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bg-background","children":["$","$L39",null,{"backHref":"/a-level/maths/aqa/questions","basePath":"/a-level/maths/aqa/e-trigonometry/questions","formulaUrl":"https://filestore.aqa.org.uk/resources/mathematics/AQA-AS-A-MATHS-FORMULAE.PDF","initialIndex":178,"initialQuestionWindow":[{"id":"7210db25-57b0-45ca-8021-438c4fc004d2","revisionId":"19bd1476-5a4b-40ec-992d-8bc65c34a68b","questionNumber":null,"questionType":"LA","specification":"Solve, for $-\\pi \\leq x \\leq \\pi$, the equation\n\n$5\\cos x + \\cot x = 0$","difficultyBand":"medium","tier":null,"totalMarks":5,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"57b7ac6f-fb26-4733-99e5-ec6aaacf1b93","label":null,"content":"Give your answers to two decimal places where appropriate.","markscheme":"- Writes $\\cot x = \\frac{\\cos x}{\\sin x}$ and factorises: $\\cos x\\left(5 + \\frac{1}{\\sin x}\\right) = 0$ M1\n- $\\cos x = 0$ gives $x = -\\frac{\\pi}{2}$ and $x = \\frac{\\pi}{2}$ A1\n- $5\\sin x + 1 = 0$, so $\\sin x = -\\frac{1}{5}$ M1\n- $x = -0.20$ A1\n- $x = -2.94$ A1\n\nAll four solutions are required. Dividing through by $\\cos x$ loses the two solutions from $\\cos x = 0$.","marks":5,"order":0,"markingRubric":null,"tableImages":[],"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"728d4ae9-23d0-424a-8951-12bc6cae52a5","revisionId":"a72d1aeb-4135-412d-acba-57ec48ae38e3","questionNumber":null,"questionType":"LA","specification":"A sub-aquatic drone's descent angle $\\theta$ is controlled by a buoyancy system. To maintain stable movement, the angle must satisfy the equilibrium equation:\n\n\\[ 5\\cot^2\\theta + 3\\text{ cosec}^2\\theta = 2\\text{ cosec }\\theta + 10 \\]","difficultyBand":"medium","tier":null,"totalMarks":7,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"fe416e8e-d690-4402-8453-81488152de75","label":"a","content":"Show that this equation can be written in the form \\[a\\text{ cosec}^2\\theta + b\\text{ cosec }\\theta + c = 0\\] where $a$, $b$, and $c$ are integers to be found.","markscheme":"- Identifies and applies the trigonometric identity $\\cot^2\\theta = \\text{cosec}^2\\theta - 1$ M1\n- Substitutes the identity into the given equation: $5(\\text{cosec}^2\\theta - 1) + 3\\text{cosec}^2\\theta = 2\\text{cosec }\\theta + 10$ A1\n- Simplifies to the form $8\\text{cosec}^2\\theta - 2\\text{cosec }\\theta - 15 = 0$, giving $a = 8, b = -2, c = -15$ A1","marks":3,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"trig_identity","label":"Trigonometric Identity Substitution","steps":[{"type":"text","order":0,"title":"Identify the relevant identity","content":"To write the given equation solely in terms of $\\csc\\theta$, we need to substitute the term $\\cot^2\\theta$.\n\nRecall the fundamental Pythagorean identity linking cosecant and cotangent from your formula booklet:\n\n$$\\cot^2\\theta = \\csc^2\\theta - 1$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$5\\htmlData{anchor=cot-term}{\\color{@sky}{\\cot^2\\theta}} + 3\\csc^2\\theta = 2\\csc\\theta + 10$"},{"latex":"$$5\\htmlData{anchor=sub-term}{\\color{@sky}{(\\csc^2\\theta - 1)}} + 3\\csc^2\\theta = 2\\csc\\theta + 10$"}],"highlights":[{"color":"sky","style":"box","anchor":"cot-term"},{"color":"sky","style":"box","anchor":"sub-term"}],"connections":[{"to":"sub-term","from":"cot-term","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute the identity","description":"We replace $\\cot^2\\theta$ with $(\\csc^2\\theta - 1)$ in the given equation. This will earn you the first method mark (M1)."