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Solutions relying entirely on calculator technology are not acceptable.","difficulty":5,"difficultyBand":"medium","tier":null,"totalMarks":9,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"fb3a7872-1c73-4afc-b648-816d50d44ae6","label":"a","content":"Two acoustic waves are superimposed such that their resulting phase angle \$\\phi\$ satisfies the equation\n\\[\\sqrt{6} \\sin(\\phi - 60^\\circ) = 2 \\cos(\\phi + 45^\\circ)\\]\n\nShow that\n\\[\\tan \\phi = \\frac{5}{2 + \\sqrt{3}}\\]\nand hence that\n\\[\\tan \\phi = 10 - 5\\sqrt{3}\\]","markscheme":"- Uses addition formulae for both sine and cosine: $\\sqrt{6}(\\sin \\phi \\cos 60^\\circ - \\cos \\phi \\sin 60^\\circ) = 2(\\cos \\phi \\cos 45^\\circ - \\sin \\phi \\sin 45^\\circ)$ M1\n- Substitutes exact values for trigonometric ratios: $\\sqrt{6}\\left(\\frac{1}{2}\\sin \\phi - \\frac{\\sqrt{3}}{2}\\cos \\phi\\right) = 2\\left(\\frac{\\sqrt{2}}{2}\\cos \\phi - \\frac{\\sqrt{2}}{2}\\sin \\phi\\right)$ M1\n- Expands and simplifies to a form with $\\sin \\phi$ and $\\cos \\phi$ terms: $\\frac{\\sqrt{6}}{2}\\sin \\phi - \\frac{3\\sqrt{2}}{2}\\cos \\phi = \\sqrt{2}\\cos \\phi - \\sqrt{2}\\sin \\phi$ M1\n- Collects terms and divides by $\\sqrt{2}\\cos \\phi$ to reach $\\tan \\phi = \\frac{1 + \\frac{3}{2}}{\\frac{\\sqrt{3}}{2} + 1}$ or $(\\sqrt{3} + 2)\\sin \\phi = 5\\cos \\phi$ M1\n- Rationalises the denominator: $\\frac{5}{2 + \\sqrt{3}} \\times \\frac{2 - \\sqrt{3}}{2 - \\sqrt{3}} = \\frac{10 - 5\\sqrt{3}}{4 - 3} = 10 - 5\\sqrt{3}$ A1","marks":5,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"algebraic_expansion","label":"Algebraic Compound Angle Expansion","steps":[{"type":"text","order":0,"title":"Apply compound angle formulae","content":"To solve the given equation, we start by expanding both sides using the compound angle formulae (also known as the addition or subtraction formulae) from your trigonometry studies:\n\n$$\\sin(A - B) = \\sin A \\cos B - \\cos A \\sin B$$\n$$\\cos(A + B) = \\cos A \\cos B - \\sin A \\sin B$$\n\nApplying these to our equation:\n\n$$\\sqrt{6}\\sin(\\phi - 60^\\circ) = 2\\cos(\\phi + 45^\\circ)$$\n\nWe get:\n\n$$\\sqrt{6}\\left(\\sin \\phi \\cos 60^\\circ - \\cos \\phi \\sin 60^\\circ\\right) = 2\\left(\\cos \\phi \\cos 45^\\circ - \\sin \\phi \\sin 45^\\circ\\right)$$\n\nThis step earns the first method mark **(M1)**."},{"type":"text","order":1,"title":"Substitute exact values","content":"Next, we substitute the exact values for the trigonometric ratios of $60^\\circ$ and $45^\\circ$:\n\n* $\\cos 60^\\circ = \\frac{1}{2}$\n* $\\sin 60^\\circ = \\frac{\\sqrt{3}}{2}$\n* $\\cos 45^\\circ = \\frac{\\sqrt{2}}{2}$\n* $\\sin 45^\\circ = \\frac{\\sqrt{2}}{2}$\n\nSubstituting these exact values into our expanded equation **(M1)**:\n\n$$\\sqrt{6}\\left(\\frac{1}{2}\\sin \\phi - \\frac{\\sqrt{3}}{2}\\cos \\phi\\right) = 2\\left(\\frac{\\sqrt{2}}{2}\\cos \\phi - \\frac{\\sqrt{2}}{2}\\sin \\phi\\right)$$"},{"type":"text","order":2,"title":"Expand and simplify terms","content":"Now, we expand both sides of the equation to simplify the surds:\n\n$$\\frac{\\sqrt{6}}{2}\\sin \\phi - \\frac{\\sqrt{18}}{2}\\cos \\phi = \\sqrt{2}\\cos \\phi - \\sqrt{2}\\sin \\phi$$\n\nSince $\\sqrt{18} = \\sqrt{9 \\times 2} = 3\\sqrt{2}$, we can rewrite the equation as:\n\n$$\\frac{\\sqrt{6}}{2}\\sin \\phi - \\frac{3\\sqrt{2}}{2}\\cos \\phi = \\sqrt{2}\\cos \\phi - \\sqrt{2}\\sin \\phi$$\n\nTo make this simpler, let's divide every term on both sides by $\\sqrt{2}$ **(M1)**:\n\n$$\\frac{\\sqrt{3}}{2}\\sin \\phi - \\frac{3}{2}\\cos \\phi = \\cos \\phi - \\sin \\phi$$"},{"type":"text","order":3,"title":"Collect terms and solve","content":"Let's group all terms involving $\\sin \\phi$ on the left-hand side and all terms involving $\\cos \\phi$ on the right-hand side:\n\n$$\\frac{\\sqrt{3}}{2}\\sin \\phi + \\sin \\phi = \\cos \\phi + \\frac{3}{2}\\cos \\phi$$\n\nFactoring out $\\sin \\phi$ and $\\cos \\phi$, we get:\n\n$$\\left(\\frac{\\sqrt{3}}{2} + 1\\right)\\sin \\phi = \\frac{5}{2}\\cos \\phi$$\n\nMultiply the entire equation by $2$ to clear the denominators:\n\n$$(\\sqrt{3} + 2)\\sin \\phi = 5\\cos \\phi$$\n\nTo find $\\tan \\phi$, we divide both sides by $\\cos \\phi$ (since $\\tan \\phi = \\frac{\\sin \\phi}{\\cos \\phi}$) and then divide by $(\\sqrt{3} + 2)$ **(M1)**:\n\n$$\\tan \\phi = \\frac{5}{2 + \\sqrt{3}}$$\n\nThis shows the first required result!"},{"type":"text","order":4,"title":"Rationalise the denominator","content":"Finally, we rationalise the denominator by multiplying both the numerator and denominator by the conjugate $(2 - \\sqrt{3})$:\n\n$$\\tan \\phi = \\frac{5}{2 + \\sqrt{3}} \\times \\frac{2 - \\sqrt{3}}{2 - \\sqrt{3}}$$\n\n$$\\tan \\phi = \\frac{5(2 - \\sqrt{3})}{(2)^2 - (\\sqrt{3})^2}$$\n\n$$\\tan \\phi = \\frac{10 - 5\\sqrt{3}}{4 - 3}$$\n\n$$\\tan \\phi = 10 - 5\\sqrt{3}$$\n\nThis completes the proof and earns the final accuracy mark **(A1)**."},{"type":"text","order":5,"title":"Link Part 1 to Part 2","content":"For Part 2, we are asked to solve the equation:\n\n$$\\sqrt{6} \\sin(3\\theta - 60^\\circ) = 2 \\cos(3\\theta + 45^\\circ)$$\n\nNotice that this equation is identical in form to the one in Part 1, but with $\\phi$ replaced by $3\\theta$.\n\nTherefore, we can use the result from Part 1 directly **(B1)**:\n\n$$\\tan(3\\theta) = 10 - 5\\sqrt{3}$$"},{"type":"text","order":6,"title":"Find the principal value","content":"First, find the principal value for $3\\theta$ by taking the inverse tangent of both sides:\n\n$$3\\theta = \\arctan(10 - 5\\sqrt{3})$$\n\nUsing a calculator in degree mode **(M1)**:\n\n$$3\\theta \\approx 53.26^\\circ$$"},{"type":"text","order":7,"title":"Find other solutions","content":"We need to find all solutions for $\\theta$ in the interval $0 \\le \\theta < 180^\\circ$. Multiplying this interval by $3$, the interval for $3\\theta$ is:\n\n$$0 \\le 3\\theta < 540^\\circ$$\n\nSince the tangent function has a period of $180^\\circ$, we find other values for $3\\theta$ by repeatedly adding $180^\\circ$ to our principal value **(M1)**:\n\n$$\\begin{aligned} 3\\theta_1 &= 53.26^\\circ \\\\ 3\\theta_2 &= 53.26^\\circ + 180^\\circ = 233.26^\\circ \\\\ 3\\theta_3 &= 53.26^\\circ + 360^\\circ = 413.26^\\circ \\end{aligned}$$\n\nNote that adding $180^\\circ$ again would result in a value outside our range of $540^\\circ$."