"LA","specification":"Solve, for $0° \\leqslant x < 360°$, the equation\n$5\\cos(x - 40°) = 2$","difficulty":null,"difficultyBand":"easy","tier":null,"totalMarks":4,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"19222f11-1190-4559-8a69-b11254cfb729","label":null,"content":"Give your answers to $2$ decimal places.","markscheme":"- Rearranges to $\\cos(x - 40°) = 0.4$ M1\n- Uses the inverse cosine and the symmetry of the cosine graph to obtain $x - 40° = 66.42°$ or $293.58°$ M1\n- Adds $40°$ to each value M1\n- $x = 106.42°$ and $x = 333.58°$ A1\n\nThe interval for $x - 40°$ is $-40° \\leqslant x - 40° < 320°$; the value $-66.42°$ lies outside it, so there are exactly two solutions. Both required.","marks":4,"order":0,"markingRubric":null,"tableImages":[],"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"36432791-e3b1-44de-8893-0fe68c790bcf","revisionId":"8d6c3877-a73a-4739-8d04-c1f9a7d52ff5","questionNumber":null,"questionType":"MC","specification":"State the number of solutions of the equation\n$\\cos x = 0.5$\nin the interval $0° \\leqslant x < 360°$.\n\nCircle your answer.","difficulty":null,"difficultyBand":"easy","tier":null,"totalMarks":1,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[],"options":[{"id":"a27149f8-8df8-473f-a4ca-2e5de03065ed","content":"$$0$","correct":false,"markscheme":null,"order":0,"metadata":null},{"id":"9e2f5bff-2c56-439d-ad7d-440071420c46","content":"$$1$","correct":false,"markscheme":null,"order":1,"metadata":null},{"id":"1a5c800f-14fb-4aaa-bb9c-256f68030293","content":"$$2$","correct":true,"markscheme":"- B1\n\n$x = 60°$ and $x = 300°$. A cosine equation with a value strictly between $-1$ and $1$ has exactly two solutions in any interval of length $360°$.","order":2,"metadata":null},{"id":"756912ef-473b-4530-aacc-6460a776b374","content":"$$4$","correct":false,"markscheme":null,"order":3,"metadata":null}],"explanation":null,"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"70499ec1-c56f-420f-adde-ce7a7491e875","revisionId":"2e8ad018-0987-4cfd-9ae2-59bd279886b1","questionNumber":null,"questionType":"LA","specification":"The vertical displacement $d$ (in mm) of a tuning fork's prong over a cycle is modeled by the function\n\\[d = \\sin \\theta^\\circ\\]\nfor \$-360 \\le \\theta \\le 360\$.","difficulty":3,"difficultyBand":"easy","tier":null,"totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"57e3f9a2-508b-47b4-bb16-5d6f9d43ceb8","label":"a","content":"A point $A$ on the displacement curve has coordinates \$(a, -\\frac{\\sqrt{3}}{2})\$.\n\n(i) Determine the value of $a$, given that \$-180 < a < -90\$.","markscheme":"- Identifies the principal value or reference angle as $60^\\circ$ (or recognises $\\sin(-60^\\circ) = -\\frac{\\sqrt{3}}{2}$) M1\n- Uses symmetry of the sine graph in the third quadrant to find $a = -180 + 60 = -120$ A1","marks":2,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"reference_angle","label":"Reference Angle and Quadrant Analysis","steps":[{"type":"text","order":0,"title":"Understand the equation","content":"We are given that the point $A\\left(a, -\\frac{\\sqrt{3}}{2}\\right)$ lies on the curve $d = \\sin \\theta^\\circ$. This means we need to solve the trigonometric equation:\n\n$$\\sin a^\\circ = -\\frac{\\sqrt{3}}{2}$$\n\nwhere the angle $a$ is constrained by the interval $-180 < a < -90$."},{"type":"text","order":1,"title":"Find the reference angle","content":"To solve $\\sin a^\\circ = -\\frac{\\sqrt{3}}{2}$ using quadrant analysis, we first find the acute **reference angle** $\\theta_{\\text{ref}}$ where the sine value is positive:\n\n$$\\sin \\theta_{\\text{ref}} = \\frac{\\sqrt{3}}{2}$$\n\nUsing our exact trigonometric values, we find:\n\n$$\\theta_{\\text{ref}} = 60^\\circ$$"},{"type":"text","order":2,"title":"Analyze the quadrant","content":"Now we determine which quadrant our angle $a$ lies in. \n\nThe given interval is $-180 < a < -90$.\n\n- Moving in the negative (clockwise) direction from the positive $x$-axis:\n - $0^\\circ$ to $-90^\\circ$ is Quadrant IV.