},{"type":"text","order":2,"title":"Expand and simplify","content":"Next, expand the bracket and group the like terms together:\n\n$$5(\\csc^2\\theta - 1) + 3\\csc^2\\theta = 2\\csc\\theta + 10$$\n\n$$5\\csc^2\\theta - 5 + 3\\csc^2\\theta = 2\\csc\\theta + 10$$\n\n$$8\\csc^2\\theta - 5 = 2\\csc\\theta + 10$$\n\nThis earns the accuracy mark (A1) for setting up the simplified substitution."},{"type":"text","order":3,"title":"Form the quadratic equation","content":"Rearrange all terms to one side of the equation to set it equal to $0$:\n\n$$8\\csc^2\\theta - 2\\csc\\theta - 5 - 10 = 0$$\n\n$$8\\csc^2\\theta - 2\\csc\\theta - 15 = 0$$\n\nComparing this to the form $a\\csc^2\\theta + b\\csc\\theta + c = 0$, we find the integer values:\n\n- $a = 8$\n- $b = -2$\n- $c = -15$\n\nThis final step completes the third mark (A1)."}]},{"id":"sine_cosine_conversion","label":"Conversion to Sine and Cosine","steps":[{"type":"text","order":0,"title":"Convert to sine and cosine","content":"To solve this using the conversion to sine and cosine approach, we first rewrite the trigonometric functions in terms of $\\sin\\theta$ and $\\cos\\theta$ using the reciprocal identities:\n\n$$\\cot\\theta = \\frac{\\cos\\theta}{\\sin\\theta} \\implies \\cot^2\\theta = \\frac{\\cos^2\\theta}{\\sin^2\\theta}$$\n\n$$\\csc\\theta = \\frac{1}{\\sin\\theta} \\implies \\csc^2\\theta = \\frac{1}{\\sin^2\\theta}$$\n\nSubstituting these into the given equation:\n\n$$5\\left(\\frac{\\cos^2\\theta}{\\sin^2\\theta}\\right) + 3\\left(\\frac{1}{\\sin^2\\theta}\\right) = 2\\left(\\frac{1}{\\sin\\theta}\\right) + 10$$"},{"type":"text","order":1,"title":"Eliminate the denominators","content":"To clear the denominators, we multiply every term in the equation by $\\sin^2\\theta$ (since $\\sin\\theta \\neq 0$):\n\n$$5\\cos^2\\theta + 3 = 2\\sin\\theta + 10\\sin^2\\theta$$"},{"type":"text","order":2,"title":"Substitute the Pythagorean identity","content":"We now use the fundamental identity $\\cos^2\\theta = 1 - \\sin^2\\theta$ to write the entire equation in terms of $\\sin\\theta$:\n\n$$5(1 - \\sin^2\\theta) + 3 = 2\\sin\\theta + 10\\sin^2\\theta$$"},{"type":"text","order":3,"title":"Simplify the quadratic equation","content":"Expand and collect all terms on one side to form a quadratic in $\\sin\\theta$:\n\n$$5 - 5\\sin^2\\theta + 3 = 2\\sin\\theta + 10\\sin^2\\theta$$\n\n$$8 - 5\\sin^2\\theta = 2\\sin\\theta + 10\\sin^2\\theta$$\n\n$$15\\sin^2\\theta + 2\\sin\\theta - 8 = 0$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$15\\left(\\frac{1}{\\htmlData{anchor=cosec-sq}{\\color{@sky}{\\csc^2\\theta}}}\\right) + 2\\left(\\frac{1}{\\htmlData{anchor=cosec}{\\color{@pink}{\\csc\\theta}}}\\right) - 8 = 0$"},{"latex":"$$15 + 2\\htmlData{anchor=cosec-term}{\\color{@pink}{\\csc\\theta}} - 8\\htmlData{anchor=cosec-sq-term}{\\color{@sky}{\\csc^2\\theta}} = 0$"}],"highlights":[{"color":"sky","style":"underline","anchor":"cosec-sq"},{"color":"pink","style":"underline","anchor":"cosec"}],"connections":[{"to":"cosec-sq-term","from":"cosec-sq","color":"sky","label":"multiply by csc^2 theta","style":"solid"},{"to":"cosec-term","from":"cosec","color":"pink","label":"multiply by csc^2 theta","style":"solid"}]},"order":4,"title":"Convert back to cosecant","description":"We substitute $\\sin\\theta = \\frac{1}{\\csc\\theta}$ back into our quadratic equation, and then multiply the entire equation by $\\csc^2\\theta$ to clear the fractions."