Trig Identities and Approximations

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 AQA Edexcel OCR

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 LevelAQAEdexcelOCR

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 LevelAQAEdexcelOCR
A Level AQA Edexcel OCR

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 LevelAQAEdexcelOCR

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 LevelAQAEdexcelOCR

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°

Related Topics

MME

Basic Trig Identities

A Level
MME

Reciprocal Trig Functions

A Level

Additional Resources

MME

Exam Tips Cheat Sheet

A Level
MME

Formula Booklet

A Level

Worksheet and Example Questions

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Small Angle Approximations

A Level

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