\n - $-90^\\circ$ to $-180^\\circ$ is Quadrant III.\n\nSince $-180 < a < -90$, the angle $a$ lies in **Quadrant III**, where the sine function is indeed negative."},{"type":"hyper_latex","block":{"nodes":[{"text":"In Quadrant III with a negative angle, the relation to the reference angle is:"},{"latex":"$$\\htmlData{anchor=ref-ang}{\\color{@amber}{\\theta_{\\text{ref}}}}$"},{"latex":"$$a = -180^\\circ + \\htmlData{anchor=ref-sub}{\\color{@amber}{\\theta_{\\text{ref}}}}$"},{"latex":"$$a = -180^\\circ + \\htmlData{anchor=val-sub}{\\color{@amber}{60^\\circ}} = \\htmlData{anchor=final-ans}{\\color{@sky}{-120^\\circ}}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"ref-ang"},{"color":"sky","style":"box","anchor":"final-ans"}],"connections":[{"to":"ref-sub","from":"ref-ang","color":"amber","label":"substitute","style":"solid"}]},"order":3,"title":"Apply symmetry rule","description":"We apply the symmetry of the sine function in the third quadrant to compute the final angle $a$ within our target interval."},{"type":"text","order":4,"title":"State the final answer","content":"The value of $a$ is:\n\n$$a = -120$$\n\n*(Note: Finding the reference angle of $60^\\circ$ earns the first mark M1, and applying the symmetry in the correct interval to obtain $-120$ earns the accuracy mark A1.)*"}]},{"id":"sine_graph_symmetry","label":"Symmetry of the Sine Graph","steps":[{"type":"text","order":0,"title":"Set up the equation","content":"Since the point $A\\left(a, -\\frac{\\sqrt{3}}{2}\\right)$ lies on the curve $d = \\sin \\theta^\\circ$, we substitute the coordinates into the equation:\n\n$$\\sin a^\\circ = -\\frac{\\sqrt{3}}{2}$$\n\nWe need to find the value of $a$ that satisfies this equation within the specific interval:\n\n$$-180 < a < -90$$"},{"type":"text","order":1,"title":"Find principal negative solution","content":"Using exact trigonometric values, we know that:\n\n$$\\sin(60^\\circ) = \\frac{\\sqrt{3}}{2}$$\n\nUsing the odd-symmetry property of the sine function, $\\sin(-\\theta) = -\\sin(\\theta)$, we can find the principal negative solution:\n\n$$\\sin(-60^\\circ) = -\\frac{\\sqrt{3}}{2}$$\n\nThis identifies the reference angle/principal value of $-60^\\circ$ M1. However, since $-60^\\circ$ does not fall within the interval $-180 < a < -90$, we need to find another symmetrical solution."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Principal value: } \\htmlData{anchor=p-val}{\\color{@sky}{-60^\\circ}}$"},{"latex":"$$\\text{Axis of symmetry: } \\htmlData{anchor=sym-axis}{\\color{@amber}{-90^\\circ}}$"},{"latex":"$$\\htmlData{anchor=final-val}{\\color{@purple}{a}} = \\color{@amber}{-90^\\circ} - (\\color{@sky}{-60^\\circ} - (\\color{@amber}{-90^\\circ}))$"},{"latex":"$$\\phantom{a} = -90^\\circ - 30^\\circ = \\color{@purple}{-120^\\circ}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"p-val"},{"color":"amber","style":"text-color","anchor":"sym-axis"},{"color":"purple","style":"text-color","anchor":"final-val"}],"connections":[{"to":"final-val","from":"p-val","color":"sky","label":"reflecting across minimum","style":"solid"},{"to":"final-val","from":"sym-axis","color":"amber","label":"local minimum","style":"solid"}]},"order":2,"title":"Apply sine curve symmetry","description":"In the negative region, the sine graph has a local minimum at $\\theta = -90^\\circ$, where $\\sin(-90^\\circ) = -1$. Because the sine curve is perfectly symmetric about the vertical line $\\theta = -90^\\circ$, we can find our target solution by reflecting the principal solution across this line."},{"type":"text","order":3,"title":"State final solution","content":"Now we check if this symmetrical solution lies within our given interval:\n\n$$-180 < -120 < -90$$\n\nSince $-120$ lies within the specified boundaries, we have found the correct value for $a$ A1.