},{"type":"text","order":5,"title":"Identify coefficients","content":"Rearrange the equation into the standard form $a\\csc^2\\theta + b\\csc\\theta + c = 0$ by multiplying by $-1$:\n\n$$8\\csc^2\\theta - 2\\csc\\theta - 15 = 0$$\n\nComparing this to the required form, we find the integer values of $a$, $b$, and $c$:\n- $a = 8$\n- $b = -2$\n- $c = -15$"}]}],"generatedAt":"2026-06-09T06:27:56.968Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"95abc8e8-8640-4dab-af31-13d270db7094","label":"b","content":"Hence, given that the drone is tilted at an obtuse angle $\\theta$ that satisfies this stability equation, determine the exact value of $\\tan \\theta$. Fully justify your answer.","markscheme":"- Solves the quadratic $8y^2 - 2y - 15 = 0$ using factorisation $(2y - 3)(4y + 5) = 0$ or the quadratic formula to find $y = \\frac{3}{2}$ or $y = -\\frac{5}{4}$ M1\n- Justifies that for an obtuse angle ($90^\\circ < \\theta < 180^\\circ$), $\\sin \\theta$ is positive, therefore $\\text{cosec }\\theta$ must be positive, so $\\text{cosec }\\theta = \\frac{3}{2}$ B1\n- Uses $\\tan^2 \\theta = \\frac{1}{\\text{cosec}^2 \\theta - 1}$ or calculates $\\sin \\theta = \\frac{2}{3}$ to find $\\cos \\theta = -\\sqrt{1 - (2/3)^2} = -\\frac{\\sqrt{5}}{3}$ M1\n- Concludes $\\tan \\theta = \\frac{2/3}{-\\sqrt{5}/3} = -\\frac{2}{\\sqrt{5}}$, which rationalises to $-\\frac{2\\sqrt{5}}{5}$, with the negative sign correctly justified by the second quadrant A1","marks":4,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"cosecant_quadratic","label":"Cosecant Quadratic Substitution","steps":[{"type":"text","order":0,"title":"Recall the trigonometric identity","content":"To rewrite the equation entirely in terms of $\\text{cosec }\\theta$, we use the standard Pythagorean identity:\n\n$$\\cot^2\\theta = \\text{cosec}^2\\theta - 1$$\n\nSubstituting this into the equilibrium equation gives:\n\n$$5(\\text{cosec}^2\\theta - 1) + 3\\text{cosec}^2\\theta = 2\\text{cosec }\\theta + 10$$","highlights":[{"text":"5\\cot^2\\theta + 3\\text{ cosec}^2\\theta = 2\\text{ cosec }\theta + 10","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$5\\htmlData{anchor=cot-sub}{(\\text{cosec}^2\\theta - 1)} + 3\\text{cosec}^2\\theta = 2\\text{cosec }\\theta + 10$"},{"latex":"$$5\\text{cosec}^2\\theta - 5 + 3\\text{cosec}^2\\theta = 2\\text{cosec }\\theta + 10$"},{"latex":"$$\\htmlData{anchor=quad-eq}{8\\text{cosec}^2\\theta - 2\\text{cosec }\theta - 15 = 0}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"cot-sub"},{"color":"sky","style":"box","anchor":"quad-eq"}],"connections":[{"to":"quad-eq","from":"cot-sub","color":"sky","label":"expand and collect terms","style":"solid"}]},"order":1,"title":"Form a quadratic equation","description":"Expand the terms and move everything to one side to set up a quadratic equation in terms of $\\text{cosec }\\theta$."