},{"type":"text","order":8,"title":"Solve for theta","content":"To find the final values of $\\theta$, we divide each of our values for $3\\theta$ by $3$:\n\n$$\\begin{aligned} \\theta_1 &= \\frac{53.26^\\circ}{3} \\approx 17.8^\\circ \\\\ \\theta_2 &= \\frac{233.26^\\circ}{3} \\approx 77.8^\\circ \\\\ \\theta_3 &= \\frac{413.26^\\circ}{3} \\approx 137.8^\\circ \\end{aligned}$$\n\nRounding to one decimal place as requested **(A1)**, we get:\n\n$$\\theta = 17.8^\\circ, \\; 77.8^\\circ, \\; 137.8^\\circ$$"}]}],"generatedAt":"2026-06-09T04:51:07.132Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"b858090b-3bda-4fc3-bc66-f2c7a4e0e03e","label":"b","content":"Hence or otherwise, solve for \$0 \\le \\theta < 180^\\circ\$,\n\\[\\sqrt{6} \\sin(3\\theta - 60^\\circ) = 2 \\cos(3\\theta + 45^\\circ)\\]\ngiving your answers to one decimal place.","markscheme":"- Recognises that the solution requires $\\tan(3\\theta) = 10 - 5\\sqrt{3}$ B1\n- Finds the principal value for $3\\theta$: $3\\theta = \\arctan(10 - 5\\sqrt{3}) \\approx 53.26^\\circ$ M1\n- Uses periodicity of tangent to find other values for $3\\theta$ in the range $0 \\le 3\\theta < 540^\\circ$: $3\\theta = 53.26^\\circ, 233.26^\\circ, 413.26^\\circ$ M1\n- Obtains final answers: $\\theta = 17.8^\\circ, 77.8^\\circ, 137.8^\\circ$ (awrt) A1","marks":4,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"compound_angle_expansion","label":"Compound Angle Expansion","steps":[{"type":"text","order":0,"title":"Recall the expansion formulas","content":"To solve this equation using the **Compound Angle Expansion** approach, we begin by recalling the sine subtraction and cosine addition formulas from our trigonometric identities:\n\n$$\\sin(A - B) = \\sin A \\cos B - \\cos A \\sin B$$\n$$\\cos(A + B) = \\cos A \\cos B - \\sin A \\sin B$$\n\nLet's apply these to each side of our given equation:\n$$\\sqrt{6} \\sin(3\\theta - 60^\\circ) = 2 \\cos(3\\theta + 45^\\circ)$$"},{"type":"text","order":1,"title":"Expand both sides","content":"Let's expand each side using the standard values for $\\sin(60^\\circ)$, $\\cos(60^\\circ)$, $\\sin(45^\\circ)$, and $\\cos(45^\\circ)$:\n- $\\cos(60^\\circ) = \\frac{1}{2}$ and $\\sin(60^\\circ) = \\frac{\\sqrt{3}}{2}$\n- $\\cos(45^\\circ) = \\frac{\\sqrt{2}}{2}$ and $\\sin(45^\\circ) = \\frac{\\sqrt{2}}{2}$\n\nExpanding the Left-Hand Side (LHS):\n$$\\sqrt{6} \\sin(3\\theta - 60^\\circ) = \\sqrt{6} \\left[ \\sin(3\\theta)\\cos(60^\\circ) - \\cos(3\\theta)\\sin(60^\\circ) \\right] = \\sqrt{6} \\left[ \\frac{1}{2}\\sin(3\\theta) - \\frac{\\sqrt{3}}{2}\\cos(3\\theta) \\right]$$\n$$\\text{LHS} = \\frac{\\sqrt{6}}{2}\\sin(3\\theta) - \\frac{3\\sqrt{2}}{2}\\cos(3\\theta)$$\n\nExpanding the Right-Hand Side (RHS):\n$$2 \\cos(3\\theta + 45^\\circ) = 2 \\left[ \\cos(3\\theta)\\cos(45^\\circ) - \\sin(3\\theta)\\sin(45^\\circ) \\right] = 2 \\left[ \\frac{\\sqrt{2}}{2}\\cos(3\\theta) - \\frac{\\sqrt{2}}{2}\\sin(3\\theta) \\right]$$\n$$\\text{RHS} = \\sqrt{2}\\cos(3\\theta) - \\sqrt{2}\\sin(3\\theta)$$"},{"type":"text","order":2,"title":"Group like terms","content":"Equating the expanded LHS and RHS:\n$$\\frac{\\sqrt{6}}{2}\\sin(3\\theta) - \\frac{3\\sqrt{2}}{2}\\cos(3\\theta) = \\sqrt{2}\\cos(3\\theta) - \\sqrt{2}\\sin(3\\theta)$$\n\nTo simplify, let's divide both sides by $\\sqrt{2}$ (noting that $\\frac{\\sqrt{6}}{\\sqrt{2}} = \\sqrt{3}$):\n$$\\frac{\\sqrt{3}}{2}\\sin(3\\theta) - \\frac{3}{2}\\cos(3\\theta) = \\cos(3\\theta) - \\sin(3\\theta)$$\n\nNow, collect all terms with $\\sin(3\\theta)$ on the left and those with $\\cos(3\\theta)$ on the right:\n$$\\left( \\frac{\\sqrt{3}}{2} + 1 \\right)\\sin(3\\theta) = \\left( 1 + \\frac{3}{2} \\right)\\cos(3\\theta)$$\n$$\\left( \\frac{\\sqrt{3} + 2}{2} \\right)\\sin(3\\theta) = \\frac{5}{2}\\cos(3\\theta)$$\n\nMultiply both sides by $2$:\n$$(\\sqrt{3} + 2)\\sin(3\\theta) = 5\\cos(3\\theta)$$"},{"type":"text","order":3,"title":"Form a tangent equation","content":"Divide both sides by $(\\sqrt{3} + 2)\\cos(3\\theta)$ to find an expression for $\\tan(3\\theta)$:\n$$\\tan(3\\theta) = \\frac{5}{2 + \\sqrt{3}}$$\n\nTo rationalize the denominator, we multiply the numerator and denominator by the conjugate $(2 - \\sqrt{3})$:\n$$\\tan(3\\theta) = \\frac{5(2 - \\sqrt{3})}{(2 + \\sqrt{3})(2 - \\sqrt{3})} = \\frac{10 - 5\\sqrt{3}}{4 - 3} = 10 - 5\\sqrt{3}$$\n\nThis confirms the key identity required by the markscheme:\n$$\\tan(3\\theta) = 10 - 5\\sqrt{3}$$"},{"type":"text","order":4,"title":"Find the interval for 3θ","content":"We need to solve for $0 \\le \\theta < 180^\\circ$.\n\nMultiplying the entire inequality by 3 gives the range for the compound angle $3\\theta$:\n$$0 \\le 3\\theta < 540^\\circ$$"},{"type":"text","order":5,"title":"Find the primary and secondary angles","content":"First, find the principal value using your calculator:\n$$3\\theta = \\arctan(10 - 5\\sqrt{3}) \\approx 53.26^\\circ$$\n\nSince the tangent function has a period of $180^\\circ$, we find all other solutions within our interval $0 \\le 3\\theta < 540^\\circ$ by repeatedly adding $180^\\circ$:\n- $3\\theta_1 = 53.26^\\circ$\n- $3\\theta_2 = 53.26^\\circ + 180^\\circ = 233.26^\\circ$\n- $3\\theta_3 = 233.26^\\circ + 180^\\circ = 413.26^\\circ$\n\nAdding another $180^\\circ$ yields $593.26^\\circ$, which is outside of our limit ($540^\\circ$)."},{"type":"text","order":6,"title":"Solve for θ","content":"Divide each angle by 3 to find the values of $\\theta$:\n- $\\theta_1 = \\frac{53.262^\\circ}{3} \\approx 17.8^\\circ$\n- $\\theta_2 = \\frac{233.262^\\circ}{3} \\approx 77.8^\\circ$\n- $\\theta_3 = \\frac{413.262^\\circ}{3} \\approx 137.8^\\circ$\n\nAll three solutions are rounded to one decimal place, which is standard as requested in the question.\n\n$$\\theta = 17.8^\\circ, 77.8^\\circ, 137.8^\\circ$$"}]}],"generatedAt":"2026-06-09T04:51:53.460Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"fb3a7872-1c73-4afc-b648-816d50d44ae6:algebraic_expansion","label":"(a) Algebraic Compound Angle Expansion","steps":[{"type":"text","order":0,"title":"Apply compound angle formulae","content":"To solve the given equation, we start by expanding both sides using the compound angle formulae (also known as the addition or subtraction formulae) from your trigonometry studies:\n\n$$\\sin(A - B) = \\sin A \\cos B - \\cos A \\sin B$$\n$$\\cos(A + B) = \\cos A \\cos B - \\sin A \\sin B$$\n\nApplying these to our equation:\n\n$$\\sqrt{6}\\sin(\\phi - 60^\\circ) = 2\\cos(\\phi + 45^\\circ)$$\n\nWe get:\n\n$$\\sqrt{6}\\left(\\sin \\phi \\cos 60^\\circ - \\cos \\phi \\sin 60^\\circ\\right) = 2\\left(\\cos \\phi \\cos 45^\\circ - \\sin \\phi \\sin 45^\\circ\\right)$$\n\nThis step earns the first method mark **(M1)**."