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

Solve the following equations in the given intervals. 

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

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

Show that $\frac{5\sin \left(\pi +x\right)}{\tan x}\equiv -5\cos xx\ne \frac{n\pi }{2}$.

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

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

Hence solve the equation $2\cos \left(x+\frac{\pi }{3}\right)=\sqrt{3}\cos x\sin x-1$ for the interval $-\pi \le x\le \pi$. Show each step of your working.

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

Give $R$  in the form $\sqrt{k}$ where $k$ is an integer.

Given that $\cos x=\frac{\sqrt{2}}{5}$ and  $\pi \le x\le 2\pi$, find the exact value of $\tan \left(2x\right)$.

Prove that $2\sin 2\theta \left(2{\cos }^2\theta -1\right)\equiv \sin 4\theta$.

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

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

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

Determine the value of the following expression

$\cos 55°\cos 5°-\cos 85°\cos 35°.$

The following diagram shows the triangle $\mathrm{ABC}$ where $\mathrm{AC}=\sqrt{3}$ , $\mathrm{A}\hat{\mathrm{B}}\mathrm{C}=135°$ and $\mathrm{B}\hat{\mathrm{C}}\mathrm{A}=15°$.

Find the length $\mathrm{AB}$ giving your answer in the form $\frac{a+\sqrt{b}}{c}$, where $a,b,c\in \mathrm{\mathbb{Z}}$.

Find the value of $\arctan \left(\frac{7}{3}\right)-\arctan \left(\frac{2}{5}\right)$.

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

By writing $\sqrt{3}\cos \theta +\sin \theta$ in the form $R\sin \left(\theta +\alpha \right)$ where $R>0$ and $0<\alpha <\frac{\pi }{2}$, or otherwise, solve $5\sqrt{3}\cos \left(-\frac{\mathrm{\pi }}{12}\right)=2x-5\sin \left(-\frac{\mathrm{\pi }}{12}\right)$.

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

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

Given that  $\sin A=-\frac{1}{\sqrt{3}}$ for $\pi \le A\le \frac{3\pi }{2}$and $\cos B=-\frac{\sqrt{2}}{3}$ for $\frac{\pi }{2}\le B\le \pi$ , find the value of $\tan \left(A+B\right)$. Give the answer in the form $\sqrt{p}\left(\sqrt{q}-r\right)$  where $p$, $q$and $r$ are prime numbers.

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

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  $-315°\le \theta \le 45°$.

The following diagram shows the triangle $\mathrm{PQR}$ where $\mathrm{PQ}=3$, $\mathrm{PR}=\sqrt{2}$ and $\mathrm{R}\hat{\mathrm{P}}\mathrm{Q}=\frac{5\pi }{12}.$

Expand and simplify $\frac{1}{2}{\left(3\sqrt{3}-1\right)}^2$ .

Show that the length of $\mathrm{QR}$ is $\frac{3\sqrt{6}-\sqrt{2}}{2}$units.