\n\nTherefore, the final answer is:\n\n$$a = -120$$"}]}],"generatedAt":"2026-06-09T07:08:14.649Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"a3608245-a346-457d-b857-042cd956b15a","label":"b","content":"(ii) State the value of \$\\sin(a^\\circ + 720^\\circ)\$.","markscheme":"- Uses the periodicity of the sine function $\\sin(\\theta + 360k) = \\sin \\theta$ to state that $\\sin(a^\\circ + 720^\\circ) = \\sin a^\\circ = -\\frac{\\sqrt{3}}{2}$ B1","marks":1,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"periodicity","label":"Periodicity of Sine","steps":[{"type":"text","order":0,"title":"Recall sine periodicity","content":"The sine function is periodic with a period of $360^\\circ$. This means that adding any multiple of $360^\\circ$ to the angle does not change the value of the sine function. \n\nWe can express this property mathematically as:\n\n$$\\sin(\\theta^\\circ + 360k^\\circ) = \\sin\\theta^\\circ$$\n\nwhere $k$ is any integer."},{"type":"text","order":1,"title":"Apply to the given angle","content":"In our case, we want to evaluate $\\sin(a^\\circ + 720^\\circ)$. \n\nNotice that $720^\\circ$ is a multiple of $360^\\circ$:\n\n$$720^\\circ = 2 \\times 360^\\circ$$\n\nThis means we can write the expression as:\n\n$$\\sin(a^\\circ + 2 \\times 360^\\circ)$$"},{"type":"text","order":2,"title":"Simplify and state the value","content":"Using the periodic property where $k = 2$:\n\n$$\\sin(a^\\circ + 720^\\circ) = \\sin a^\\circ$$\n\nFrom the previous part of the question, we know that $\\sin a^\\circ = -\\frac{\\sqrt{3}}{2}$. Therefore:\n\n$$\\sin(a^\\circ + 720^\\circ) = -\\frac{\\sqrt{3}}{2}$$"}]}],"generatedAt":"2026-06-09T07:08:48.270Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"05a3ee70-c8b5-473a-84f4-a8efbf177707","label":"c","content":"A second point $B$ on the curve has coordinates \$(b, \\frac{3}{7})\$, where \$0 < b < 90\$.\n\n(i) Find the exact value of \$\\sin(b^\\circ - 180^\\circ)\$.","markscheme":"- Uses the identity $\\sin(\\theta - 180^\\circ) = -\\sin \\theta$ or refers to the graph's rotational symmetry to obtain $-\\frac{3}{7}$ B1","marks":1,"order":2,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"trig_identities","label":"Using Trigonometric Identities","steps":[{"type":"text","order":0,"title":"Identify the given value","content":"Since the point $B\\left(b, \\frac{3}{7}\\right)$ lies on the curve $d = \\sin \\theta^\\circ$, we can substitute the coordinates into the equation:\n\n$$\\sin(b^\\circ) = \\frac{3}{7}$$","highlights":[{"text":"B","location":"specification"},{"text":"b, \\frac{3}{7}","location":"specification"}]},{"type":"text","order":1,"title":"Recall the trigonometric identity","content":"We want to find the value of $\\sin(b^\\circ - 180^\\circ)$. We can use the standard trigonometric shift identity:\n\n$$\\sin(\\theta - 180^\\circ) = -\\sin\\theta$$\n\nAlternatively, we can derive this using the subtraction formula for sine, $\\sin(A - B) = \\sin A\\cos B - \\cos A\\sin B$:\n\n$$\\begin{aligned} \\sin(b^\\circ - 180^\\circ) &= \\sin(b^\\circ)\\cos(180^\\circ) - \\cos(b^\\circ)\\sin(180^\\circ) \\\\ &= \\sin(b^\\circ)(-1) - \\cos(b^\\circ)(0) \\\\ &= -\\sin(b^\\circ) \\end{aligned}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\sin(b^\\circ - 180^\\circ) = -\\sin(b^\\circ)$"},{"latex":"$$\\text{Since } \\sin(b^\\circ) = \\htmlData{anchor=given-val}{\\color{@sky}{\\frac{3}{7}}}$"},{"latex":"$$\\sin(b^\\circ - 180^\\circ) = \\htmlData{anchor=ans-val}{-\\color{@sky}{\\frac{3}{7}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"given-val"},{"color":"sky","style":"text-color","anchor":"ans-val"}],"connections":[{"to":"ans-val","from":"given-val","color":"sky","label":"substitution","style":"solid"}]},"order":2,"title":"Substitute and evaluate","description":"Substitute the value of $\\sin(b^circ)$ into our identity to get the final exact value."