},{"type":"text","order":1,"title":"Substitute exact values","content":"Next, we substitute the exact values for the trigonometric ratios of $60^\\circ$ and $45^\\circ$:\n\n* $\\cos 60^\\circ = \\frac{1}{2}$\n* $\\sin 60^\\circ = \\frac{\\sqrt{3}}{2}$\n* $\\cos 45^\\circ = \\frac{\\sqrt{2}}{2}$\n* $\\sin 45^\\circ = \\frac{\\sqrt{2}}{2}$\n\nSubstituting these exact values into our expanded equation **(M1)**:\n\n$$\\sqrt{6}\\left(\\frac{1}{2}\\sin \\phi - \\frac{\\sqrt{3}}{2}\\cos \\phi\\right) = 2\\left(\\frac{\\sqrt{2}}{2}\\cos \\phi - \\frac{\\sqrt{2}}{2}\\sin \\phi\\right)$$"},{"type":"text","order":2,"title":"Expand and simplify terms","content":"Now, we expand both sides of the equation to simplify the surds:\n\n$$\\frac{\\sqrt{6}}{2}\\sin \\phi - \\frac{\\sqrt{18}}{2}\\cos \\phi = \\sqrt{2}\\cos \\phi - \\sqrt{2}\\sin \\phi$$\n\nSince $\\sqrt{18} = \\sqrt{9 \\times 2} = 3\\sqrt{2}$, we can rewrite the equation as:\n\n$$\\frac{\\sqrt{6}}{2}\\sin \\phi - \\frac{3\\sqrt{2}}{2}\\cos \\phi = \\sqrt{2}\\cos \\phi - \\sqrt{2}\\sin \\phi$$\n\nTo make this simpler, let's divide every term on both sides by $\\sqrt{2}$ **(M1)**:\n\n$$\\frac{\\sqrt{3}}{2}\\sin \\phi - \\frac{3}{2}\\cos \\phi = \\cos \\phi - \\sin \\phi$$"},{"type":"text","order":3,"title":"Collect terms and solve","content":"Let's group all terms involving $\\sin \\phi$ on the left-hand side and all terms involving $\\cos \\phi$ on the right-hand side:\n\n$$\\frac{\\sqrt{3}}{2}\\sin \\phi + \\sin \\phi = \\cos \\phi + \\frac{3}{2}\\cos \\phi$$\n\nFactoring out $\\sin \\phi$ and $\\cos \\phi$, we get:\n\n$$\\left(\\frac{\\sqrt{3}}{2} + 1\\right)\\sin \\phi = \\frac{5}{2}\\cos \\phi$$\n\nMultiply the entire equation by $2$ to clear the denominators:\n\n$$(\\sqrt{3} + 2)\\sin \\phi = 5\\cos \\phi$$\n\nTo find $\\tan \\phi$, we divide both sides by $\\cos \\phi$ (since $\\tan \\phi = \\frac{\\sin \\phi}{\\cos \\phi}$) and then divide by $(\\sqrt{3} + 2)$ **(M1)**:\n\n$$\\tan \\phi = \\frac{5}{2 + \\sqrt{3}}$$\n\nThis shows the first required result!"},{"type":"text","order":4,"title":"Rationalise the denominator","content":"Finally, we rationalise the denominator by multiplying both the numerator and denominator by the conjugate $(2 - \\sqrt{3})$:\n\n$$\\tan \\phi = \\frac{5}{2 + \\sqrt{3}} \\times \\frac{2 - \\sqrt{3}}{2 - \\sqrt{3}}$$\n\n$$\\tan \\phi = \\frac{5(2 - \\sqrt{3})}{(2)^2 - (\\sqrt{3})^2}$$\n\n$$\\tan \\phi = \\frac{10 - 5\\sqrt{3}}{4 - 3}$$\n\n$$\\tan \\phi = 10 - 5\\sqrt{3}$$\n\nThis completes the proof and earns the final accuracy mark **(A1)**."},{"type":"text","order":5,"title":"Link Part 1 to Part 2","content":"For Part 2, we are asked to solve the equation:\n\n$$\\sqrt{6} \\sin(3\\theta - 60^\\circ) = 2 \\cos(3\\theta + 45^\\circ)$$\n\nNotice that this equation is identical in form to the one in Part 1, but with $\\phi$ replaced by $3\\theta$.\n\nTherefore, we can use the result from Part 1 directly **(B1)**:\n\n$$\\tan(3\\theta) = 10 - 5\\sqrt{3}$$"},{"type":"text","order":6,"title":"Find the principal value","content":"First, find the principal value for $3\\theta$ by taking the inverse tangent of both sides:\n\n$$3\\theta = \\arctan(10 - 5\\sqrt{3})$$\n\nUsing a calculator in degree mode **(M1)**:\n\n$$3\\theta \\approx 53.26^\\circ$$"},{"type":"text","order":7,"title":"Find other solutions","content":"We need to find all solutions for $\\theta$ in the interval $0 \\le \\theta < 180^\\circ$. Multiplying this interval by $3$, the interval for $3\\theta$ is:\n\n$$0 \\le 3\\theta < 540^\\circ$$\n\nSince the tangent function has a period of $180^\\circ$, we find other values for $3\\theta$ by repeatedly adding $180^\\circ$ to our principal value **(M1)**:\n\n$$\\begin{aligned} 3\\theta_1 &= 53.26^\\circ \\\\ 3\\theta_2 &= 53.26^\\circ + 180^\\circ = 233.26^\\circ \\\\ 3\\theta_3 &= 53.26^\\circ + 360^\\circ = 413.26^\\circ \\end{aligned}$$\n\nNote that adding $180^\\circ$ again would result in a value outside our range of $540^\\circ$."},{"type":"text","order":8,"title":"Solve for theta","content":"To find the final values of $\\theta$, we divide each of our values for $3\\theta$ by $3$:\n\n$$\\begin{aligned} \\theta_1 &= \\frac{53.26^\\circ}{3} \\approx 17.8^\\circ \\\\ \\theta_2 &= \\frac{233.26^\\circ}{3} \\approx 77.8^\\circ \\\\ \\theta_3 &= \\frac{413.26^\\circ}{3} \\approx 137.8^\\circ \\end{aligned}$$\n\nRounding to one decimal place as requested **(A1)**, we get:\n\n$$\\theta = 17.8^\\circ, \\; 77.8^\\circ, \\; 137.8^\\circ$$"}]},{"id":"b858090b-3bda-4fc3-bc66-f2c7a4e0e03e:compound_angle_expansion","label":"(b) Compound Angle Expansion","steps":[{"type":"text","order":0,"title":"Recall the expansion formulas","content":"To solve this equation using the **Compound Angle Expansion** approach, we begin by recalling the sine subtraction and cosine addition formulas from our trigonometric identities:\n\n$$\\sin(A - B) = \\sin A \\cos B - \\cos A \\sin B$$\n$$\\cos(A + B) = \\cos A \\cos B - \\sin A \\sin B$$\n\nLet's apply these to each side of our given equation:\n$$\\sqrt{6} \\sin(3\\theta - 60^\\circ) = 2 \\cos(3\\theta + 45^\\circ)$$"},{"type":"text","order":1,"title":"Expand both sides","content":"Let's expand each side using the standard values for $\\sin(60^\\circ)$, $\\cos(60^\\circ)$, $\\sin(45^\\circ)$, and $\\cos(45^\\circ)$:\n- $\\cos(60^\\circ) = \\frac{1}{2}$ and $\\sin(60^\\circ) = \\frac{\\sqrt{3}}{2}$\n- $\\cos(45^\\circ) = \\frac{\\sqrt{2}}{2}$ and $\\sin(45^\\circ) = \\frac{\\sqrt{2}}{2}$\n\nExpanding the Left-Hand Side (LHS):\n$$\\sqrt{6} \\sin(3\\theta - 60^\\circ) = \\sqrt{6} \\left[ \\sin(3\\theta)\\cos(60^\\circ) - \\cos(3\\theta)\\sin(60^\\circ) \\right] = \\sqrt{6} \\left[ \\frac{1}{2}\\sin(3\\theta) - \\frac{\\sqrt{3}}{2}\\cos(3\\theta) \\right]$$\n$$\\text{LHS} = \\frac{\\sqrt{6}}{2}\\sin(3\\theta) - \\frac{3\\sqrt{2}}{2}\\cos(3\\theta)$$\n\nExpanding the Right-Hand Side (RHS):\n$$2 \\cos(3\\theta + 45^\\circ) = 2 \\left[ \\cos(3\\theta)\\cos(45^\\circ) - \\sin(3\\theta)\\sin(45^\\circ) \\right] = 2 \\left[ \\frac{\\sqrt{2}}{2}\\cos(3\\theta) - \\frac{\\sqrt{2}}{2}\\sin(3\\theta) \\right]$$\n$$\\text{RHS} = \\sqrt{2}\\cos(3\\theta) - \\sqrt{2}\\sin(3\\theta)$$"},{"type":"text","order":2,"title":"Group like terms","content":"Equating the expanded LHS and RHS:\n$$\\frac{\\sqrt{6}}{2}\\sin(3\\theta) - \\frac{3\\sqrt{2}}{2}\\cos(3\\theta) = \\sqrt{2}\\cos(3\\theta) - \\sqrt{2}\\sin(3\\theta)$$\n\nTo simplify, let's divide both sides by $\\sqrt{2}$ (noting that $\\frac{\\sqrt{6}}{\\sqrt{2}} = \\sqrt{3}$):\n$$\\frac{\\sqrt{3}}{2}\\sin(3\\theta) - \\frac{3}{2}\\cos(3\\theta) = \\cos(3\\theta) - \\sin(3\\theta)$$\n\nNow, collect all terms with $\\sin(3\\theta)$ on the left and those with $\\cos(3\\theta)$ on the right:\n$$\\left( \\frac{\\sqrt{3}}{2} + 1 \\right)\\sin(3\\theta) = \\left( 1 + \\frac{3}{2} \\right)\\cos(3\\theta)$$\n$$\\left( \\frac{\\sqrt{3} + 2}{2} \\right)\\sin(3\\theta) = \\frac{5}{2}\\cos(3\\theta)$$\n\nMultiply both sides by $2$:\n$$(\\sqrt{3} + 2)\\sin(3\\theta) = 5\\cos(3\\theta)$$"},{"type":"text","order":3,"title":"Form a tangent equation","content":"Divide both sides by $(\\sqrt{3} + 2)\\cos(3\\theta)$ to find an expression for $\\tan(3\\theta)$:\n$$\\tan(3\\theta) = \\frac{5}{2 + \\sqrt{3}}$$\n\nTo rationalize the denominator, we multiply the numerator and denominator by the conjugate $(2 - \\sqrt{3})$:\n$$\\tan(3\\theta) = \\frac{5(2 - \\sqrt{3})}{(2 + \\sqrt{3})(2 - \\sqrt{3})} = \\frac{10 - 5\\sqrt{3}}{4 - 3} = 10 - 5\\sqrt{3}$$\n\nThis confirms the key identity required by the markscheme:\n$$\\tan(3\\theta) = 10 - 5\\sqrt{3}$$"},{"type":"text","order":4,"title":"Find the interval for 3θ","content":"We need to solve for $0 \\le \\theta < 180^\\circ$.\n\nMultiplying the entire inequality by 3 gives the range for the compound angle $3\\theta$:\n$$0 \\le 3\\theta < 540^\\circ$$"},{"type":"text","order":5,"title":"Find the primary and secondary angles","content":"First, find the principal value using your calculator:\n$$3\\theta = \\arctan(10 - 5\\sqrt{3}) \\approx 53.26^\\circ$$\n\nSince the tangent function has a period of $180^\\circ$, we find all other solutions within our interval $0 \\le 3\\theta < 540^\\circ$ by repeatedly adding $180^\\circ$:\n- $3\\theta_1 = 53.26^\\circ$\n- $3\\theta_2 = 53.26^\\circ + 180^\\circ = 233.26^\\circ$\n- $3\\theta_3 = 233.26^\\circ + 180^\\circ = 413.26^\\circ$\n\nAdding another $180^\\circ$ yields $593.26^\\circ$, which is outside of our limit ($540^\\circ$)."},{"type":"text","order":6,"title":"Solve for θ","content":"Divide each angle by 3 to find the values of $\\theta$:\n- $\\theta_1 = \\frac{53.262^\\circ}{3} \\approx 17.8^\\circ$\n- $\\theta_2 = \\frac{233.262^\\circ}{3} \\approx 77.8^\\circ$\n- $\\theta_3 = \\frac{413.262^\\circ}{3} \\approx 137.8^\\circ$\n\nAll three solutions are rounded to one decimal place, which is standard as requested in the question.\n\n$$\\theta = 17.8^\\circ, 77.8^\\circ, 137.8^\\circ$$"}]}],"generatedAt":"2026-06-09T04:52:07.597Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"331a938b-350b-447e-8049-d0431335442d","revisionId":"0a59f189-c2ff-4efd-995a-b40c59c52609","questionNumber":null,"questionType":"LA","specification":"Solve, for $0° \\leq \\theta < 360°$, the equation","difficulty":null,"difficultyBand":"medium","tier":null,"totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"7067c788-59e7-4c0e-abb0-d3306f1fbd9c","label":null,"content":"$$\\sin(\\theta - 45°) = \\cos(\\theta + 30°)$","markscheme":"- Uses $\\cos X \\equiv \\sin(90° - X)$: $\\cos(\\theta + 30°) = \\sin(60° - \\theta)$ M1\n- $\\sin(\\theta - 45°) = \\sin(60° - \\theta)$ A1\n- $\\theta - 45° = 60° - \\theta + 360n$ M1\n- $2\\theta = 105° + 360n$ A1\n- $\\theta = 52.5°$ A1\n- $\\theta = 232.5°$ A1\n\nThe alternative branch $\\theta - 45° = 180° - (60° - \\theta)$ gives $-45° = 120°$, which has no solution, so there are exactly two values. Expanding both sides with the compound-angle formulae and forming $\\tan\\theta = \\frac{\\sqrt{3} + \\sqrt{2}}{\\sqrt{2} + 1}$ is an equally valid but longer route; award the marks for a correct method.","marks":6,"order":0,"markingRubric":null,"tableImages":[],"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"38c05923-2889-40f5-996b-faab30fc7f8b","revisionId":"978c1da1-428c-44cc-825e-96d442599bd2","questionNumber":null,"questionType":"LA","specification":"The angular displacement, $\\phi$, of a high-precision robotic arm relative to its equilibrium position is modelled by the equilibrium state equation:\n\n\\[11 \\sec \\phi = 32 \\sin \\phi\\]","difficulty":3,"difficultyBand":"medium","tier":null,"totalMarks":7,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"82da30a0-202b-4412-901f-1b12ce93e798","label":"a","content":"Show that this equation can be expressed in the form $\\sin 2\\phi = k$, where $k$ is a constant to be determined.","markscheme":"- Uses the identity $\\sec \\phi = \\frac{1}{\\cos \\phi}$ to rewrite the equation as $11 = 32 \\sin \\phi \\cos \\phi$ M1\n- Recognises that the right hand side can be written in terms of the double angle identity $\\sin 2\\phi = 2 \\sin \\phi \\cos \\phi$ M1\n- Substitutes $32 \\sin \\phi \\cos \\phi = 16(2 \\sin \\phi \\cos \\phi) = 16 \\sin 2\\phi$ M1\n- Rearranges to find $\\sin 2\\phi = \\frac{11}{16}$ or $\\sin 2\\phi = 0.6875$ A1","marks":4,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"double_angle_sine","label":"Double-Angle Sine Identity","steps":[{"type":"text","order":0,"title":"Use the reciprocal identity","content":"To begin, we recall the reciprocal trigonometric identity for the secant function:\n\n$$\\sec \\phi = \\frac{1}{\\cos \\phi}$$\n\nSubstituting this into the given equilibrium state equation, we obtain:\n\n$$11 \\left(\\frac{1}{\\cos \\phi}\\right) = 32 \\sin \\phi$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$11 \\left(\\frac{1}{\\htmlData{anchor=cos-left}{\\color{@sky}{\\cos \\phi}}}\\right) = 32 \\sin \\phi$"},{"latex":"$$11 = 32 \\sin \\phi \\htmlData{anchor=cos-right}{\\color{@sky}{\\cos \\phi}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"cos-left"},{"color":"sky","style":"text-color","anchor":"cos-right"}],"connections":[{"to":"cos-right","from":"cos-left","color":"sky","label":"multiply both sides by cos φ","style":"solid"}]},"order":1,"title":"Rearrange the equation","description":"We multiply both sides of the equation by $\\cos \\phi$ to clear the fraction, which groups the sine and cosine terms together."