Compound angles
Trigonometry
- A2
- Non-calculator
- 8 questions
- 37 marks
These are original practice questions for the Edexcel A-level Pure content, not a past paper. This site is independent and not affiliated with Pearson.
-
Use a compound-angle formula to find the exact value of $\sin 75^\circ$.
3 marks
-
$\cos \theta = \dfrac{3}{5}$ and $\theta$ is acute. Find the exact value of $\cos 2\theta$.
3 marks
-
$\sin A = \dfrac{3}{5}$, $\cos A = \dfrac{4}{5}$, $\sin B = \dfrac{5}{13}$ and $\cos B = \dfrac{12}{13}$, with $A$ and $B$ acute. Find $\sin(A - B)$.
4 marks
-
Solve $\sin 2x = \cos x$ for $0 \le x < 2\pi$.
4 marks
-
Express $5\cos x - 12\sin x$ in the form $R\cos(x + \alpha)$, where $R > 0$ and $0 < \alpha < \dfrac{\pi}{2}$.
5 marks
-
Solve $\cos x + \sin x = 1$ for $0 \le x < 2\pi$.
5 marks
-
Solve $\cos x + \sqrt{3}\sin x = 1$ for $0 \le x < 2\pi$.
6 marks
-
Find the maximum value of $5\cos x - 12\sin x$, and the smallest positive value of $x$ at which that maximum occurs.
7 marks
Mark scheme
-
Answer. $\dfrac{\sqrt{6} + \sqrt{2}}{4}$
- $\sin 75^\circ = \sin(45^\circ + 30^\circ)$.
- $\dfrac{\sqrt{2}}{2} \times \dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{2}}{2} \times \dfrac{1}{2} = \dfrac{\sqrt{6} + \sqrt{2}}{4}$.
-
Answer. $-\dfrac{7}{25}$
- Use $\cos 2\theta = 2\cos^2 \theta - 1$.
- $2 \times \dfrac{9}{25} - 1 = -\dfrac{7}{25}$.
-
Answer. $\dfrac{16}{65}$
- $\sin(A - B) = \sin A \cos B - \cos A \sin B$.
- $\dfrac{3}{5} \times \dfrac{12}{13} - \dfrac{4}{5} \times \dfrac{5}{13} = \dfrac{16}{65}$.
-
Answer. $x = \dfrac{\pi}{6}$, $\dfrac{\pi}{2}$, $\dfrac{5\pi}{6}$ or $\dfrac{3\pi}{2}$
- $2\sin x \cos x - \cos x = 0$.
- $\cos x(2\sin x - 1) = 0$.
- $\cos x = 0$ or $\sin x = \dfrac{1}{2}$.
-
Answer. $13\cos(x + \alpha)$, where $\cos \alpha = \dfrac{5}{13}$ and $\sin \alpha = \dfrac{12}{13}$.
- $R = \sqrt{5^2 + 12^2} = 13$.
- $R\cos(x + \alpha) = R\cos x \cos \alpha - R\sin x \sin \alpha$.
- $R\cos \alpha = 5$ and $R\sin \alpha = 12$.
-
Answer. $x = 0$ or $x = \dfrac{\pi}{2}$
- Write the left side as $\sqrt{2}\cos\left(x - \dfrac{\pi}{4}\right)$.
- $\cos\left(x - \dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}$.
- The solutions in the interval are $x = 0$ and $x = \dfrac{\pi}{2}$.
-
Answer. $x = 0$ or $x = \dfrac{2\pi}{3}$
- The left side is $2\sin\left(x + \dfrac{\pi}{6}\right)$.
- $\sin\left(x + \dfrac{\pi}{6}\right) = \dfrac{1}{2}$.
- $x + \dfrac{\pi}{6} = \dfrac{\pi}{6}$ or $\dfrac{5\pi}{6}$.
- $x = 0$ or $x = \dfrac{2\pi}{3}$.
-
Answer. The maximum is $13$, when $x = 2\pi - \arctan \dfrac{12}{5}$.
- $5\cos x - 12\sin x = 13\cos(x + \alpha)$, where $\tan \alpha = \dfrac{12}{5}$.
- The maximum value is $13$.
- This occurs when $x + \alpha = 0$, so the smallest positive solution is $x = 2\pi - \alpha$.