}]},{"id":"graphical_symmetry","label":"Using Graphical Symmetry","steps":[{"type":"text","order":0,"title":"Identify point coordinates","content":"Since the point $B\\left(b, \\frac{3}{7}\\right)$ lies on the curve $d = \\sin \\theta^\\circ$, we can substitute the coordinates into the equation to find:\n\n$$\\sin(b^\\circ) = \\frac{3}{7}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=sin-b}{\\color{@sky}{\\sin(b^\\circ)}} = \\htmlData{anchor=val}{\\color{@orange}{\\frac{3}{7}}}$"},{"latex":"$$\\sin(b^\\circ - 180^\\circ) = -\\htmlData{anchor=sin-b-ref}{\\color{@sky}{\\sin(b^\\circ)}}$"},{"latex":"$$\\phantom{\\sin(b^\\circ - 180^\\circ)} = -\\htmlData{anchor=val-ref}{\\color{@orange}{\\frac{3}{7}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sin-b"},{"color":"orange","style":"text-color","anchor":"val"}],"connections":[{"to":"sin-b-ref","from":"sin-b","color":"sky","label":"substitute","style":"solid"},{"to":"val-ref","from":"val","color":"orange","label":"value","style":"solid"}]},"order":1,"title":"Apply graphical symmetry","description":"The sine curve is periodic and symmetric. A horizontal shift of $180^\\circ$ in either direction results in a vertical reflection of the curve, changing the sign of the corresponding $y$-value."},{"type":"text","order":2,"title":"State the final answer","content":"Therefore, the exact value of $\\sin(b^\\circ - 180^\\circ)$ is:\n\n$$-\\frac{3}{7}$$"}]}],"generatedAt":"2026-06-11T03:38:20.678Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"5f27374c-2b57-4323-b1c8-f85e465fd711","label":"d","content":"(ii) Find the exact value of \$\\cos b^\\circ\$.","markscheme":"- Employs the identity $\\cos^2 b + \\sin^2 b = 1$ to write $\\cos^2 b = 1 - (\\frac{3}{7})^2$ M1\n- Calculates $\\cos^2 b = \\frac{40}{49}$ and deduces $\\cos b = \\frac{\\sqrt{40}}{7} = \\frac{2\\sqrt{10}}{7}$ (must be positive since $0 < b < 90$) A1","marks":2,"order":3,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"pythagorean_identity","label":"Pythagorean Identity","steps":[{"type":"text","order":0,"title":"Recall the identity","content":"To find the exact value of $\\cos b^\\circ$ when we know $\\sin b^\\circ = \\frac{3}{7}$, we can use the fundamental Pythagorean trigonometric identity:\n\n$$\\cos^2 b^\\circ + \\sin^2 b^\\circ = 1$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\sin b^\\circ = \\htmlData{anchor=sin-val}{\\color{@sky}{\\frac{3}{7}}}$"},{"latex":"$$\\cos^2 b^\\circ + \\left(\\htmlData{anchor=sin-sub}{\\color{@sky}{\\frac{3}{7}}}\\right)^2 = 1$"},{"latex":"$$\\cos^2 b^\\circ = 1 - \\htmlData{anchor=sq-val}{\\color{@amber}{\\frac{9}{49}}}$"},{"latex":"$$\\cos^2 b^\\circ = \\htmlData{anchor=cos2-val}{\\color{@purple}{\\frac{40}{49}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sin-val"},{"color":"amber","style":"text-color","anchor":"sq-val"},{"color":"purple","style":"text-color","anchor":"cos2-val"}],"connections":[{"to":"sin-sub","from":"sin-val","color":"sky","label":"substitute","style":"solid"},{"to":"sq-val","from":"sin-sub","color":"amber","label":"square fraction","style":"solid"},{"to":"cos2-val","from":"sq-val","color":"purple","label":"subtract","style":"solid"}]},"order":1,"title":"Substitute and solve","description":"Substitute $\\sin b^\\circ = \\frac{3}{7}$ into the Pythagorean identity and rearrange to find $\\cos^2 b^\\circ$."},{"type":"text","order":2,"title":"Determine the sign","content":"To solve for $\\cos b^\\circ$, we take the square root of both sides. Since $b$ is in the first quadrant ($0 < b < 90$), the cosine of $b^\\circ$ must be positive:\n\n$$\\cos b^\\circ = \\sqrt{\\frac{40}{49}}$$\n\nSimplifying this expression gives:\n\n$$\\cos b^\\circ = \\frac{\\sqrt{40}}{\\sqrt{49}} = \\frac{2\\sqrt{10}}{7}$$"}]},{"id":"right_triangle","label":"Right-Angled Triangle Method","steps":[{"type":"text","order":0,"title":"Set up the triangle","content":"From the previous part of the question, we know that \n\n$$\\sin b^\\circ = \\frac{3}{7}$$\n\nSince $\\sin \\theta = \\frac{\\text{Opposite}}{\\text{Hypotenuse}}$, we can represent the angle $b^\\circ$ in a right-angled triangle where:\n\n* The **opposite** side is $3$\n* The **hypotenuse** is $7$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$a^2 + b^2 = c^2$"},{"latex":"$$\\text{Adjacent}^2 + \\htmlData{anchor=opp-val}{3}^2 = \\htmlData{anchor=hyp-val}{7}^2$"},{"latex":"$$\\text{Adjacent}^2 + 9 = 49$"},{"latex":"$$\\text{Adjacent}^2 = 40$"},{"latex":"$$\\text{Adjacent} = \\sqrt{40} = \\htmlData{anchor=adj-val}{2\\sqrt{10}}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"opp-val"},{"color":"sky","style":"text-color","anchor":"hyp-val"},{"color":"green","style":"text-color","anchor":"adj-val"}],"connections":[{"to":"adj-val","from":"opp-val","color":"amber","label":"Pythagorean theorem","style":"solid"}]},"order":1,"title":"Find the adjacent side","description":"We apply the Pythagorean theorem to find the length of the adjacent side."