},{"type":"text","order":2,"title":"Apply double-angle identity","content":"Next, recall the double-angle identity for sine:\n\n$$\\sin 2\\phi = 2 \\sin \\phi \\cos \\phi$$\n\nWe can rewrite the right-hand side of our equation, $32 \\sin \\phi \\cos \\phi$, to feature this identity directly by factoring out $16$:\n\n$$11 = 16(2 \\sin \\phi \\cos \\phi)$$\n\nNow we substitute $\\sin 2\\phi$ for $2 \\sin \\phi \\cos \\phi$:\n\n$$11 = 16 \\sin 2\\phi$$"},{"type":"text","order":3,"title":"Solve for sin 2φ","content":"Finally, isolate $\\sin 2\\phi$ by dividing both sides of the equation by $16$:\n\n$$\\sin 2\\phi = \\frac{11}{16}$$\n\nIn decimal form, this is:\n\n$$\\sin 2\\phi = 0.6875$$\n\nThus, the equation is in the form $\\sin 2\\phi = k$, where $k = \\frac{11}{16} = 0.6875$."},{"type":"text","order":4,"title":"Find principal value","content":"To solve for $\\phi$ in Part 2, we first find the principal value for $2\\phi$ by taking the inverse sine of both sides:\n\n$$2\\phi = \\arcsin(0.6875)$$\n\nUsing a calculator in degree mode:\n\n$$2\\phi \\approx 43.4325^{\\circ}$$"},{"type":"text","order":5,"title":"Determine smallest positive phi","content":"Divide by $2$ to find the smallest positive value of $\\phi$:\n\n$$\\phi \\approx \\frac{43.4325^{\\circ}}{2} \\approx 21.716^{\\circ}$$\n\nRounding to one decimal place, we get:\n\n$$\\phi = 21.7^{\\circ}$$"}]},{"id":"tangent_identities","label":"Tangent Identity Substitution","steps":[{"type":"text","order":0,"title":"Multiply by secant","content":"To solve this using tangent substitution, let's start with our equilibrium equation:\n\n$$11 \\sec \\phi = 32 \\sin \\phi$$\n\nMultiply both sides of the equation by $\\sec \\phi$ to introduce the $\\sec^2 \\phi$ term:\n\n$$11 \\sec^2 \\phi = 32 \\sin \\phi \\sec \\phi$$"},{"type":"text","order":1,"title":"Use tangent identity","content":"Recall that $\\sec \\phi = \\frac{1}{\\cos \\phi}$, so we can rewrite the product on the right-hand side as a tangent:\n\n$$\\sin \\phi \\sec \\phi = \\frac{\\sin \\phi}{\\cos \\phi} = \\tan \\phi$$\n\nSubstituting this back into our equation gives:\n\n$$11 \\sec^2 \\phi = 32 \\tan \\phi$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=sec-id}{\\sec^2 \\phi} = \\htmlData{anchor=tan-id}{1 + \\tan^2 \\phi}$"},{"latex":"$$11(\\htmlData{anchor=subbed-tan}{1 + \\tan^2 \\phi}) = 32 \\tan \\phi$"}],"highlights":[{"color":"sky","style":"underline","anchor":"sec-id"},{"color":"sky","style":"box","anchor":"tan-id"}],"connections":[{"to":"subbed-tan","from":"tan-id","color":"sky","label":"substitute","style":"solid"}]},"order":2,"title":"Substitute secant squared","description":"Next, we use the fundamental trigonometric identity $\\sec^2 \\phi = 1 + \\tan^2 \\phi$ to write the entire equation in terms of $\\tan \\phi$."},{"type":"text","order":3,"title":"Rearrange the equation","content":"Now, let's rearrange our equation to isolate a rational expression in terms of $\\tan \\phi$:\n\n$$\\frac{11}{32} = \\frac{\\tan \\phi}{1 + \\tan^2 \\phi}$$"},{"type":"hyper_latex","block":{"nodes":[{"text":"The double-angle identity for sine in terms of tangent is defined as:"},{"latex":"$$\\htmlData{anchor=double-sin}{\\sin 2\\phi} = \\htmlData{anchor=double-tan}{\\frac{2 \\tan \\phi}{1 + \\tan^2 \\phi}}$"},{"text":"Multiplying our rearranged equation by 2 gives:"},{"latex":"$$2 \\left(\\frac{11}{32}\\right) = \\htmlData{anchor=target-expr}{\\frac{2 \\tan \\phi}{1 + \\tan^2 \\phi}}$"}],"highlights":[{"color":"purple","style":"box","anchor":"double-sin"}],"connections":[{"to":"target-expr","from":"double-tan","color":"purple","label":"match","style":"dashed"}]},"order":4,"title":"Apply double-angle identity","description":"We can relate this directly to the double-angle identity for sine."},{"type":"text","order":5,"title":"State the final answer","content":"Simplifying the fraction on the left-hand side, we get:\n\n$$\\sin 2\\phi = \\frac{11}{16} = 0.6875$$\n\nThus, $k = \\frac{11}{16} = 0.6875$."},{"type":"text","order":6,"title":"Solve for smallest positive phi","content":"Now we find the smallest positive value of $\\phi$ in degrees as required by Part 2:\n\n$$2\\phi = \\arcsin(0.6875) \\approx 43.4325^{\\circ}$$\n\nDividing by $2$:\n\n$$\\phi \\approx 21.716^{\\circ}$$\n\nTo one decimal place, this is $\\phi = 21.7^{\\circ}$."}]}],"generatedAt":"2026-06-09T04:18:13.119Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"62f74a6c-eb18-4f6e-927a-1360d312fff5","label":"b","content":"Hence, find the smallest positive value of $\\phi$, in degrees, that satisfies this equilibrium state. Give your answer to one decimal place.","markscheme":"- Calculates the principal value $2\\phi = \\arcsin(0.6875) \\approx 43.432...^\\circ$ M1\n- Divides by $2$ to find the smallest positive solution $\\phi \\approx 21.716...^\\circ$ M1\n- States $\\phi = 21.7^\\circ$ to one decimal place A1","marks":3,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"double_angle_identity","label":"Double-Angle Identity","steps":[{"type":"text","order":0,"title":"Rewrite secant using cosine","content":"To begin solving the equilibrium equation, we can rewrite the secant function using its reciprocal identity. Recall that:\n\n$$\\sec \\phi = \\frac{1}{\\cos \\phi}$$\n\nSubstituting this into the equation gives us:\n\n$$11 \\left(\\frac{1}{\\cos \\phi}\\right) = 32 \\sin \\phi$$","highlights":[{"text":"11 \\sec \\phi = 32 \\sin \\phi","location":"specification"}]},{"type":"text","order":1,"title":"Rearrange the equation","content":"Next, multiply both sides of the equation by $\\cos \\phi$ to remove the fraction. This gives us a product of $\\sin \\phi$ and $\\cos \\phi$:\n\n$$11 = 32 \\sin \\phi \\cos \\phi$$"},{"type":"text","order":2,"title":"Apply double-angle identity","content":"Now we recognize a double-angle identity. From the formula booklet, the double-angle identity for sine is:\n\n$$\\sin(2\\phi) = 2 \\sin \\phi \\cos \\phi$$\n\nWe can rewrite the right-hand side of our equation, $32 \\sin \\phi \\cos \\phi$, to use this identity:\n\n$$32 \\sin \\phi \\cos \\phi = 16 (2 \\sin \\phi \\cos \\phi) = 16 \\sin(2\\phi)$$\n\nSubstituting this back into our equation yields:\n\n$$11 = 16 \\sin(2\\phi)$$"},{"type":"text","order":3,"title":"Isolate the sine term","content":"To solve for $\\phi$, isolate the sine expression by dividing both sides by $16$:\n\n$$\\sin(2\\phi) = \\frac{11}{16}$$\n\n$$\\sin(2\\phi) = 0.6875$$"},{"type":"text","order":4,"title":"Find the principal value","content":"Take the inverse sine ($\\arcsin$) of both sides to find the principal angle in degrees:\n\n$$2\\phi = \\arcsin(0.6875)$$\n\n$$2\\phi \\approx 43.4325^\\circ$$\n\n*Note: This earns the first method mark M1.