# Trig Identities and Approximations

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

## Trig Identities and Approximations

From our first identity $\textcolor{blue}{\sin ^2 \theta} + \textcolor{limegreen}{\cos ^2 \theta} \equiv 1$, we have two new identities.

Make sure you are happy with the following topics before continuing.

A Level

## Two New Identities

So, we started out in the Basic Trig Identities section by forming the identity $\textcolor{blue}{\sin ^2 \theta} + \textcolor{limegreen}{\cos ^2 \theta} \equiv 1$. From there, we can derive two new ones.

Identity 1:

$\dfrac{\textcolor{blue}{\sin ^2 \theta} + \textcolor{limegreen}{\cos ^2 \theta} \equiv 1}{\textcolor{blue}{\sin ^2 \theta}}$

gives

$\dfrac{\textcolor{blue}{\sin ^2 \theta}}{\textcolor{blue}{\sin ^2 \theta}} + \dfrac{\textcolor{limegreen}{\cos ^2 \theta}}{\textcolor{blue}{\sin ^2 \theta}} \equiv \dfrac{1}{\textcolor{blue}{\sin ^2 \theta}}$

or,

$1 + \cot ^2 \theta \equiv \cosec ^2 \theta$

Identity 2:

$\dfrac{\textcolor{blue}{\sin ^2 \theta} + \textcolor{limegreen}{\cos ^2 \theta} \equiv 1}{\textcolor{limegreen}{\cos ^2 \theta}}$

gives

$\dfrac{\textcolor{blue}{\sin ^2 \theta}}{\textcolor{limegreen}{\cos ^2 \theta}} + \dfrac{\textcolor{limegreen}{\cos ^2 \theta}}{\textcolor{limegreen}{\cos ^2 \theta}} \equiv \dfrac{1}{\textcolor{limegreen}{\cos ^2 \theta}}$

or,

$\textcolor{red}{\tan ^2 \theta} + 1 \equiv \sec ^2 \theta$

A Level

## Small Angle Approximations

We also have approximations for $\textcolor{blue}{\sin \theta}$, $\textcolor{limegreen}{\cos \theta}$ and $\textcolor{red}{\tan \theta}$ is small (i.e. smaller than $1$, in radians):

$\textcolor{blue}{\sin \theta} \approx \theta$

$\textcolor{limegreen}{\cos \theta} \approx 1 - \dfrac{1}{2}\theta ^2$

$\textcolor{red}{\tan \theta} \approx \theta$

We can use these approximations to find rough values for complicated functions, in order to simplify them.

The expressions do extend to multiples of $\theta$, also, but only when the product is less than $1$.

So, our expressions are:

$\textcolor{blue}{\sin n\theta} \approx n\theta$

$\textcolor{limegreen}{\cos n\theta} \approx 1 - \dfrac{1}{2}(n\theta) ^2$

$\textcolor{red}{\tan n\theta} \approx n\theta$

A Level
A Level

## Example 1: Using Approximations

For small values of $\theta$, find an approximation for $\dfrac{1}{2}\textcolor{blue}{\sin \theta} + 2 \textcolor{limegreen}{\cos \theta} - 2$, and find any value of $\theta$ where the expression is $0$.

[3 marks]

The expression $\dfrac{1}{2}\textcolor{blue}{\sin \theta} + 2 \textcolor{limegreen}{\cos \theta} - 2$ can be replaced by our small angle approximations to

$\dfrac{1}{2}\theta + 2 - \theta ^2 - 2$

which can be simplified to

$\dfrac{1}{2}\theta - \theta ^2$

When this expression is equal to zero, we have $\theta ^2 - \dfrac{1}{2}\theta = 0$, which has roots $\theta = 0, \dfrac{1}{2}$.

A Level

## Example 2: Using Identities

Solve $2\sec ^2 \theta = 1 + 3\textcolor{red}{\tan ^2 \theta}$ for all values $-360° \leq x \leq 360°$.

[3 marks]

First, we need to convert all trig functions to be of one form.

So,

$2 + 2\textcolor{red}{\tan ^2 \theta} = 1 + 3\textcolor{red}{\tan ^2 \theta}$

rearranges to

$1 = \textcolor{red}{\tan ^2 \theta}$

so,

$\textcolor{red}{\tan \theta} = ±1$

or,

$\theta = -315°, -225°, -135°, -45°, 45°, 135°, 225°, 315°$

A Level

## Example Questions

$\cos \theta = 1 - \dfrac{1}{2}\theta ^2$

So, $\cos \dfrac{\pi}{4} = 1 - \left( \dfrac{1}{2} \times \dfrac{\pi ^2}{16}\right) = 1 - \dfrac{\pi ^2}{32}$

$\sin (3\theta ) + 2\cos \theta$ $= 3\theta + 2 - \theta ^2$

Using a value of $\theta = 0.2$, we have $= 3\theta + 2 - \theta ^2 = (3 \times 0.2) + 2 - 0.2^2 = 0.6 + 2 - 0.04 = 2.56$

The maximum value of $\theta$ is $\dfrac{1}{3}$.

$\sin ^2 \theta + \cos ^2 \theta \equiv 1$

so

$\dfrac{\sin ^2 \theta}{\sin ^2 \theta} + \dfrac{\cos ^2 \theta}{\sin ^2 \theta} \equiv \dfrac{1}{\sin ^2 \theta}$

meaning

$\cosec ^2 \theta \equiv 1 + \cot ^2 \theta$

$\sin ^2 \theta + \cos ^2 \theta \equiv 1$

so

$\dfrac{\sin ^2 \theta}{\cos ^2 \theta} + \dfrac{\cos ^2 \theta}{\cos ^2 \theta} \equiv \dfrac{1}{\cos ^2 \theta}$

meaning

$\sec ^2 \theta \equiv 1 + \tan ^2 \theta$

$\tan ^4 \theta - \sec ^2 \theta = 5$

$(\sec ^2 \theta - 1)^2 - \sec ^2 \theta = 5$

$\sec ^4 \theta - 3\sec ^2 \theta - 4 = 0$

Let $x = \sec ^2 \theta$.

Then we have

$x^2 - 3x - 4 = 0$

which has solutions $x = -1, 4$.

Since $\sec ^2 \theta \neq -1$, we have $\sec ^2 \theta = 4$, or $\sec \theta = ±2$.

$\sec \theta = ±2$ is equivalent to $\cos \theta = ±\dfrac{1}{2}$

So, we have

$\theta = 60°, 120°, 240°, 300°$

A Level

A Level

A Level

A Level

A Level

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