},{"type":"text","order":2,"title":"Calculate cos b","content":"Now we use the definition of cosine in a right-angled triangle:\n\n$$\\cos b^\\circ = \\frac{\\text{Adjacent}}{\\text{Hypotenuse}}$$\n\nSubstituting our values in gives:\n\n$$\\cos b^\\circ = \\frac{2\\sqrt{10}}{7}$$\n\nSince $0 < b < 90$, the angle lies in the first quadrant where the cosine value must be positive. Therefore, the exact value is:\n\n$$\\cos b^\\circ = \\frac{2\\sqrt{10}}{7}$$"}]}],"generatedAt":"2026-06-11T03:38:33.704Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"57e3f9a2-508b-47b4-bb16-5d6f9d43ceb8:reference_angle","label":"(a) Reference Angle and Quadrant Analysis","steps":[{"type":"text","order":0,"title":"Understand the equation","content":"We are given that the point $A\\left(a, -\\frac{\\sqrt{3}}{2}\\right)$ lies on the curve $d = \\sin \\theta^\\circ$. This means we need to solve the trigonometric equation:\n\n$$\\sin a^\\circ = -\\frac{\\sqrt{3}}{2}$$\n\nwhere the angle $a$ is constrained by the interval $-180 < a < -90$."},{"type":"text","order":1,"title":"Find the reference angle","content":"To solve $\\sin a^\\circ = -\\frac{\\sqrt{3}}{2}$ using quadrant analysis, we first find the acute **reference angle** $\\theta_{\\text{ref}}$ where the sine value is positive:\n\n$$\\sin \\theta_{\\text{ref}} = \\frac{\\sqrt{3}}{2}$$\n\nUsing our exact trigonometric values, we find:\n\n$$\\theta_{\\text{ref}} = 60^\\circ$$"},{"type":"text","order":2,"title":"Analyze the quadrant","content":"Now we determine which quadrant our angle $a$ lies in. \n\nThe given interval is $-180 < a < -90$.\n\n- Moving in the negative (clockwise) direction from the positive $x$-axis:\n - $0^\\circ$ to $-90^\\circ$ is Quadrant IV.\n - $-90^\\circ$ to $-180^\\circ$ is Quadrant III.\n\nSince $-180 < a < -90$, the angle $a$ lies in **Quadrant III**, where the sine function is indeed negative."},{"type":"hyper_latex","block":{"nodes":[{"text":"In Quadrant III with a negative angle, the relation to the reference angle is:"},{"latex":"$$\\htmlData{anchor=ref-ang}{\\color{@amber}{\\theta_{\\text{ref}}}}$"},{"latex":"$$a = -180^\\circ + \\htmlData{anchor=ref-sub}{\\color{@amber}{\\theta_{\\text{ref}}}}$"},{"latex":"$$a = -180^\\circ + \\htmlData{anchor=val-sub}{\\color{@amber}{60^\\circ}} = \\htmlData{anchor=final-ans}{\\color{@sky}{-120^\\circ}}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"ref-ang"},{"color":"sky","style":"box","anchor":"final-ans"}],"connections":[{"to":"ref-sub","from":"ref-ang","color":"amber","label":"substitute","style":"solid"}]},"order":3,"title":"Apply symmetry rule","description":"We apply the symmetry of the sine function in the third quadrant to compute the final angle $a$ within our target interval."},{"type":"text","order":4,"title":"State the final answer","content":"The value of $a$ is:\n\n$$a = -120$$\n\n*(Note: Finding the reference angle of $60^\\circ$ earns the first mark M1, and applying the symmetry in the correct interval to obtain $-120$ earns the accuracy mark A1.)