*","calculator":[{"gdc":{"expression":"\\sin^{-1}(0.6875)","instructions":"Set angle units to Degrees. Enter asin(0.6875) in the run menu."},"ti84":{"expression":"\\sin^{-1}(0.6875)","instructions":"Press [MODE] and ensure 'DEGREE' is selected. Press [2nd] [SIN] 0.6875 [ENTER]."},"desmos":{"expression":"\\arcsin(0.6875)","instructions":"Click the wrench icon to switch to degree mode. Type 'arcsin(0.6875)' or 'sin^{-1}(0.6875)'."},"description":"Calculate the inverse sine of 0.6875 in degree mode."}]},{"type":"text","order":5,"title":"Solve for phi","content":"Divide by $2$ to calculate the value of $\\phi$:\n\n$$\\phi \\approx \\frac{43.4325^\\circ}{2} \\approx 21.716...^\\circ$$\n\n*Note: Finding this value earns the second method mark M1.*"},{"type":"text","order":6,"title":"State the final answer","content":"Rounding to one decimal place as requested, we obtain:\n\n$$\\phi = 21.7^\\circ$$\n\n*Note: Stating the correct final rounded answer earns the accuracy mark A1.*","highlights":[{"text":"Give your answer to one decimal place.","location":"specification"}]}]},{"id":"graphical_calculator","label":"Using Graphing Calculator","steps":[{"type":"text","order":0,"title":"Formulate the equations","content":"To find the smallest positive value of $\\phi$ that satisfies the equilibrium state, we can find where the graphs of the functions on the left-hand side and right-hand side of the equation intersect.\n\nLet's define the two functions in terms of $x$ (representing $\\phi$):\n\n$$y_1 = 11 \\sec(x) = \\frac{11}{\\cos(x)}$$\n$$y_2 = 32 \\sin(x)$$\n\nWe need to find the $x$-coordinate of the first intersection point for $x > 0$ when the calculator is set to **Degree** mode.","highlights":[{"text":"11 \\sec \\phi = 32 \\sin \\phi","location":"specification"}]},{"type":"text","order":1,"title":"Configure the calculator","content":"Let's enter these two functions into your graphing calculator and search for their intersection.","calculator":[{"gdc":{"expression":"f1(x) = 11 / cos(x), f2(x) = 32 sin(x)","instructions":"1. Ensure your calculator is set to **Degree** mode.\n2. Open the graph application.\n3. Enter the first relation as \$y1 = 11 / \\cos(x)\$.\n4. Enter the second relation as \$y2 = 32 \\sin(x)\$.\n5. Adjust the window settings so you can see the interval \$0 < x < 90\$.\n6. Use the **Intersection** tool (usually under Graph Analysis) to find the coordinate where the curves meet."},"ti84":{"expression":"Y_1 = 11 / \\cos(X), Y_2 = 32 \\sin(X)","instructions":"1. Press **MODE** and ensure **DEGREE** is selected.\n2. Press **Y=**.\n3. Enter \$Y_1 = 11 / \\cos(X)\$.\n4. Enter \$Y_2 = 32 \\sin(X)\$.\n5. Press **WINDOW** and set \$X_{\\min} = 0\$, \$X_{\\max} = 90\$, \$Y_{\\min} = 0\$, and \$Y_{\\max} = 40\$.\n6. Press **GRAPH**.\n7. Press **2nd** then **TRACE** (CALC) and select **5: intersect**.\n8. Press **ENTER** for the First curve, Second curve, and Guess."},"desmos":{"expression":"y = 11/cos(x), y = 32*sin(x)","instructions":"1. Click the wrench icon (Settings) in the top-right and select **Degrees**.\n2. Type `y = 11 / cos(x)` in the first expression bar.\n3. Type `y = 32 * sin(x)` in the second expression bar.\n4. Click on the first intersection point that occurs to the right of the y-axis (where \$x > 0\$) to display its coordinates."},"description":"Graph the two equations and find their first positive intersection."}]},{"type":"text","order":2,"title":"Find the intersection point","content":"From the calculator graph, we observe the first intersection point in the first quadrant where $x > 0$:\n\n* **$x$-coordinate:** $\\phi \\approx 21.716...^{\\circ}$\n* **$y$-coordinate:** $y \\approx 11.838...$\n\nRounding the $x$-value (which represents $\\phi$) to one decimal place, we obtain:\n\n$$\\phi \\approx 21.7^{\\circ}$$","highlights":[{"text":"smallest positive value of \\phi","location":"specification"},{"text":"one decimal place","location":"specification"}]}]}],"generatedAt":"2026-06-11T03:16:35.907Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"82da30a0-202b-4412-901f-1b12ce93e798:double_angle_sine","label":"(a) Double-Angle Sine Identity","steps":[{"type":"text","order":0,"title":"Use the reciprocal identity","content":"To begin, we recall the reciprocal trigonometric identity for the secant function:\n\n$$\\sec \\phi = \\frac{1}{\\cos \\phi}$$\n\nSubstituting this into the given equilibrium state equation, we obtain:\n\n$$11 \\left(\\frac{1}{\\cos \\phi}\\right) = 32 \\sin \\phi$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$11 \\left(\\frac{1}{\\htmlData{anchor=cos-\\left}{\\color{@sky}{\\cos \\phi}}}\\right) = 32 \\sin \\phi$"},{"latex":"$$11 = 32 \\sin \\phi \\htmlData{anchor=cos-\\right}{\\color{@sky}{\\cos \\phi}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"cos-left"},{"color":"sky","style":"text-color","anchor":"cos-right"}],"connections":[{"to":"cos-right","from":"cos-left","color":"sky","label":"multiply both sides by cos φ","style":"solid"}]},"order":1,"title":"Rearrange the equation","description":"We multiply both sides of the equation by $\\cos \\phi$ to clear the fraction, which groups the sine and cosine terms together."},{"type":"text","order":2,"title":"Apply double-angle identity","content":"Next, recall the double-angle identity for sine:\n\n$$\\sin 2\\phi = 2 \\sin \\phi \\cos \\phi$$\n\nWe can rewrite the right-hand side of our equation, $32 \\sin \\phi \\cos \\phi$, to feature this identity directly by factoring out $16$:\n\n$$11 = 16(2 \\sin \\phi \\cos \\phi)$$\n\nNow we substitute $\\sin 2\\phi$ for $2 \\sin \\phi \\cos \\phi$:\n\n$$11 = 16 \\sin 2\\phi$$"},{"type":"text","order":3,"title":"Solve for sin 2φ","content":"Finally, isolate $\\sin 2\\phi$ by dividing both sides of the equation by $16$:\n\n$$\\sin 2\\phi = \\frac{11}{16}$$\n\nIn decimal form, this is:\n\n$$\\sin 2\\phi = 0.6875$$\n\nThus, the equation is in the form $\\sin 2\\phi = k$, where $k = \\frac{11}{16} = 0.6875$."