Trigonometric Proof & Equation Strategies (DP IB Analysis & Approaches (AA): HL): Exam Questions

Consider the equation cos(x − 45°) = 2 sin x in the interval 0° ≤ x ≤ 360°.

Find an exact value for $\tan x$.

Express $\cos 4\theta$ in terms of $\cos 2\theta$.

Hence, show that $\cos 4\theta =8{\cos }^2\theta ({\cos }^2\theta -1)+1$.

Given that $\tan A=\frac{\sqrt{3}}{2}$, solve the equation $\tan \left(A+x\right)=\frac{4}{5}$ in the interval 0° ≤ x ≤ 360°, by first finding the exact value of $\tan x$.

Prove that $\cos 3x\equiv 4{\cos }^{3}x-3\cos x$.

Find the exact values for $\tan x$ given that ${\tan }^{2}x+4\tan x+1=0$

Hence, solve the equation $\frac{\tan x}{2\tan x+1}=\tan 2x$ algebraically for the interval $0\le x\le 2\pi .$

Show that $\frac{5\sin \left(\pi +x\right)}{\tan x}\equiv -5\cos x$, for $x\ne \frac{n\pi }{2}$, where $n\in \mathrm{\mathbb{Z}}$.

Hence find the exact value of $\sin x$ when $\frac{5\sin \left(\pi +x\right)}{\tan x}=4$, for $\frac{\pi }{2}\le x\le \pi$.

Express $2\cos \theta -5\sin \theta$ in the form $R\cos \left(\theta +\alpha \right)$, where $R>0$ and $0<\alpha <\frac{\pi }{2}$.

Give $R$ as an exact value.

Solve the equation $\frac{1}{{\cos }^{2}2x}=1+\tan 2x$ for the interval $-\pi <x<0$.

By writing $\sqrt{3}\cos \theta +\sin \theta$ in the form $R\sin \left(\theta +\alpha \right)$, where $R>0$ and $0<\alpha <\frac{\pi }{2}$, find the exact value of $x$ such that

$5\sqrt{3}\cos \left(-\frac{\pi }{12}\right)+5\sin \left(-\frac{\pi }{12}\right)=2x$.

Solve the following equations in the given intervals.

(i) ${\cos }^{2}x-{\sin }^{2}x=\frac{1}{2}$, for $-\pi \le x\le \pi$

(ii) $\frac{2\tan x}{1-{\tan }^{2}x}=-\sqrt{3}$, for $0\le x\le \pi$

Show that $\cos x-\sqrt{3}\sin x\equiv 2\cos \left(x+\frac{\pi }{3}\right)$.

Hence, by factorising, solve the equation $2\cos \left(x+\frac{\pi }{3}\right)=\sqrt{3}\cos x\sin x-1$ for $-\pi \le x\le \pi$.

Prove the identity $\frac{1-\sin 2\theta }{2\cos 2\theta }\equiv \frac{1-\tan \theta }{2\left(1+\tan \theta \right)}$.

Hence solve the equation $\frac{1-\sin 2\theta }{2\cos 2\theta }=2\sqrt{2}$ for $-\pi \le \theta \le \pi$.

Solve $\tan 2x=\frac{\tan x}{12+5\tan x}$ algebraically for $-\frac{3}{2}\pi <x<\frac{\pi }{2}$.

By writing $A=\frac{A+B}{2}+\frac{A-B}{2}$ and finding a similar expression for $B$, or otherwise, show that

$\sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$

Hence show that $\sin \frac{7\pi }{12}+\sin \frac{\pi }{12}=\frac{\sqrt{6}}{2}$.

Prove that $\frac{\cos 3\theta -\cos \theta }{\sin 3\theta -\sin \theta }\equiv -\tan 2\theta$.

Solve the equation $\sin 2x-\cos 2x=\frac{\sin x+\cos x}{2}-1$ for the interval $-\pi <x<0.$

Show that $1-\cos 2x=2-2{\cos }^{2}x$

Show that $\frac{1}{\cos 2x}-\tan 2x=\frac{\cos x-\sin x}{\cos x+\sin x}$

Let $f\left(x\right)=\tan \left(2x+\pi \right)\sin \left(x-\pi \right)$, where $0<x<\frac{\pi }{2}$ and $x\ne \frac{\pi }{4}$.

Express $f\left(x\right)$ in terms of $\sin x$ only.

By writing $A=\frac{A+B}{2}+\frac{A-B}{2}$ and finding a similar expression for $B$, or otherwise, show that

$\cos A-\cos B=-2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)$

Hence solve $4\sin \left(\frac{75+\theta }{2}\right)\sin \left(\frac{75-\theta }{2}\right)=1-\frac{\sqrt{6}-\sqrt{2}}{2}$ for  $-315°\le \theta \le 45°$.