Trigonometric Equations (Edexcel IGCSE Further Pure Maths): Exam Questions

Using formulae (opens in a new tab) , show that

(i) $\sin 2A=2\sin A\cos A$

(ii) $\cos 2A=2{\cos }^{2}A-1$

$\mathrm{f}\left(\theta \right)=\frac{2\tan \theta }{1+{\tan }^2\theta }$

Show that $\mathrm{f}\left(\theta \right)=\sin 2\theta$

Solve, in radians to 3 significant figures, for $-\frac{\pi }{2}⩽x⩽\frac{\pi }{2}$, the equation

$5\tan \left(x+\frac{\pi }{6}\right)=\left[1+{\tan }^2\left(x+\frac{\pi }{6}\right)\right]\left[1-2{\cos }^2\left(x+\frac{\pi }{6}\right)\right]$

Using calculus, find the exact value of

${\int }_0^{\frac{\pi }{2}}\left(\frac{4\tan \theta }{1+{\tan }^2\theta }-\cos 5\theta +2\right) \mathrm{d}\theta$

Using a formula (opens in a new tab), show that

$\tan 2A=\frac{2\tan A}{1-{\tan }^{2}A}$

Hence, solve the equation

$\tan A°-\tan 2A°=0$ for $0⩽A⩽180$

Using a formula given on page 2, solve, giving your solutions as exact values

$\cos \left(x-\frac{\pi }{6}\right)=\sin x$ for $-\pi ⩽x⩽2\pi$

Using formulae (opens in a new tab), show that

(i) $\cos 2A=2{\cos }^{2}A-1$

(3)

(ii) $\sin 2A=2\sin A\cos A$

(1)

Show that ${\cos }^{3}A=\frac{\cos 3A+3\cos A}{4}$

Hence, or otherwise, solve, giving exact values in terms of $\pi$

$8{\cos }^3\left(\frac{\theta }{2}\right)-6\cos \left(\frac{\theta }{2}\right)-1=0$ for $0⩽\theta ⩽2\mathrm{\pi }$

Use algebraic integration to find the exact value of

${\int }_0^{\frac{\mathrm{\pi }}{6}}\left(4{\cos }^3\theta -\sin 2\theta \right)\mathrm{d}\theta$

Using the formulae (opens in a new tab), show that

(i) $\sin 2\theta =2\sin \theta \cos \theta$

[2]

(ii) $\cos 2\theta =2{\cos }^2\theta -1$

[3]

Given that $\theta \ne (90°+180°n)$ where $n\in \mathrm{\mathbb{Z}}$

use the results from part (a) to show that $\sin 2\theta -\tan \theta$ can be written as $\tan \theta \cos 2\theta$

Solve for $0<x<360$

$\sin 2x°-\tan x°=0$

Using the formulae (opens in a new tab) , show that

(i) ${\cos }^{2}A=\frac{\cos 2A+1}{2}$

(ii) ${\sin }^{2}A=\frac{1-\cos 2A}{2}$

Show that

$(2\sin x-\cos x)(\sin x-3\cos x)=\frac{1}{2}(\cos 2x-7\sin 2x+5)$

$y=(2\sin x-\cos x)(\sin x-3\cos x)$

Solve, for $0°⩽x⩽180°$ the equation, $\frac{\mathrm{d}y}{\mathrm{d}x}=0$

Give your answers to the nearest whole number.