},{"type":"text","order":2,"title":"Solve the quadratic equation","content":"Let $y = \\text{cosec }\\theta$. Our equation is:\n\n$$8y^2 - 2y - 15 = 0$$\n\nWe can solve this by factorisation:\n\n$$(2y - 3)(4y + 5) = 0$$\n\nThis gives two possible solutions:\n\n$$y = \\frac{3}{2} \\quad \\text{or} \\quad y = -\\frac{5}{4}$$\n\nSo,\n\n$$\\text{cosec }\\theta = \\frac{3}{2} \\quad \\text{or} \\quad \\text{cosec }\\theta = -\\frac{5}{4}$$"},{"type":"text","order":3,"title":"Apply the obtuse constraint","content":"We are given that $\\theta$ is an **obtuse angle**, which means:\n\n$$90^\\circ < \\theta < 180^\\circ \\quad \\left(\\text{or } \\frac{\\pi}{2} < \\theta < \\pi\\right)$$\n\nIn the second quadrant, the sine function is positive:\n\n$$\\sin\\theta > 0 \\implies \\text{cosec }\\theta = \\frac{1}{\\sin\\theta} > 0$$\n\nTherefore, we reject the negative solution and conclude:\n\n$$\\text{cosec }\\theta = \\frac{3}{2}$$","highlights":[{"text":"obtuse angle \\theta","location":"specification"}]},{"type":"text","order":4,"title":"Find the exact tan value","content":"Since $\\text{cosec }\\theta = \\frac{3}{2}$, we have:\n\n$$\\sin \\theta = \\frac{2}{3}$$\n\nUsing the Pythagorean identity $\\cos^2\\theta = 1 - \\sin^2\\theta$:\n\n$$\\cos^2\\theta = 1 - \\left(\\frac{2}{3}\\right)^2 = 1 - \\frac{4}{9} = \\frac{5}{9}$$\n\nSince $\\theta$ is obtuse (in the second quadrant), the cosine must be negative:\n\n$$\\cos \\theta = -\\frac{\\sqrt{5}}{3}$$\n\nNow, we find $\\tan\\theta$:\n\n$$\\tan\\theta = \\frac{\\sin\\theta}{\\cos\\theta} = \\frac{\\frac{2}{3}}{-\\frac{\\sqrt{5}}{3}} = -\\frac{2}{\\sqrt{5}}$$\n\nRationalising the denominator, we get:\n\n$$\\tan\\theta = -\\frac{2\\sqrt{5}}{5}$$","highlights":[{"text":"exact value of \\tan \\theta","location":"specification"}]}]},{"id":"sine_quadratic","label":"Sine Quadratic Conversion","steps":[{"type":"text","order":0,"title":"Express terms in sine and cosine","content":"To solve using the **Sine Quadratic Conversion** approach, we rewrite each trigonometric term in the equation in terms of $\\sin\\theta$ and $\\cos\\theta$ using the reciprocal and quotient definitions:\n\n$$\\cot^2\\theta = \\frac{\\cos^2\\theta}{\\sin^2\\theta}, \\quad \\csc^2\\theta = \\frac{1}{\\sin^2\\theta}, \\quad \\csc\\theta = \\frac{1}{\\sin\\theta}$$\n\nSubstituting these into the given equilibrium equation:\n\n$$5\\left(\\frac{\\cos^2\\theta}{\\sin^2\\theta}\\right) + 3\\left(\\frac{1}{\\sin^2\\theta}\\right) = \\frac{2}{\\sin\\theta} + 10$$","highlights":[{"text":"5\\cot^2\\theta + 3\\text{ cosec}^2\\theta = 2\\text{ cosec }\theta + 10","location":"specification"}]},{"type":"text","order":1,"title":"Form a quadratic in sine","content":"To clear the denominators, we multiply every term by $\\sin^2\\theta$ (which is non-zero because $\\theta$ is an obtuse angle):\n\n$$5\\cos^2\\theta + 3 = 2\\sin\\theta + 10\\sin^2\\theta$$\n\nNext, we use the fundamental trigonometric identity $\\cos^2\\theta = 1 - \\sin^2\\theta$ to express the entire equation in terms of $\\sin\\theta$:\n\n$$5(1 - \\sin^2\\theta) + 3 = 2\\sin\\theta + 10\\sin^2\\theta$$\n\n$$5 - 5\\sin^2\\theta + 3 = 2\\sin\\theta + 10\\sin^2\\theta$$\n\nRearranging all terms to one side yields the quadratic equation:\n\n$$15\\sin^2\\theta + 2\\sin\\theta - 8 = 0$$"},{"type":"text","order":2,"title":"Solve for sine","content":"We can solve the quadratic equation $15\\sin^2\\theta + 2\\sin\\theta - 8 = 0$ by factoring. We look for two numbers that multiply to $15 \\times (-8) = -120$ and add to $2$. These numbers are $12$ and $-10$:\n\n$$15\\sin^2\\theta + 12\\sin\\theta - 10\\sin\\theta - 8 = 0$$\n\n$$3\\sin\\theta(5\\sin\\theta + 4) - 2(5\\sin\\theta + 4) = 0$$\n\n$$(3\\sin\\theta - 2)(5\\sin\\theta + 4) = 0$$\n\nThis gives two possible solutions for $\\sin\\theta$:\n\n$$\\sin\\theta = \\frac{2}{3} \\quad \\text{or} \\quad \\sin\\theta = -\\frac{4}{5}$$"},{"type":"text","order":3,"title":"Analyze the obtuse angle condition","content":"We are given that $\\theta$ is an **obtuse angle**, meaning:\n\n$$90^\\circ < \\theta < 180^\\circ$$\n\nIn the second quadrant, the sine function is positive ($ \\sin\\theta > 0 $). Therefore, we discard the negative root and choose:\n\n$$\\sin\\theta = \\frac{2}{3}$$","highlights":[{"text":"obtuse angle \\theta","location":"specification"}]},{"type":"text","order":4,"title":"Determine cosine and tangent","content":"To find $\\tan\\theta$, we first find $\\cos\\theta$ using the Pythagorean identity $\\cos^2\\theta = 1 - \\sin^2\\theta$. Since $\\theta$ is in the second quadrant, cosine must be negative:\n\n$$\\cos\\theta = -\\sqrt{1 - \\sin^2\\theta} = -\\sqrt{1 - \\left(\\frac{2}{3}\\right)^2} = -\\sqrt{1 - \\frac{4}{9}} = -\\frac{\\sqrt{5}}{3}$$\n\nNow, we calculate the exact value of $\\tan\\theta$ using the identity $\\tan\\theta = \\frac{\\sin\\theta}{\\cos\\theta}$:\n\n$$\\tan\\theta = \\frac{\\frac{2}{3}}{-\\frac{\\sqrt{5}}{3}} = -\\frac{2}{\\sqrt{5}}$$\n\nRationalising the denominator:\n\n$$\\tan\\theta = -\\frac{2\\sqrt{5}}{5}$$","highlights":[{"text":"determine the exact value of \\tan \\theta","location":"specification"}]}]}],"generatedAt":"2026-06-11T03:48:38.489Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"fe416e8e-d690-4402-8453-81488152de75:trig_identity","label":"(a) Trigonometric Identity Substitution","steps":[{"type":"text","order":0,"title":"Identify the relevant identity","content":"To write the given equation solely in terms of $\\csc\\theta$, we need to substitute the term $\\cot^2\\theta$.\n\nRecall the fundamental Pythagorean identity linking cosecant and cotangent from your formula booklet:\n\n$$\\cot^2\\theta = \\csc^2\\theta - 1$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$5\\htmlData{anchor=cot-term}{\\color{@sky}{\\cot^2\\theta}} + 3\\csc^2\\theta = 2\\csc\\theta + 10$"},{"latex":"$$5\\htmlData{anchor=sub-term}{\\color{@sky}{(\\csc^2\\theta - 1)}} + 3\\csc^2\\theta = 2\\csc\\theta + 10$"}],"highlights":[{"color":"sky","style":"box","anchor":"cot-term"},{"color":"sky","style":"box","anchor":"sub-term"}],"connections":[{"to":"sub-term","from":"cot-term","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute the identity","description":"We replace $\\cot^2\\theta$ with $(\\csc^2\\theta - 1)$ in the given equation. This will earn you the first method mark (M1)."},{"type":"text","order":2,"title":"Expand and simplify","content":"Next, expand the bracket and group the like terms together:\n\n$$5(\\csc^2\\theta - 1) + 3\\csc^2\\theta = 2\\csc\\theta + 10$$\n\n$$5\\csc^2\\theta - 5 + 3\\csc^2\\theta = 2\\csc\\theta + 10$$\n\n$$8\\csc^2\\theta - 5 = 2\\csc\\theta + 10$$\n\nThis earns the accuracy mark (A1) for setting up the simplified substitution."