*"}]},{"id":"57e3f9a2-508b-47b4-bb16-5d6f9d43ceb8:sine_graph_symmetry","label":"(a) Symmetry of the Sine Graph","steps":[{"type":"text","order":0,"title":"Set up the equation","content":"Since the point $A\\left(a, -\\frac{\\sqrt{3}}{2}\\right)$ lies on the curve $d = \\sin \\theta^\\circ$, we substitute the coordinates into the equation:\n\n$$\\sin a^\\circ = -\\frac{\\sqrt{3}}{2}$$\n\nWe need to find the value of $a$ that satisfies this equation within the specific interval:\n\n$$-180 < a < -90$$"},{"type":"text","order":1,"title":"Find principal negative solution","content":"Using exact trigonometric values, we know that:\n\n$$\\sin(60^\\circ) = \\frac{\\sqrt{3}}{2}$$\n\nUsing the odd-symmetry property of the sine function, $\\sin(-\\theta) = -\\sin(\\theta)$, we can find the principal negative solution:\n\n$$\\sin(-60^\\circ) = -\\frac{\\sqrt{3}}{2}$$\n\nThis identifies the reference angle/principal value of $-60^\\circ$ M1. However, since $-60^\\circ$ does not fall within the interval $-180 < a < -90$, we need to find another symmetrical solution."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Principal value: } \\htmlData{anchor=p-val}{\\color{@sky}{-60^\\circ}}$"},{"latex":"$$\\text{Axis of symmetry: } \\htmlData{anchor=sym-axis}{\\color{@amber}{-90^\\circ}}$"},{"latex":"$$\\htmlData{anchor=final-val}{\\color{@purple}{a}} = \\color{@amber}{-90^\\circ} - (\\color{@sky}{-60^\\circ} - (\\color{@amber}{-90^\\circ}))$"},{"latex":"$$\\phantom{a} = -90^\\circ - 30^\\circ = \\color{@purple}{-120^\\circ}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"p-val"},{"color":"amber","style":"text-color","anchor":"sym-axis"},{"color":"purple","style":"text-color","anchor":"final-val"}],"connections":[{"to":"final-val","from":"p-val","color":"sky","label":"reflecting across minimum","style":"solid"},{"to":"final-val","from":"sym-axis","color":"amber","label":"local minimum","style":"solid"}]},"order":2,"title":"Apply sine curve symmetry","description":"In the negative region, the sine graph has a local minimum at $\\theta = -90^\\circ$, where $\\sin(-90^\\circ) = -1$. Because the sine curve is perfectly symmetric about the vertical line $\\theta = -90^\\circ$, we can find our target solution by reflecting the principal solution across this line."},{"type":"text","order":3,"title":"State final solution","content":"Now we check if this symmetrical solution lies within our given interval:\n\n$$-180 < -120 < -90$$\n\nSince $-120$ lies within the specified boundaries, we have found the correct value for $a$ A1.\n\nTherefore, the final answer is:\n\n$$a = -120$$"}]},{"id":"a3608245-a346-457d-b857-042cd956b15a:periodicity","label":"(b) Periodicity of Sine","steps":[{"type":"text","order":0,"title":"Recall sine periodicity","content":"The sine function is periodic with a period of $360^\\circ$. This means that adding any multiple of $360^\\circ$ to the angle does not change the value of the sine function. \n\nWe can express this property mathematically as:\n\n$$\\sin(\\theta^\\circ + 360k^\\circ) = \\sin\\theta^\\circ$$\n\nwhere $k$ is any integer."},{"type":"text","order":1,"title":"Apply to the given angle","content":"In our case, we want to evaluate $\\sin(a^\\circ + 720^\\circ)$. \n\nNotice that $720^\\circ$ is a multiple of $360^\\circ$:\n\n$$720^\\circ = 2 \\times 360^\\circ$$\n\nThis means we can write the expression as:\n\n$$\\sin(a^\\circ + 2 \\times 360^\\circ)$$"},{"type":"text","order":2,"title":"Simplify and state the value","content":"Using the periodic property where $k = 2$:\n\n$$\\sin(a^\\circ + 720^\\circ) = \\sin a^\\circ$$\n\nFrom the previous part of the question, we know that $\\sin a^\\circ = -\\frac{\\sqrt{3}}{2}$. Therefore:\n\n$$\\sin(a^\\circ + 720^\\circ) = -\\frac{\\sqrt{3}}{2}$$"}]},{"id":"05a3ee70-c8b5-473a-84f4-a8efbf177707:trig_identities","label":"(c) Using Trigonometric Identities","steps":[{"type":"text","order":0,"title":"Identify the given value","content":"Since the point $B\\left(b, \\frac{3}{7}\\right)$ lies on the curve $d = \\sin \\theta^\\circ$, we can substitute the coordinates into the equation:\n\n$$\\sin(b^\\circ) = \\frac{3}{7}$$","highlights":[{"text":"B","location":"specification"},{"text":"b, \\frac{3}{7}","location":"specification"}]},{"type":"text","order":1,"title":"Recall the trigonometric identity","content":"We want to find the value of $\\sin(b^\\circ - 180^\\circ)$. We can use the standard