Compound and double angle formulae; R cos(θ ± α) – A-level Maths exam questions with mark scheme

1StandardCalculator6 marks

Given that $\sin A=\tfrac35$, where $A$ is obtuse, and $\cos B=\tfrac{12}{13}$, where $B$ is acute,

(a) write down the exact value of $\cos A$. [1]

(b) Find the exact value of $\sin(A+B)$. [3]

(a) $\cos A=-\tfrac45$B1
(b) $\sin B=\tfrac{5}{13}$B1
Uses $\sin(A+B)=\sin A\cos B+\cos A\sin B$ with their valuesM1
$\tfrac35\cdot\tfrac{12}{13}-\tfrac45\cdot\tfrac{5}{13}=\tfrac{16}{65}$A1
(c) Uses $\tan2A=\dfrac{2\tan A}{1-\tan^{2}A}$ with $\tan A=-\tfrac34$M1
$-\dfrac{24}{7}$A1
Answer: (a) $-\tfrac45$ (b) $\tfrac{16}{65}$ (c) $-\tfrac{24}{7}$
Examiner tip: Obtuse means the second quadrant, where cosine and tangent are negative – a positive $\cos A$ loses marks in every part.

2HardCalculator10 marks

(a) Express $5\cos\theta-12\sin\theta$ in the form $R\cos(\theta+\alpha)$, where $R>0$ and $0<\alpha<\dfrac{\pi}{2}$. Give the value of $\alpha$ to 4 significant figures. [3]

(b) Hence solve, for $0\leqslant\theta<2\pi$, $$5\cos\theta-12\sin\theta=6.5$$ giving your answers to 3 decimal places. [4]

(c) Find the maximum value of $$\frac{1}{20+5\cos\theta-12\sin\theta}$$ and the smallest positive value of $\theta$ at which it occurs, to 3 decimal places. [3]

(a) $R=13$B1
$\tan\alpha=\pm\tfrac{12}{5}$ (or $\sin\alpha=\pm\tfrac{12}{13}$, $\cos\alpha=\pm\tfrac{5}{13}$)M1
$\alpha=1.176$A1
(b) $\cos(\theta+\alpha)=\tfrac{6.5}{13}=\tfrac12$M1
Finds a correct value of $\theta+\alpha$ in the range $[\alpha,2\pi+\alpha)$, e.g. $\tfrac{5\pi}{3}$ or $\tfrac{7\pi}{3}$, and subtracts $\alpha$M1
One of $\theta=4.060$ or $\theta=6.154$A1
Both, and no others in range (note $\tfrac{\pi}{3}-\alpha<0$)A1
(c) Maximum value $\dfrac{1}{20-13}=\dfrac17$B1
$\cos(\theta+\alpha)=-1\Rightarrow\theta+\alpha=\pi$M1
$\theta=1.966$A1
Answer: (a) $13\cos(\theta+1.176)$ (b) $\theta=4.060,\ 6.154$ (c) $\tfrac17$ at $\theta=1.966$
Examiner tip: Adjust the interval for $\theta+\alpha$ before solving – otherwise you either miss $\tfrac{7\pi}{3}-\alpha$ or keep the negative value $\tfrac{\pi}{3}-\alpha$.

3HardCalculator6 marks

(a) Prove that $$\operatorname{cosec}2x+\cot2x\equiv\cot x,\qquad x\neq\frac{n\pi}{2},\ n\in\mathbb{Z}$$ [4]

(b) Hence solve, for $0[2]

(a) Writes LHS as a single fraction in terms of $\sin2x$ and $\cos2x$: $\dfrac{1+\cos2x}{\sin2x}$ (allow equivalent, e.g. writing both terms directly over the common denominator $2\sin x\cos x$)M1
Uses $\sin2x=2\sin x\cos x$ and a correct form of $\cos2x$M1
Obtains $\dfrac{2\cos^{2}x}{2\sin x\cos x}$, e.g. from $\dfrac{1+2\cos^{2}x-1}{2\sin x\cos x}$ or $\dfrac{1+\cos^{2}x-\sin^{2}x}{2\sin x\cos x}$A1
$=\dfrac{\cos x}{\sin x}=\cot x$ csoA1*
(b) $\cot x=3\Rightarrow\tan x=\tfrac13$, finds one value of $x$ (awrt $0.322$, or $18.4^\circ$ seen)M1
$x=0.322,\ 3.46$ (awrt) and no other answers in $0A1
Answer: (b) $x=0.322,\ 3.46$
Examiner tip: Write the left-hand side as a single fraction $\dfrac{1+\cos2x}{\sin2x}$ and use $\cos2x=2\cos^{2}x-1$ so the $1$ cancels straight away; in (b) remember $\tan x$ has period $\pi$, so there are two answers in $0

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4StretchCalculator7 marks

(a) Using $\sin3x=\sin(2x+x)$, prove that $$\sin3x\equiv3\sin x-4\sin^{3}x$$ [4]

(b) Hence solve, for $0\leqslant x<\pi$, the equation $$\sin3x=2\sin x$$ giving your answers as exact multiples of $\pi$. [3]

(a) Uses $\sin(2x+x)=\sin2x\cos x+\cos2x\sin x$M1
Uses $\sin2x=2\sin x\cos x$ and a correct form of $\cos2x$, e.g. $\cos2x=1-2\sin^{2}x$M1
Uses $\cos^{2}x=1-\sin^{2}x$ (where needed) to obtain an expression in $\sin x$ only, e.g. $2\sin x(1-\sin^{2}x)+\sin x-2\sin^{3}x$M1
$3\sin x-4\sin^{3}x$ csoA1*
(b) Uses (a) to form an equation in $\sin x$ only and solves it, e.g. $\sin x(1-4\sin^{2}x)=0\Rightarrow\sin x=0$ or $\sin x=\pm\tfrac12$ (allow $\sin^{2}x=\tfrac14$ after dividing by $\sin x$)M1
$x=\dfrac{\pi}{6},\ \dfrac{5\pi}{6}$A1
$x=0$ (from $\sin x=0$, so not available to a candidate who divides by $\sin x$), and no other answers in range ($\sin x=-\tfrac12$ gives none; $x=\pi$ excluded)A1
Answer: (b) $x=0,\ \dfrac{\pi}{6},\ \dfrac{5\pi}{6}$
Examiner tip: Factorise out $\sin x$ rather than dividing by it – dividing loses the solution $x=0$.

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