},{"type":"text","order":3,"title":"Form the quadratic equation","content":"Rearrange all terms to one side of the equation to set it equal to $0$:\n\n$$8\\csc^2\\theta - 2\\csc\\theta - 5 - 10 = 0$$\n\n$$8\\csc^2\\theta - 2\\csc\\theta - 15 = 0$$\n\nComparing this to the form $a\\csc^2\\theta + b\\csc\\theta + c = 0$, we find the integer values:\n\n- $a = 8$\n- $b = -2$\n- $c = -15$\n\nThis final step completes the third mark (A1)."}]},{"id":"fe416e8e-d690-4402-8453-81488152de75:sine_cosine_conversion","label":"(a) Conversion to Sine and Cosine","steps":[{"type":"text","order":0,"title":"Convert to sine and cosine","content":"To solve this using the conversion to sine and cosine approach, we first rewrite the trigonometric functions in terms of $\\sin\\theta$ and $\\cos\\theta$ using the reciprocal identities:\n\n$$\\cot\\theta = \\frac{\\cos\\theta}{\\sin\\theta} \\implies \\cot^2\\theta = \\frac{\\cos^2\\theta}{\\sin^2\\theta}$$\n\n$$\\csc\\theta = \\frac{1}{\\sin\\theta} \\implies \\csc^2\\theta = \\frac{1}{\\sin^2\\theta}$$\n\nSubstituting these into the given equation:\n\n$$5\\left(\\frac{\\cos^2\\theta}{\\sin^2\\theta}\\right) + 3\\left(\\frac{1}{\\sin^2\\theta}\\right) = 2\\left(\\frac{1}{\\sin\\theta}\\right) + 10$$"},{"type":"text","order":1,"title":"Eliminate the denominators","content":"To clear the denominators, we multiply every term in the equation by $\\sin^2\\theta$ (since $\\sin\\theta \\neq 0$):\n\n$$5\\cos^2\\theta + 3 = 2\\sin\\theta + 10\\sin^2\\theta$$"},{"type":"text","order":2,"title":"Substitute the Pythagorean identity","content":"We now use the fundamental identity $\\cos^2\\theta = 1 - \\sin^2\\theta$ to write the entire equation in terms of $\\sin\\theta$:\n\n$$5(1 - \\sin^2\\theta) + 3 = 2\\sin\\theta + 10\\sin^2\\theta$$"},{"type":"text","order":3,"title":"Simplify the quadratic equation","content":"Expand and collect all terms on one side to form a quadratic in $\\sin\\theta$:\n\n$$5 - 5\\sin^2\\theta + 3 = 2\\sin\\theta + 10\\sin^2\\theta$$\n\n$$8 - 5\\sin^2\\theta = 2\\sin\\theta + 10\\sin^2\\theta$$\n\n$$15\\sin^2\\theta + 2\\sin\\theta - 8 = 0$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$15\\left(\\frac{1}{\\htmlData{anchor=cosec-sq}{\\color{@sky}{\\csc^2\\theta}}}\\right) + 2\\left(\\frac{1}{\\htmlData{anchor=cosec}{\\color{@pink}{\\csc\\theta}}}\\right) - 8 = 0$"},{"latex":"$$15 + 2\\htmlData{anchor=cosec-term}{\\color{@pink}{\\csc\\theta}} - 8\\htmlData{anchor=cosec-sq-term}{\\color{@sky}{\\csc^2\\theta}} = 0$"}],"highlights":[{"color":"sky","style":"underline","anchor":"cosec-sq"},{"color":"pink","style":"underline","anchor":"cosec"}],"connections":[{"to":"cosec-sq-term","from":"cosec-sq","color":"sky","label":"multiply by csc^2 theta","style":"solid"},{"to":"cosec-term","from":"cosec","color":"pink","label":"multiply by csc^2 theta","style":"solid"}]},"order":4,"title":"Convert back to cosecant","description":"We substitute $\\sin\\theta = \\frac{1}{\\csc\\theta}$ back into our quadratic equation, and then multiply the entire equation by $\\csc^2\\theta$ to clear the fractions."