},{"type":"text","order":4,"title":"Find principal value","content":"To solve for $\\phi$ in Part 2, we first find the principal value for $2\\phi$ by taking the inverse sine of both sides:\n\n$$2\\phi = \\arcsin(0.6875)$$\n\nUsing a calculator in degree mode:\n\n$$2\\phi \\approx 43.4325^{\\circ}$$"},{"type":"text","order":5,"title":"Determine smallest positive phi","content":"Divide by $2$ to find the smallest positive value of $\\phi$:\n\n$$\\phi \\approx \\frac{43.4325^{\\circ}}{2} \\approx 21.716^{\\circ}$$\n\nRounding to one decimal place, we get:\n\n$$\\phi = 21.7^{\\circ}$$"}]},{"id":"82da30a0-202b-4412-901f-1b12ce93e798:tangent_identities","label":"(a) Tangent Identity Substitution","steps":[{"type":"text","order":0,"title":"Multiply by secant","content":"To solve this using tangent substitution, let's start with our equilibrium equation:\n\n$$11 \\sec \\phi = 32 \\sin \\phi$$\n\nMultiply both sides of the equation by $\\sec \\phi$ to introduce the $\\sec^2 \\phi$ term:\n\n$$11 \\sec^2 \\phi = 32 \\sin \\phi \\sec \\phi$$"},{"type":"text","order":1,"title":"Use tangent identity","content":"Recall that $\\sec \\phi = \\frac{1}{\\cos \\phi}$, so we can rewrite the product on the right-hand side as a tangent:\n\n$$\\sin \\phi \\sec \\phi = \\frac{\\sin \\phi}{\\cos \\phi} = \\tan \\phi$$\n\nSubstituting this back into our equation gives:\n\n$$11 \\sec^2 \\phi = 32 \\tan \\phi$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=sec-id}{\\sec^2 \\phi} = \\htmlData{anchor=tan-id}{1 + \\tan^2 \\phi}$"},{"latex":"$$11(\\htmlData{anchor=subbed-tan}{1 + \\tan^2 \\phi}) = 32 \\tan \\phi$"}],"highlights":[{"color":"sky","style":"underline","anchor":"sec-id"},{"color":"sky","style":"box","anchor":"tan-id"}],"connections":[{"to":"subbed-tan","from":"tan-id","color":"sky","label":"substitute","style":"solid"}]},"order":2,"title":"Substitute secant squared","description":"Next, we use the fundamental trigonometric identity $\\sec^2 \\phi = 1 + \\tan^2 \\phi$ to write the entire equation in terms of $\\tan \\phi$."},{"type":"text","order":3,"title":"Rearrange the equation","content":"Now, let's rearrange our equation to isolate a rational expression in terms of $\\tan \\phi$:\n\n$$\\frac{11}{32} = \\frac{\\tan \\phi}{1 + \\tan^2 \\phi}$$"},{"type":"hyper_latex","block":{"nodes":[{"text":"The double-angle identity for sine in terms of tangent is defined as:"},{"latex":"$$\\htmlData{anchor=double-sin}{\\sin 2\\phi} = \\htmlData{anchor=double-tan}{\\frac{2 \\tan \\phi}{1 + \\tan^2 \\phi}}$"},{"text":"Multiplying our rearranged equation by 2 gives:"},{"latex":"$$2 \\left(\\frac{11}{32}\\right) = \\htmlData{anchor=target-expr}{\\frac{2 \\tan \\phi}{1 + \\tan^2 \\phi}}$"}],"highlights":[{"color":"purple","style":"box","anchor":"double-sin"}],"connections":[{"to":"target-expr","from":"double-tan","color":"purple","label":"match","style":"dashed"}]},"order":4,"title":"Apply double-angle identity","description":"We can relate this directly to the double-angle identity for sine."},{"type":"text","order":5,"title":"State the final answer","content":"Simplifying the fraction on the left-hand side, we get:\n\n$$\\sin 2\\phi = \\frac{11}{16} = 0.6875$$\n\nThus, $k = \\frac{11}{16} = 0.6875$."},{"type":"text","order":6,"title":"Solve for smallest positive phi","content":"Now we find the smallest positive value of $\\phi$ in degrees as required by Part 2:\n\n$$2\\phi = \\arcsin(0.6875) \\approx 43.4325^{\\circ}$$\n\nDividing by $2$:\n\n$$\\phi \\approx 21.716^{\\circ}$$\n\nTo one decimal place, this is $\\phi = 21.7^{\\circ}$."}]},{"id":"62f74a6c-eb18-4f6e-927a-1360d312fff5:double_angle_identity","label":"(b) Double-Angle Identity","steps":[{"type":"text","order":0,"title":"Rewrite secant using cosine","content":"To begin solving the equilibrium equation, we can rewrite the secant function using its reciprocal identity. Recall that:\n\n$$\\sec \\phi = \\frac{1}{\\cos \\phi}$$\n\nSubstituting this into the equation gives us:\n\n$$11 \\left(\\frac{1}{\\cos \\phi}\\right) = 32 \\sin \\phi$$","highlights":[{"text":"11 \\sec \\phi = 32 \\sin \\phi","location":"specification"}]},{"type":"text","order":1,"title":"Rearrange the equation","content":"Next, multiply both sides of the equation by $\\cos \\phi$ to remove the fraction. This gives us a product of $\\sin \\phi$ and $\\cos \\phi$:\n\n$$11 = 32 \\sin \\phi \\cos \\phi$$"},{"type":"text","order":2,"title":"Apply double-angle identity","content":"Now we recognize a double-angle identity. From the formula booklet, the double-angle identity for sine is:\n\n$$\\sin(2\\phi) = 2 \\sin \\phi \\cos \\phi$$\n\nWe can rewrite the right-hand side of our equation, $32 \\sin \\phi \\cos \\phi$, to use this identity:\n\n$$32 \\sin \\phi \\cos \\phi = 16 (2 \\sin \\phi \\cos \\phi) = 16 \\sin(2\\phi)$$\n\nSubstituting this back into our equation yields:\n\n$$11 = 16 \\sin(2\\phi)$$"},{"type":"text","order":3,"title":"Isolate the sine term","content":"To solve for $\\phi$, isolate the sine expression by dividing both sides by $16$:\n\n$$\\sin(2\\phi) = \\frac{11}{16}$$\n\n$$\\sin(2\\phi) = 0.6875$$"},{"type":"text","order":4,"title":"Find the principal value","content":"Take the inverse sine ($\\arcsin$) of both sides to find the principal angle in degrees:\n\n$$2\\phi = \\arcsin(0.6875)$$\n\n$$2\\phi \\approx 43.4325^\\circ$$\n\n*Note: This earns the first method mark M1.*","calculator":[{"gdc":{"expression":"\\sin^{-1}(0.6875)","instructions":"Set angle units to Degrees. Enter asin(0.6875) in the run menu."},"ti84":{"expression":"\\sin^{-1}(0.6875)","instructions":"Press [MODE] and ensure 'DEGREE' is selected. Press [2nd] [SIN] 0.6875 [ENTER]."},"desmos":{"expression":"\\arcsin(0.6875)","instructions":"Click the wrench icon to switch to degree mode. Type 'arcsin(0.6875)' or 'sin^{-1}(0.6875)'."},"description":"Calculate the inverse sine of 0.6875 in degree mode."}]},{"type":"text","order":5,"title":"Solve for phi","content":"Divide by $2$ to calculate the value of $\\phi$:\n\n$$\\phi \\approx \\frac{43.4325^\\circ}{2} \\approx 21.716...^\\circ$$\n\n*Note: Finding this value earns the second method mark M1.*"},{"type":"text","order":6,"title":"State the final answer","content":"Rounding to one decimal place as requested, we obtain:\n\n$$\\phi = 21.7^\\circ$$\n\n*Note: Stating the correct final rounded answer earns the accuracy mark A1.