Trigonometric Proof (OCR A Level Maths A: Pure): Exam Questions

Given the identity

sin² $\theta +$cos² $\theta \equiv 1$

prove the following identities:

(i) sec² $\theta \equiv 1+$tan² $\theta$

(ii) cosec² $\theta \equiv 1+$cot² $\theta$

(i) By using the double angle formula for cosine, prove the identity

cos $4\theta \equiv 8$ cos⁴ $\theta -8$ cos² $\theta +1$

(ii) Show by counter-example that

sin $4\theta \not\equiv 8$ sin⁴ $\theta -8$ sin² $\theta +1$

Given that $\theta$ is small, and that terms involving ${\theta }^3$ or higher powers of $\theta$ can be ignored, use an appropriate approximation to show that

$4$cos $4\theta -2$cos² $2\theta \approx 2-24{\theta }^2$

Show that the result in part (a) gives a percentage error of 0.583%, to 3 significant figures, when used to approximate

$4$cos$\frac{\mathrm{\pi }}{6}-2$cos² $\frac{\mathrm{\pi }}{12}$

Prove the identity

$\frac{4{\sin }^4\theta }{{\sin }^{2}2\theta }\equiv {\tan }^2\theta \theta \ne \mathrm{k\pi }$

Show that

sin $\theta \left({\mathrm{cosec}}^2\theta -2\right)\equiv \frac{\cos 2\theta }{\sin \theta }$

Use the quotient rule to show that

$\frac{d}{dx}\left[\tan x\right]={\sec }^{2}x$

Show that

sin $3\theta +$sin $\theta \equiv 4$ sin $\theta -4$sin³ $\theta$

Prove the identity

$\frac{4\cot x\cos 2x}{\sin 4x}\equiv {\mathrm{cosec}}^{2}xx\ne \frac{\mathrm{k\pi }}{4}$

Show that

sin$\left(\arccos \left(-\frac{1}{2}\right)\right)=\sqrt{3}\sin \left(\frac{\mathrm{\pi }}{6}\right)$

Show that

$\arcsin \left(\cos \frac{3\mathrm{\pi }}{4}\right)=-\arcsin \left(\cos \frac{\mathrm{\pi }}{4}\right)$

Show that

$\sqrt{2}\sin \left(\theta -\frac{\mathrm{\pi }}{4}\right)\equiv \sin \theta -\cos \theta$

Given the identity

cos$\left(A+B\right)=$cos $A$ cos $B$$-$sin $A$ sin $B$

prove the following identities:

(i) cos $2\theta \equiv$cos² $\theta -$sin² $\theta$

(ii) cos $2\theta \equiv 1-2$sin² $\theta$

(iii) cos $2\theta \equiv 2$cos² $\theta -1$

(i) Prove the identity

sin $3\theta \equiv 3$ sin $\theta -4$ sin³ $\theta$

(ii) Show by counter-example that

cos $3\theta \not\equiv 3$ cos $\theta -$$4$ cos³ $\theta$

Given that $\theta$ is small, and that terms involving ${\theta }^3$ or higher powers of $\theta$ can be ignored, show that

$\frac{1}{{\mathrm{cosec}}^2\left({\displaystyle \frac{\theta }{2}}\right)}+\frac{1}{{\sec }^2\left({\displaystyle \frac{\theta }{4}}\right)}\approx 1+\frac{3}{16}{\theta }^2$

Determine the percentage error when the result in part (a) is used to approximate

$\frac{1}{{\mathrm{cosec}}^2\left({\displaystyle \frac{7}{20}}\right)}+\frac{1}{{\sec }^2\left({\displaystyle \frac{7}{40}}\right)}$

giving your answer correct to 3 significant figures.

Show that

cos $4\theta +$cos $\frac{\mathrm{\pi }}{3}\equiv 8$ sin⁴ $\theta -8$ sin² $\theta +\frac{3}{2}$

Prove that

cot² $\theta -$tan² $\theta \equiv 4$ cot $2\theta$ cosec $2\theta$.