},{"type":"text","order":5,"title":"Identify coefficients","content":"Rearrange the equation into the standard form $a\\csc^2\\theta + b\\csc\\theta + c = 0$ by multiplying by $-1$:\n\n$$8\\csc^2\\theta - 2\\csc\\theta - 15 = 0$$\n\nComparing this to the required form, we find the integer values of $a$, $b$, and $c$:\n- $a = 8$\n- $b = -2$\n- $c = -15$"}]},{"id":"95abc8e8-8640-4dab-af31-13d270db7094:cosecant_quadratic","label":"(b) Cosecant Quadratic Substitution","steps":[{"type":"text","order":0,"title":"Recall the trigonometric identity","content":"To rewrite the equation entirely in terms of $\\text{cosec }\\theta$, we use the standard Pythagorean identity:\n\n$$\\cot^2\\theta = \\text{cosec}^2\\theta - 1$$\n\nSubstituting this into the equilibrium equation gives:\n\n$$5(\\text{cosec}^2\\theta - 1) + 3\\text{cosec}^2\\theta = 2\\text{cosec }\\theta + 10$$","highlights":[{"text":"5\\cot^2\\theta + 3\\text{ cosec}^2\\theta = 2\\text{ cosec }\theta + 10","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$5\\htmlData{anchor=cot-sub}{(\\text{cosec}^2\\theta - 1)} + 3\\text{cosec}^2\\theta = 2\\text{cosec }\\theta + 10$"},{"latex":"$$5\\text{cosec}^2\\theta - 5 + 3\\text{cosec}^2\\theta = 2\\text{cosec }\\theta + 10$"},{"latex":"$$\\htmlData{anchor=quad-eq}{8\\text{cosec}^2\\theta - 2\\text{cosec }\\theta - 15 = 0}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"cot-sub"},{"color":"sky","style":"box","anchor":"quad-eq"}],"connections":[{"to":"quad-eq","from":"cot-sub","color":"sky","label":"expand and collect terms","style":"solid"}]},"order":1,"title":"Form a quadratic equation","description":"Expand the terms and move everything to one side to set up a quadratic equation in terms of $\\text{cosec }\\theta$."},{"type":"text","order":2,"title":"Solve the quadratic equation","content":"Let $y = \\text{cosec }\\theta$. Our equation is:\n\n$$8y^2 - 2y - 15 = 0$$\n\nWe can solve this by factorisation:\n\n$$(2y - 3)(4y + 5) = 0$$\n\nThis gives two possible solutions:\n\n$$y = \\frac{3}{2} \\quad \\text{or} \\quad y = -\\frac{5}{4}$$\n\nSo,\n\n$$\\text{cosec }\\theta = \\frac{3}{2} \\quad \\text{or} \\quad \\text{cosec }\\theta = -\\frac{5}{4}$$"},{"type":"text","order":3,"title":"Apply the obtuse constraint","content":"We are given that $\\theta$ is an **obtuse angle**, which means:\n\n$$90^\\circ < \\theta < 180^\\circ \\quad \\left(\\text{or } \\frac{\\pi}{2} < \\theta < \\pi\\right)$$\n\nIn the second quadrant, the sine function is positive:\n\n$$\\sin\\theta > 0 \\implies \\text{cosec }\\theta = \\frac{1}{\\sin\\theta} > 0$$\n\nTherefore, we reject the negative solution and conclude:\n\n$$\\text{cosec }\\theta = \\frac{3}{2}$$","highlights":[{"text":"obtuse angle \\theta","location":"specification"}]},{"type":"text","order":4,"title":"Find the exact tan value","content":"Since $\\text{cosec }\\theta = \\frac{3}{2}$, we have:\n\n$$\\sin \\theta = \\frac{2}{3}$$\n\nUsing the Pythagorean identity $\\cos^2\\theta = 1 - \\sin^2\\theta$:\n\n$$\\cos^2\\theta = 1 - \\left(\\frac{2}{3}\\right)^2 = 1 - \\frac{4}{9} = \\frac{5}{9}$$\n\nSince $\\theta$ is obtuse (in the second quadrant), the cosine must be negative:\n\n$$\\cos \\theta = -\\frac{\\sqrt{5}}{3}$$\n\nNow, we find $\\tan\\theta$:\n\n$$\\tan\\theta = \\frac{\\sin\\theta}{\\cos\\theta} = \\frac{\\frac{2}{3}}{-\\frac{\\sqrt{5}}{3}} = -\\frac{2}{\\sqrt{5}}$$\n\nRationalising the denominator, we get:\n\n$$\\tan\\theta = -\\frac{2\\sqrt{5}}{5}$$","highlights":[{"text":"exact value of \\tan \\theta","location":"specification"}]}]},{"id":"95abc8e8-8640-4dab-af31-13d270db7094:sine_quadratic","label":"(b) Sine Quadratic Conversion","steps":[{"type":"text","order":0,"title":"Express terms in sine and