# Cos 2A Formula – Proof and Examples

Here you will learn what is the formula of cos 2A in terms of sin and cos and also in terms of tan with proof and examples.

Let’s begin –

## Cos 2A Formula :

### (i) In Terms of Cos and Sin

Given below are all the formulas for cos 2A.

(i) cos 2A = $$cos^2 A$$ – $$sin^2 A$$

(ii) cos 2A = $$2cos^2 A – 1$$    or,    1 + cos 2A = $$2cos^2 A$$

(iii) cos 2A = $$1 – 2sin^2 A$$    or,    1 – cos 2A = $$2sin^2 A$$

Proof :

(i) We have,

Cos (A + B) = cos A cos B – sin A sin B

Replacing B by A,

$$\implies$$ cos 2A = cos A cos A + sin A sin A

$$\implies$$ cos 2A = $$cos^2 A$$ – $$sin^2 A$$

(ii) We have,

cos 2A = $$cos^2 A$$ – $$sin^2 A$$

$$\implies$$ cos 2A = $$cos^2 A$$ – $$1 – cos^2 A$$

$$\implies$$ cos 2A = $$2cos^2 A – 1$$

Again, cos 2A = $$2cos^2 A – 1$$

$$\implies$$ 1 + cos 2A = $$2cos^2 A$$

(iii) We have,

cos 2A = $$cos^2 A$$ – $$sin^2 A$$

$$\implies$$ cos 2A = $$1 – sin^2 A$$ – $$sin^2 A$$

$$\implies$$ cos 2A = $$1- 2sin^2 A$$

Again, cos 2A = $$1- 2sin^2 A$$

$$\implies$$ 1 – cos 2A = $$2ain^2 A$$

We can also write above relation in terms of angle A/2, just replace A by A/2, we get

(i) cos A = $$cos^2 ({A\over 2})$$ – $$sin^2 ({A\over 2})$$

(ii) cos A = $$2cos^2 ({A\over 2}) – 1$$    or,    1 + cos A = $$2cos^2 ({A\over 2})$$

(iii) cos A = $$1 – 2sin^2 ({A\over 2})$$    or,    1 – cos A = $$2sin^2 ({A\over 2})$$

### (ii) Cos 2A Formula in Terms of Tan :

Cos 2A = $$1 – tan^2 A\over 1 + tan^2 A$$

Proof :

We have,

cos 2A = $$cos^2 A$$ – $$sin^2 A$$

$$\implies$$ cos 2A = $$cos^2 A – sin^2 A\over sin^2 A + cos^2 A$$

[ $$\because$$  $$sin^2 A + cos^2 A$$ = 1 ]

Now, Dividing numerator and denominator by $$cos^2 A$$,

$$\implies$$  cos 2A = $${cos^2 A – sin^2 A\over cos^2 A}\over {sin^2 A + cos^2 A\over cos^2 A}$$

$$\implies$$ cos 2A = $$1 – tan^2 A\over 1 + tan^2 A$$

We can also write above relation in terms of angle A/2, just replace A by A/2, we get

cos A = $$1 – tan^2 ({A\over 2})\over 1 + tan^2 ({A\over 2})$$

Example : Find the value of Cos 120 ?

Solution : We Know that sin 60 = $$\sqrt{3}\over 2$$ and cos 60 = $$1\over 2$$

By using above formula,

cos 120 = $$cos^2 60$$ – $$sin^2 60$$ = $$1\over 4$$ – $$3\over 4$$

$$\implies$$  cos 120 = $$-1\over 2$$

Example : If sin A = $$3\over 5$$, where 0 < A < 90, find the value of cos 2A ?

Solution : We have,

sin A = $$3\over 5$$ where 0 < A < 90 degrees

$$\therefore$$ $$cos^2 A$$ = 1 – $$sin^2 A$$

$$\implies$$ cos A = $$\sqrt{1 – sin^2 A}$$ = $$\sqrt{1 – {9\over 25}}$$ = $$4\over 5$$

By using above formula,

cos 2A = $$cos^2 A$$ – $$sin^2 A$$ = $$16\over 25$$ – $$9\over 25$$

$$\implies$$  cos 2A = $$7\over 25$$