# Proving Trigonometrical Identities (A level)

## Proving Trigonometrical Identities (A level)

Proving trigonometrical identities requires you to use mathematical and logical steps to prove that one side of the identity can be simplified into the other side of the equation. Logically you cannot work from both sides of the identity at the same time.

Since we will be simplifying one side to prove that it is equal to the other side, it is helpful to assign names to the sides. The “left-hand side” of an identity is denoted by LHS, and the “right-hand side” is denoted as RHS.

### Question 1

$latex cos^2 \theta – sin^2 \theta \equiv 2cos^2 \theta – 1$

### Solution 1

$latex cos^2 \theta – sin^2 \theta \equiv 2cos^2 \theta – 1$

Take LHS

- $latex cos^2 \theta – sin^2 \theta$

*However we know that $latex cos^2 \theta + sin^2 \theta \equiv 1$ and we can transpose it to $latex sin^2 \theta \equiv 1 – cos^2 \theta $. By substituting this value of $latex sin^2 \theta$ into our working we get:*

- $latex cos^2 \theta – (1 – cos^2 \theta)$
- $latex cos^2 \theta – 1 + cos^2 \theta$
- $latex 2cos^2 \theta – 1$

### Question 2

$latex (\frac{cosec \theta + cot \theta}{tan \theta + sec \theta})tan \theta \equiv \frac{1 + cos \theta}{sin \theta + 1}$

### Solution 2

$latex (\frac{cosec \theta + cot \theta}{tan \theta + sec \theta})tan \theta \equiv \frac{1 + cos \theta}{sin \theta + 1}$

Take the LHS

- $latex (\frac{cosec \theta + cot \theta}{tan \theta + sec \theta})tan \theta $
- $latex (\frac{ \frac{1}{sin \theta} +\frac{cos \theta}{sin \theta}}{\frac{sin \theta}{cos \theta} + \frac{1}{cos \theta}})\frac{sin \theta}{cos \theta} $
- $latex (\frac{ \frac{1 + cos \theta}{sin \theta} }{ \frac{sin \theta + 1}{cos \theta}})\frac{sin \theta}{cos \theta} $
- $latex ( \frac{1 + cos \theta}{sin \theta} \div \frac{sin \theta + 1}{cos \theta})\frac{sin \theta}{cos \theta} $
- $latex ( \frac{1 + cos \theta}{sin \theta} \times \frac{cos \theta}{sin \theta + 1})\frac{sin \theta}{cos \theta} $
- $latex \frac{1 + cos \theta}{sin \theta} \times \frac{cos \theta}{sin \theta + 1} \times \frac{sin \theta}{cos \theta} $
- $latex \frac{1 + cos \theta}{sin \theta + 1} $

### Question 3

$latex \frac{1 + tan^2 \theta}{1 + cot^2 \theta } \equiv tan^2 \theta$

### Solution 3

$latex \frac{1 + tan^2 \theta}{1 + cot^2 \theta } \equiv tan^2 \theta$

Take the LHS

- $latex \frac{1 + tan^2 \theta}{1 + cot^2 \theta }$
- $latex \frac{1 + tan^2 \theta}{1 + \frac{1}{tan^2 \theta} }$
- $latex \frac{1 + tan^2 \theta}{\frac{tan^2 \theta +1}{tan^2 \theta} }$
- $latex ( 1 + tan^2 \theta ) \div \frac{tan^2 \theta +1}{tan^2 \theta}$
- $latex ( 1 + tan^2 \theta ) \times \frac{tan^2 \theta}{tan^2 \theta + 1}$
- $latex tan^2 \theta$

### Question 4

$latex (1 – cos \theta) (1 + sec \theta) \equiv sin \theta tan \theta$

### Solution 4

$latex (1 – cos \theta) (1 + sec \theta) \equiv sin \theta tan \theta$

Take the LHS

- $latex (1 – cos \theta) (1 + sec \theta)$
- $latex 1 + sec \theta – cos \theta – cos \theta sec \theta$
- $latex 1 + \frac{1}{cos \theta} – cos \theta – cos \theta \frac{1}{cos \theta}$
- $latex 1 + \frac{1}{cos \theta} – cos \theta – 1$
- $latex \frac{1}{cos \theta} – cos \theta $
- $latex \frac{1 – cos^2 \theta}{cos \theta} $

*However $latex sin^2 \theta \equiv 1 – cos^2 \theta $*

- $latex \frac{sin^2 \theta}{cos \theta} $
- $latex \frac{sin \theta sin \theta}{cos \theta} $
- $latex sin \theta tan \theta$

### Question 5

$latex \frac{1+cot\ y}{1+tan\ y} \equiv cot\ y$

### Solution 5

$latex \frac{1+cot\ y}{1+tan\ y} \equiv cot\ y$

Take the LHS

- $latex \frac{1+cot\ y}{1+tan\ y}$
- $latex (1+cot\ y) \div (1+tan\ y)$
- $latex (1+\frac{cos\ y}{sin\ y}) \div (1+\frac{sin\ y}{cos\ y})$
- $latex (\frac{sin\ y + cos\ y}{sin\ y}) \div (\frac{cos\ y + sin\ y}{cos\ y})$
- $latex (\frac{sin\ y + cos\ y}{sin\ y}) \times (\frac{cos\ y}{cos\ y + sin\ y})$
- $latex \frac{cos\ y}{sin\ y}$
- $latex cot\ y$