*","highlights":[{"text":"Give your answer to one decimal place.","location":"specification"}]}]},{"id":"62f74a6c-eb18-4f6e-927a-1360d312fff5:graphical_calculator","label":"(b) Using Graphing Calculator","steps":[{"type":"text","order":0,"title":"Formulate the equations","content":"To find the smallest positive value of $\\phi$ that satisfies the equilibrium state, we can find where the graphs of the functions on the left-hand side and right-hand side of the equation intersect.\n\nLet's define the two functions in terms of $x$ (representing $\\phi$):\n\n$$y_1 = 11 \\sec(x) = \\frac{11}{\\cos(x)}$$\n$$y_2 = 32 \\sin(x)$$\n\nWe need to find the $x$-coordinate of the first intersection point for $x > 0$ when the calculator is set to **Degree** mode.","highlights":[{"text":"11 \\sec \\phi = 32 \\sin \\phi","location":"specification"}]},{"type":"text","order":1,"title":"Configure the calculator","content":"Let's enter these two functions into your graphing calculator and search for their intersection.","calculator":[{"gdc":{"expression":"f1(x) = 11 / cos(x), f2(x) = 32 sin(x)","instructions":"1. Ensure your calculator is set to **Degree** mode.\n2. Open the graph application.\n3. Enter the first relation as \$y1 = 11 / \\cos(x)\$.\n4. Enter the second relation as \$y2 = 32 \\sin(x)\$.\n5. Adjust the window settings so you can see the interval \$0 < x < 90\$.\n6. Use the **Intersection** tool (usually under Graph Analysis) to find the coordinate where the curves meet."},"ti84":{"expression":"Y_1 = 11 / \\cos(X), Y_2 = 32 \\sin(X)","instructions":"1. Press **MODE** and ensure **DEGREE** is selected.\n2. Press **Y=**.\n3. Enter \$Y_1 = 11 / \\cos(X)\$.\n4. Enter \$Y_2 = 32 \\sin(X)\$.\n5. Press **WINDOW** and set \$X_{\\min} = 0\$, \$X_{\\max} = 90\$, \$Y_{\\min} = 0\$, and \$Y_{\\max} = 40\$.\n6. Press **GRAPH**.\n7. Press **2nd** then **TRACE** (CALC) and select **5: intersect**.\n8. Press **ENTER** for the First curve, Second curve, and Guess."},"desmos":{"expression":"y = 11/cos(x), y = 32*sin(x)","instructions":"1. Click the wrench icon (Settings) in the top-right and select **Degrees**.\n2. Type `y = 11 / cos(x)` in the first expression bar.\n3. Type `y = 32 * sin(x)` in the second expression bar.\n4. Click on the first intersection point that occurs to the right of the y-axis (where \$x > 0\$) to display its coordinates."},"description":"Graph the two equations and find their first positive intersection."}]},{"type":"text","order":2,"title":"Find the intersection point","content":"From the calculator graph, we observe the first intersection point in the first quadrant where $x > 0$:\n\n* **$x$-coordinate:** $\\phi \\approx 21.716...^{\\circ}$\n* **$y$-coordinate:** $y \\approx 11.838...$\n\nRounding the $x$-value (which represents $\\phi$) to one decimal place, we obtain:\n\n$$\\phi \\approx 21.7^{\\circ}$$","highlights":[{"text":"smallest positive value of \\phi","location":"specification"},{"text":"one decimal place","location":"specification"}]}]}],"generatedAt":"2026-06-11T03:16:36.111Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths__ccea","summaries":[{"id":"65b33456-ebd4-4af9-b7b3-c47294de42db","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"94057f61-abbc-4527-807f-573c6d2bb3ef","difficultyBand":"easy","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"e8ba8a76-b67a-41d7-97e1-02d637d03671","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"0aec2d07-eeb2-4854-99e5-78c5846bc99a","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"11a10575-0a46-4902-8659-9ef5676fb790","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"1c3539bd-2b7b-42e2-96cd-ad5ecc1a9b85","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"3155a703-639c-42e9-a2a0-065a726bb8c6","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"47a37e7b-a87b-4583-8b17-28f02146334b","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"51aea9f2-3e1a-41e2-b43a-1b5737a1c0fc","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"71c4bd57-dd52-4e28-94ac-f95b2dc4dc4a","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"938f7bbf-f10d-4cd5-a0b0-f8e7ed8cfd61","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"953a2097-8503-4a13-8b3c-315e5e488039","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"e016be92-1d7c-4924-b219-abd4a2c0b1d2","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"e0cd8580-436f-4f08-8dee-284b953c607a","difficultyBand":"easy","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"f2b54dcf-595e-401a-b92f-9a3bba892d0d","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"f4923309-8d0b-4a75-bdf8-bf6885937510","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"f72cfefd-990a-475c-8fde-0ec350de12d9","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"04df7d63-f5cb-4485-9727-ce5baaaf8bdc","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"acd6f9f4-de00-431c-a1f3-5c3fddbc8625","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"b9b10d51-18c3-4a4b-9a9b-6f4296660855","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"009035ab-b95e-4a9f-9087-d0a651547233","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"048c3a3e-e04e-4be6-b422-8029fdb4fa0c","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"0721a4ee-e4ab-467f-97e9-5fda7c60e630","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"0e1af8ff-3856-482d-bf8c-2e012d25c122","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"14274a7a-233a-4bf1-bc53-389b53840114","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"327076de-b61b-400a-bde2-937a30472d46","difficultyBand":"easy","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"4ed7094e-cf03-4e1e-8691-1d3d0175a970","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"719c5525-de5d-44ce-b58b-8c4a302df638","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"78a118f7-2344-4c79-b026-e5f3ada312c2","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"8714d6b0-9b11-45c2-b4a2-7e1eab0cf698","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"a2f875e1-470c-47b6-a47c-8dd83e0d38e0","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"c6241b26-4040-4dd8-8d9e-7a596676f9b1","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculat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Trigonometry (A-level only)","topicId":"97cb7c2e-f32f-45ff-ada1-651a91e4cf22","topicSidebar":"$L3c","topicSlug":"trigonometry-8","topicTitle":"3.4 Trigonometry (A-level only)"}]}],"$L3d","$L3e"]}]

3.4 Trigonometry (A-level only)

Question 72

3.4 Trigonometry (A-level only) Questions

3.4 Trigonometry (A-level only)

Question 72

3.4 Trigonometry (A-level only) Questions