trigonometric shift identity:\n\n$$\\sin(\\theta - 180^\\circ) = -\\sin\\theta$$\n\nAlternatively, we can derive this using the subtraction formula for sine, $\\sin(A - B) = \\sin A\\cos B - \\cos A\\sin B$:\n\n$$\\begin{aligned} \\sin(b^\\circ - 180^\\circ) &= \\sin(b^\\circ)\\cos(180^\\circ) - \\cos(b^\\circ)\\sin(180^\\circ) \\\\ &= \\sin(b^\\circ)(-1) - \\cos(b^\\circ)(0) \\\\ &= -\\sin(b^\\circ) \\end{aligned}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\sin(b^\\circ - 180^\\circ) = -\\sin(b^\\circ)$"},{"latex":"$$\\text{Since } \\sin(b^\\circ) = \\htmlData{anchor=given-val}{\\color{@sky}{\\frac{3}{7}}}$"},{"latex":"$$\\sin(b^\\circ - 180^\\circ) = \\htmlData{anchor=ans-val}{-\\color{@sky}{\\frac{3}{7}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"given-val"},{"color":"sky","style":"text-color","anchor":"ans-val"}],"connections":[{"to":"ans-val","from":"given-val","color":"sky","label":"substitution","style":"solid"}]},"order":2,"title":"Substitute and evaluate","description":"Substitute the value of $\\sin(b^\\circ)$ into our identity to get the final exact value."}]},{"id":"05a3ee70-c8b5-473a-84f4-a8efbf177707:graphical_symmetry","label":"(c) Using Graphical Symmetry","steps":[{"type":"text","order":0,"title":"Identify point coordinates","content":"Since the point $B\\left(b, \\frac{3}{7}\\right)$ lies on the curve $d = \\sin \\theta^\\circ$, we can substitute the coordinates into the equation to find:\n\n$$\\sin(b^\\circ) = \\frac{3}{7}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=sin-b}{\\color{@sky}{\\sin(b^\\circ)}} = \\htmlData{anchor=val}{\\color{@orange}{\\frac{3}{7}}}$"},{"latex":"$$\\sin(b^\\circ - 180^\\circ) = -\\htmlData{anchor=sin-b-ref}{\\color{@sky}{\\sin(b^\\circ)}}$"},{"latex":"$$\\phantom{\\sin(b^\\circ - 180^\\circ)} = -\\htmlData{anchor=val-ref}{\\color{@orange}{\\frac{3}{7}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sin-b"},{"color":"orange","style":"text-color","anchor":"val"}],"connections":[{"to":"sin-b-ref","from":"sin-b","color":"sky","label":"substitute","style":"solid"},{"to":"val-ref","from":"val","color":"orange","label":"value","style":"solid"}]},"order":1,"title":"Apply graphical symmetry","description":"The sine curve is periodic and symmetric. A horizontal shift of $180^\\circ$ in either direction results in a vertical reflection of the curve, changing the sign of the corresponding $y$-value."},{"type":"text","order":2,"title":"State the final answer","content":"Therefore, the exact value of $\\sin(b^\\circ - 180^\\circ)$ is:\n\n$$-\\frac{3}{7}$$"}]},{"id":"5f27374c-2b57-4323-b1c8-f85e465fd711:pythagorean_identity","label":"(d) Pythagorean Identity","steps":[{"type":"text","order":0,"title":"Recall the identity","content":"To find the exact value of $\\cos b^\\circ$ when we know $\\sin b^\\circ = \\frac{3}{7}$, we can use the fundamental Pythagorean trigonometric identity:\n\n$$\\cos^2 b^\\circ + \\sin^2 b^\\circ = 1$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\sin b^\\circ = \\htmlData{anchor=sin-val}{\\color{@sky}{\\frac{3}{7}}}$"},{"latex":"$$\\cos^2 b^\\circ + \\left(\\htmlData{anchor=sin-sub}{\\color{@sky}{\\frac{3}{7}}}\\right)^2 = 1$"},{"latex":"$$\\cos^2 b^\\circ = 1 - \\htmlData{anchor=sq-val}{\\color{@amber}{\\frac{9}{49}}}$"},{"latex":"$$\\cos^2 b^\\circ = \\htmlData{anchor=cos2-val}{\\color{@purple}{\\frac{40}{49}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sin-val"},{"color":"amber","style":"text-color","anchor":"sq-val"},{"color":"purple","style":"text-color","anchor":"cos2-val"}],"connections":[{"to":"sin-sub","from":"sin-val","color":"sky","label":"substitute","style":"solid"},{"to":"sq-val","from":"sin-sub","color":"amber","label":"square fraction","style":"solid"},{"to":"cos2-val","from":"sq-val","color":"purple","label":"subtract","style":"solid"}]},"order":1,"title":"Substitute and solve","description":"Substitute $\\sin b^\\circ = \\frac{3}{7}$ into the Pythagorean identity and rearrange to find $\\cos^2 b^\\circ$."