Prove the identity

$\frac{1-{\tan }^{2}x}{\cos 2x}\equiv {\sec }^{2}xx\ne \frac{2\mathrm{k}+1}{4}\mathrm{\pi }$

Prove the identity

cosec $x\equiv \frac{{\displaystyle \frac{1}{2}}{\sec }^2{\displaystyle \frac{x}{2}}}{\tan {\displaystyle \frac{x}{2}}}$

Show that

tan$\frac{x}{2}\equiv \frac{1}{\mathrm{cosec}x+\cot x}x\ne 2\mathrm{k\pi }$

Given that $y=\arcsin \left(kx\right)$, where $k$ is a constant, show that   $x=\frac{1}{k}\cos \left(\frac{\mathrm{\pi }}{2}-y\right)$ .

Hence show that the value of $\arcsin kx+\arccos kx$ is constant and independent of $k$. Find the value of this constant.

Show that

$\frac{10}{4\cos \theta +3\sin \theta }\equiv 2\sec \left(\theta -\mathrm{\alpha }\right)$

where

$\alpha =\arctan \left(\frac{3}{4}\right)$

Consider the three triangles, all of height 1, as shown below.

By applying the area of a triangle formula $A=\frac{1}{2}\mathrm{ab}\sin C$  to each one, prove that,

sin$\left(A+B\right)\equiv$sin $A$ cos $B+$sin $B$ cos $A$

Briefly explain why this only proves the result for $A$ and $B$ being acute angles.

Prove the identity

$\tan 4\theta \equiv \frac{4\tan \theta \left(1-{\tan }^2\theta \right)}{1-6{\tan }^2\theta +{\tan }^4\theta }$

Use the small angle approximations for sine and cosine to confirm the following two limit results:

$\underset{\mathrm{h}\to 0}{\lim }\frac{\sin h}{h}=1\underset{\mathrm{h}\to 0}{\lim }\frac{\cos h-1}{h}=0$

Be sure to explain why use of the small angle approximations is justified here.

Hence prove from first principles that

$\frac{\mathrm{d}}{dx}\left[\sin x\right]=\cos x$

Prove the identity

$-16$ cot $2\theta$ cosec³ $2\theta \equiv$sec⁴ $\theta -$cosec⁴ $\theta$

Show that

$\frac{\sqrt{2}\cos \left(\theta +{\displaystyle \frac{\mathrm{\pi }}{4}}\right)}{\sin \left(\theta -{\displaystyle \frac{\mathrm{\pi }}{2}}\right)}\equiv \tan \theta -1$

Show that

sin $3\theta \equiv 3$ sin $\mathrm{\theta }$ cos² $\theta -$sin³ $\theta$

Hence, or otherwise, show that

$\frac{\cos 3\theta -\cos \theta }{\sin 3\theta \sin \theta }\equiv \frac{4\cos \theta }{1-4{\cos }^2\theta }\theta \ne \mathrm{k\pi }$

Show that

$4$cos² $\left(x-\frac{\mathrm{\pi }}{6}\right)\equiv 3-2$sin² $x+\sqrt{3}$ sin $2x$

Show that

tan$\left(\frac{2x+\mathrm{\pi }}{4}\right)\equiv$sec $x+$tan $x$

On the axes below sketch the graphs of

$y=\arccos \left(-x\right)$  and  $y=\left|\arcsin x\right|$

With the help of your sketch, determine the exact solution(s) to the equation

$\arccos \left(-x\right)=\left|\arcsin x\right|$

What can you say about the solution(s) to the equation

$\left|\arccos x\right|=\arcsin \left(-x\right)?$

Justify your answer.

Show that

$\frac{1}{{\left({\displaystyle \frac{\sqrt{3}}{2}}\cos \theta -{\displaystyle \frac{1}{2}}\sin \theta \right)}^2}+\frac{1}{{\left({\displaystyle \frac{\sqrt{3}}{2}}\sin \theta +{\displaystyle \frac{1}{2}}\cos \theta \right)}^2}\equiv 4{\mathrm{cosec}}^2\left(2\theta +\frac{\mathrm{\pi }}{3}\right)$