# Trig Proofs

A LevelAQAEdexcelOCREdexcel 2022OCR 2022

## Trig Proofs

In this section, we’ll be taking a look at some standard proof methods, involving all of the techniques we’ve learned so far.

A Level

## Example 1

Prove that $1 + \cot ^2 A \equiv \cosec ^2 A$.

[3 marks]

$1 + \cot ^2 A$

$= \dfrac{\textcolor{blue}{\sin} ^2 \textcolor{purple}{A}}{\textcolor{blue}{\sin} ^2 \textcolor{purple}{A}} + \dfrac{\textcolor{limegreen}{\cos} ^2 \textcolor{purple}{A}}{\textcolor{blue}{\sin} ^2 \textcolor{purple}{A}}$

$= \dfrac{\textcolor{blue}{\sin} ^2 \textcolor{purple}{A} + \textcolor{limegreen}{\cos} ^2 \textcolor{purple}{A}}{\textcolor{blue}{\sin} ^2 \textcolor{purple}{A}}$

$= \dfrac{1}{\textcolor{blue}{\sin} ^2 \textcolor{purple}{A}}$

$= \cosec ^2 \textcolor{purple}{A}$

A Level

## Example 2

Show that $\sqrt{\dfrac{1 - \textcolor{limegreen}{\cos 2x}}{1 + \textcolor{limegreen}{\cos} 2x}} = \textcolor{red}{\tan} x$.

[4 marks]

By $\textcolor{limegreen}{\cos} 2A \equiv 2\textcolor{limegreen}{\cos} ^2 A - 1 \equiv 1 - 2\textcolor{blue}{\sin} ^2 A$,

$\sqrt{\dfrac{1 - \textcolor{limegreen}{\cos} 2x}{1 + \textcolor{limegreen}{\cos} 2x}}$

$= \sqrt{\dfrac{2\textcolor{blue}{\sin} ^2 x}{2\textcolor{limegreen}{\cos} ^2 x}}$

$= \sqrt{\textcolor{red}{\tan} ^2 x}$

$= \textcolor{red}{\tan} x$

A Level

## Example 3

Prove that $\textcolor{limegreen}{\cos} x = \textcolor{limegreen}{\cos} (-x)$.

[2 marks]

$\textcolor{limegreen}{\cos} (-x)$

$= \textcolor{limegreen}{\cos} (0 - x)$

$= \textcolor{limegreen}{\cos} 0 \textcolor{limegreen}{\cos} x + \textcolor{blue}{\sin} 0 \textcolor{blue}{\sin} x$

$= (1 \times \textcolor{limegreen}{\cos} x) + (0 \times \textcolor{blue}{\sin} x)$

$= \textcolor{limegreen}{\cos} x$

A Level

## Example 4

Show that $\textcolor{red}{\tan} \theta + \cot \theta = 2\cosec 2 \theta$.

[3 marks]

$\textcolor{red}{\tan} \theta + \cot \theta$

$= \dfrac{\textcolor{blue}{\sin} \theta}{\textcolor{limegreen}{\cos} \theta} + \dfrac{\textcolor{limegreen}{\cos} \theta}{\textcolor{blue}{\sin} \theta}$

$= \dfrac{\textcolor{blue}{\sin} ^2 \theta}{\textcolor{blue}{\sin} \theta \textcolor{limegreen}{\cos} \theta} + \dfrac{\textcolor{limegreen}{\cos} ^2 \theta}{\textcolor{blue}{\sin} \theta \textcolor{limegreen}{\cos} \theta}$

$= \dfrac{\textcolor{blue}{\sin} ^2 \theta + \textcolor{limegreen}{\cos} ^2 \theta}{\textcolor{blue}{\sin} \theta \textcolor{limegreen}{\cos} \theta}$

$= \dfrac{1}{\textcolor{blue}{\sin} \theta \textcolor{limegreen}{\cos} \theta}$

$= \dfrac{2}{\textcolor{blue}{\sin} 2\theta}$

$= 2\cosec 2 \theta$

A Level

## Example Questions

$\dfrac{3 - 3(1 - 2x^2)}{4x^2} = \dfrac{6x^2}{4x^2} = \dfrac{3}{2}$

• $\sin x = \dfrac{\text{opp}}{\text{hyp}}$

• $\cos x = \dfrac{\text{adj}}{\text{hyp}}$

• $\tan x = \dfrac{\text{opp}}{\text{adj}}$

$\dfrac{\sin x}{\cos x} = \dfrac{\left( \dfrac{\text{opp}}{\text{hyp}}\right) }{\left( \dfrac{\text{adj}}{\text{hyp}}\right) } = \dfrac{\text{opp}}{\text{adj}} = \tan x$

\begin{aligned}\cot ^2 x + \sin ^2 x&= \dfrac{\cos ^2 x}{\sin ^2 x} + \sin ^2 x\\[1.2em]&= \dfrac{\cos ^2 x}{\sin ^2 x} + 1 - \cos ^2 x\\[1.2em]&= \dfrac{\cos ^2 x}{\sin ^2 x} + \dfrac{\sin ^2 x}{\sin ^2 x} - \cos ^2 x\\[1.2em]&= \dfrac{\cos ^2 x + \sin ^2 x}{\sin ^2 x} - \cos ^2 x\\[1.2em]&=\dfrac{1}{\sin ^2 x} - \cos ^2 x\\[1.2em]&= \cosec ^2 x - \cos ^2 x\end{aligned}

A Level

A Level

A Level

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