},{"type":"text","order":2,"title":"Determine the sign","content":"To solve for $\\cos b^\\circ$, we take the square root of both sides. Since $b$ is in the first quadrant ($0 < b < 90$), the cosine of $b^\\circ$ must be positive:\n\n$$\\cos b^\\circ = \\sqrt{\\frac{40}{49}}$$\n\nSimplifying this expression gives:\n\n$$\\cos b^\\circ = \\frac{\\sqrt{40}}{\\sqrt{49}} = \\frac{2\\sqrt{10}}{7}$$"}]},{"id":"5f27374c-2b57-4323-b1c8-f85e465fd711:right_triangle","label":"(d) Right-Angled Triangle Method","steps":[{"type":"text","order":0,"title":"Set up the triangle","content":"From the previous part of the question, we know that \n\n$$\\sin b^\\circ = \\frac{3}{7}$$\n\nSince $\\sin \\theta = \\frac{\\text{Opposite}}{\\text{Hypotenuse}}$, we can represent the angle $b^\\circ$ in a right-angled triangle where:\n\n* The **opposite** side is $3$\n* The **hypotenuse** is $7$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$a^2 + b^2 = c^2$"},{"latex":"$$\\text{Adjacent}^2 + \\htmlData{anchor=opp-val}{3}^2 = \\htmlData{anchor=hyp-val}{7}^2$"},{"latex":"$$\\text{Adjacent}^2 + 9 = 49$"},{"latex":"$$\\text{Adjacent}^2 = 40$"},{"latex":"$$\\text{Adjacent} = \\sqrt{40} = \\htmlData{anchor=adj-val}{2\\sqrt{10}}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"opp-val"},{"color":"sky","style":"text-color","anchor":"hyp-val"},{"color":"green","style":"text-color","anchor":"adj-val"}],"connections":[{"to":"adj-val","from":"opp-val","color":"amber","label":"Pythagorean theorem","style":"solid"}]},"order":1,"title":"Find the adjacent side","description":"We apply the Pythagorean theorem to find the length of the adjacent side."},{"type":"text","order":2,"title":"Calculate cos b","content":"Now we use the definition of cosine in a right-angled triangle:\n\n$$\\cos b^\\circ = \\frac{\\text{Adjacent}}{\\text{Hypotenuse}}$$\n\nSubstituting our values in gives:\n\n$$\\cos b^\\circ = \\frac{2\\sqrt{10}}{7}$$\n\nSince $0 < b < 90$, the angle lies in the first quadrant where the cosine value must be positive. Therefore, the exact value is:\n\n$$\\cos b^\\circ = \\frac{2\\sqrt{10}}{7}$$"}]}],"generatedAt":"2026-06-11T03:38:33.766Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths__ocr","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":"148b7a0b-9bdb-4027-864c-b263feb3a11f","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"16625c96-dbb8-4a12-aaca-d77403e7dbac","difficultyBand":"hard","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"1e025dde-5149-4829-8b94-187e72194654","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"1eeb206a-a348-4e3a-a08f-bbe35d6be702","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"21b4f24d-1066-4db1-87e6-25a1d3e10f8c","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"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":"36432791-e3b1-44de-8893-0fe68c790bcf","difficultyBand":"easy","calculatorUse":null,"totalMarks":1,"questionType":"MC","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":"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":"5dbd6a16-7dfb-44ce-aece-fe9cee1738bd","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"5e53ca22-d935-483a-9685-aa174bd8da42","difficultyBand":"hard","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":false},{"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":"70499ec1-c56f-420f-adde-ce7a7491e875","difficultyBand":"easy","calculatorUse":null,"totalMarks":6,"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":"8a7720af-6125-4af5-9cb7-5fe14aa7b0c4","difficultyBand":"hard","calculatorUse":null,"totalMarks":8,"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},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Trigonometric equations","topicId":"a94326e7-4351-4f54-959c-b7b59d51356e","topicSidebar":"$L3c","topicSlug":"trigonometric-equations","topicTitle":"1.5.15 Trigonometric equations"}]}],"$L3d","$L3e"]}]

1.5.15 Trigonometric equations

Question 2

1.5.15 Trigonometric equations Questions

1.5.15 Trigonometric equations

Question 2

1.5.15 Trigonometric equations Questions