Further trigonometry – Year 13 A level maths test answers to download as PDF

← Back to the test

1. Answer to Question 1 (4 marks)

   Method: the small-angle approximations only work in radians: \(\sin\theta\approx\theta\), \(\tan\theta\approx\theta\), \(\cos\theta\approx1-\dfrac{\theta^2}{2}\). Replace each function, simplify, and the powers of \(\theta\) usually cancel. Mark scheme: 1 mark for each approximation in part a), 1 mark for the value and 1 mark for the percentage in part b).

   1. \(\theta\tan2\theta\approx2\theta^2\) and \(1-\cos3\theta\approx\dfrac{9\theta^2}2\), so the expression \(\approx\dfrac{2\theta^2}{9\theta^2/2}=\dfrac49\).
   2. \(\dfrac{0.1\tan0.2}{1-\cos0.3}=\dfrac{0.020271}{0.044664}=0.4539\). Difference from \(\tfrac49=0.4444\): \(\dfrac{0.4539-0.4444}{0.4539}=2.1\%\).

   Answer: the expression is approximately \(\tfrac49\), and for \(\theta=0.1\) the true value differs from it by about 2.1%.

2. Answer to Question 2 (4 marks)

   Method: replace \(\tan^2x\) by \(\sec^2x-1\) to get a quadratic in \(\sec x\), factorise, then change each \(\sec x\) value into a cosine.

   1. Use \(\tan^2x=\sec^2x-1\): \(2\sec^2x-2+\sec x=1\), so \(2\sec^2x+\sec x-3=0\) [1 mark]. Factorise: \((2\sec x+3)(\sec x-1)=0\) [1 mark]. \(\sec x=1\) gives \(x=0^\circ\). \(\sec x=-\tfrac32\) gives \(\cos x=-\tfrac23\), so \(x=131.8^\circ\) or \(228.2^\circ\) [2 marks]. Solutions: 0°, 131.8°, 228.2°.

   Check \(x=0^\circ\): \(2\times0+1=1\) ✓. The value \(360^\circ\) is excluded by the strict inequality in the interval.

3. Answer to Question 3 (4 marks)

   Method: use the compound angle formulae \(\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B\), \(\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B\) and \(\tan(A\pm B)=\dfrac{\tan A\pm\tan B}{1\mp\tan A\tan B}\), with exact values of \(30^\circ\), \(45^\circ\) and \(60^\circ\). Mark scheme: 2 marks for each part.

   1. Write \(105^\circ=60^\circ+45^\circ\). \(\sin105^\circ=\sin(60^\circ+45^\circ)=\dfrac{\sqrt3}2\cdot\dfrac{\sqrt2}2+\dfrac12\cdot\dfrac{\sqrt2}2=\dfrac{\sqrt6+\sqrt2}4\). So \(\sin105^\circ=\dfrac{\sqrt6+\sqrt2}4\approx0.966\).
   2. Both angles are acute, so \(\cos A\) and \(\sin B\) are positive; use Pythagoras to find them. \(\cos A=\dfrac{12}{13}\), \(\sin B=\dfrac35\). \(\cos(A-B)=\cos A\cos B+\sin A\sin B=\dfrac{12}{13}\cdot\dfrac45+\dfrac5{13}\cdot\dfrac35=\dfrac{48+15}{65}=\dfrac{63}{65}\). So \(\cos(A-B)=\dfrac{63}{65}\).

4. Answer to Question 4 (4 marks)

   Method: replace \(\cos2x\) by the form that only contains \(\sin x\), so that the equation becomes a quadratic in \(\sin x\).

   Use \(\cos2x=1-2\sin^2x\): \(2-4\sin^2x+3\sin x=1\), so \(4\sin^2x-3\sin x-1=0\) [2 marks]. Factorise: \((4\sin x+1)(\sin x-1)=0\) [1 mark]. \(\sin x=1\) gives \(x=90^\circ\). \(\sin x=-\tfrac14\) gives \(x=194.5^\circ\) or \(345.5^\circ\) [1 mark]. Solutions: 90°, 194.5°, 345.5°.

   Check \(x=90^\circ\): \(2\cos180^\circ+3=-2+3=1\) ✓. For \(\sin x=-\tfrac14\), the principal value is \(-14.5^\circ\), so the answers in range are \(180^\circ+14.5^\circ\) and \(360^\circ-14.5^\circ\).

5. Answer to Question 5 (4 marks)

   Method: expand \(R\cos(x-\alpha)\), compare coefficients of \(\cos x\) and \(\sin x\), then use \(R=\sqrt{a^2+b^2}\) and \(\tan\alpha=\dfrac ba\). Mark scheme: 1 mark for \(R\), 1 mark for \(\alpha\), 2 marks for the solutions.

   1. Compare \(R\cos\alpha=8\) and \(R\sin\alpha=15\). \(R=\sqrt{64+225}=17\). \(R\cos(x-\alpha)=R\cos x\cos\alpha+R\sin x\sin\alpha\), so \(\tan\alpha=\dfrac{15}8\), \(\alpha=61.93^\circ\). Expression \(=17\cos(x-61.93^\circ)\). Check: \(17\cos61.93^\circ=8.00\).
   2. Divide by 17 and use the inverse cosine; cosine is positive in the first and fourth quadrants. \(\cos(x-61.93^\circ)=\dfrac{10}{17}=0.5882\). \(x-61.93^\circ=\pm53.97^\circ\), so \(x=115.9^\circ\) or \(x=7.96^\circ\). Solutions: 8.0°, 115.9°. Both values lie in the required interval.

Total : 20 marks