cosine","content":"To solve using the **Sine Quadratic Conversion** approach, we rewrite each trigonometric term in the equation in terms of $\\sin\\theta$ and $\\cos\\theta$ using the reciprocal and quotient definitions:\n\n$$\\cot^2\\theta = \\frac{\\cos^2\\theta}{\\sin^2\\theta}, \\quad \\csc^2\\theta = \\frac{1}{\\sin^2\\theta}, \\quad \\csc\\theta = \\frac{1}{\\sin\\theta}$$\n\nSubstituting these into the given equilibrium equation:\n\n$$5\\left(\\frac{\\cos^2\\theta}{\\sin^2\\theta}\\right) + 3\\left(\\frac{1}{\\sin^2\\theta}\\right) = \\frac{2}{\\sin\\theta} + 10$$","highlights":[{"text":"5\\cot^2\\theta + 3\\text{ cosec}^2\\theta = 2\\text{ cosec }\theta + 10","location":"specification"}]},{"type":"text","order":1,"title":"Form a quadratic in sine","content":"To clear the denominators, we multiply every term by $\\sin^2\\theta$ (which is non-zero because $\\theta$ is an obtuse angle):\n\n$$5\\cos^2\\theta + 3 = 2\\sin\\theta + 10\\sin^2\\theta$$\n\nNext, we use the fundamental trigonometric identity $\\cos^2\\theta = 1 - \\sin^2\\theta$ to express the entire equation in terms of $\\sin\\theta$:\n\n$$5(1 - \\sin^2\\theta) + 3 = 2\\sin\\theta + 10\\sin^2\\theta$$\n\n$$5 - 5\\sin^2\\theta + 3 = 2\\sin\\theta + 10\\sin^2\\theta$$\n\nRearranging all terms to one side yields the quadratic equation:\n\n$$15\\sin^2\\theta + 2\\sin\\theta - 8 = 0$$"},{"type":"text","order":2,"title":"Solve for sine","content":"We can solve the quadratic equation $15\\sin^2\\theta + 2\\sin\\theta - 8 = 0$ by factoring. We look for two numbers that multiply to $15 \\times (-8) = -120$ and add to $2$. These numbers are $12$ and $-10$:\n\n$$15\\sin^2\\theta + 12\\sin\\theta - 10\\sin\\theta - 8 = 0$$\n\n$$3\\sin\\theta(5\\sin\\theta + 4) - 2(5\\sin\\theta + 4) = 0$$\n\n$$(3\\sin\\theta - 2)(5\\sin\\theta + 4) = 0$$\n\nThis gives two possible solutions for $\\sin\\theta$:\n\n$$\\sin\\theta = \\frac{2}{3} \\quad \\text{or} \\quad \\sin\\theta = -\\frac{4}{5}$$"},{"type":"text","order":3,"title":"Analyze the obtuse angle condition","content":"We are given that $\\theta$ is an **obtuse angle**, meaning:\n\n$$90^\\circ < \\theta < 180^\\circ$$\n\nIn the second quadrant, the sine function is positive ($ \\sin\\theta > 0 $). Therefore, we discard the negative root and choose:\n\n$$\\sin\\theta = \\frac{2}{3}$$","highlights":[{"text":"obtuse angle \\theta","location":"specification"}]},{"type":"text","order":4,"title":"Determine cosine and tangent","content":"To find $\\tan\\theta$, we first find $\\cos\\theta$ using the Pythagorean identity $\\cos^2\\theta = 1 - \\sin^2\\theta$. Since $\\theta$ is in the second quadrant, cosine must be negative:\n\n$$\\cos\\theta = -\\sqrt{1 - \\sin^2\\theta} = -\\sqrt{1 - \\left(\\frac{2}{3}\\right)^2} = -\\sqrt{1 - \\frac{4}{9}} = -\\frac{\\sqrt{5}}{3}$$\n\nNow, we calculate the exact value of $\\tan\\theta$ using the identity $\\tan\\theta = \\frac{\\sin\\theta}{\\cos\\theta}$:\n\n$$\\tan\\theta = \\frac{\\frac{2}{3}}{-\\frac{\\sqrt{5}}{3}} = -\\frac{2}{\\sqrt{5}}$$\n\nRationalising the denominator:\n\n$$\\tan\\theta = -\\frac{2\\sqrt{5}}{5}$$","highlights":[{"text":"determine the exact value of \\tan 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1.8 E: Trigonometry

Question 109

1.8 E: Trigonometry Questions

1.8 E: Trigonometry

Question 109

1.8 E: Trigonometry Questions