# Prove the following identity: cos4x - sin4xcot2x = -1

*print*Print*list*Cite

### 2 Answers

We have to prove that cos 4x - sin 4x * cot 2x = -1

cos 4x - sin 4x * cot 2x = -1

use cos 2x = (cos x)^2 - (sin x)^2 and sin 2x = 2 sin x cos x and cot x = (cos x)/(sin x)

=> (cos 2x)^2 - (sin x)^2 - 2*(sin 2x)*(cos 2x)*(cos 2x)/(sin 2x)

=>(cos 2x)^2 - (sin x)^2 - 2*(cos 2x)*(cos 2x)

=> ( cos 2x)^2 - 2 ( cos 2x)^2 - ( sin 2x)^2

=> - (cos 2x)^2 - (sin 2x)^2

=> -1*[(cos 2x)^2 + (sin 2x)^2]

As (cos x)^2 + (sin x)^2 = 1

=> -1

Therefore we proved that

**cos 4x - sin 4x * cot 2x = -1**

We know that cos 4x = cos 2*(2x) = 2(cos 2x)^2 - 1

We'll write sin 4x = sin 2(2x) = 2 sin (2x)*cos (2x)

We'll write cot 2x = cos 2x/sin 2x

We'll re-write the identity:

2(cos 2x)^2 - 1 - 2 sin (2x)*cos (2x)*cos 2x/sin 2x = -1

We'll simplify and we'll get:

2(cos 2x)^2 - 1 - 2*cos 2x*cos 2x = -1

2(cos 2x)^2 - 1 - 2(cos 2x)^2 = -1

We'll addÂ 2(cos 2x)^2 both sides:

**2(cos 2x)^2 - 1 = 2(cos 2x)^2 - 1 q.e.d.**