# Prove: cos^2 x cos^2 y + sin^2 x sin^2 y + sin^2 x cos^2 y + sin^2 y cos^2 x = 1

*print*Print*list*Cite

### 2 Answers

We have to prove that (cos x)^2 * (cos y)^2 + (sin x)^2 * (sin y) ^2 + (sin x)^2 * (cos y)^2 + (sin y)^2 * (cos x)^2 = 1

Now sin (x+y) = (sin x)*(cos y) + (cos x)*(sin y) and cos (x+ y) = (cos x)*(cos y) - (sin x)*(sin y)

(cos x)^2 * (cos y)^2 + (sin x)^2 * (sin y) ^2 + (sin x)^2 * (cos y)^2 + (sin y)^2 * (cos x)^2

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

Now we use the relation a^2 + b^2 = (a+b)^2 - 2ab

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

We know (cos x)^2 + ( sin x)^2 = 1

=> [cos (x+y)]^2 + [sin (x+y)]^2

=> 1

**Therefore (cos x)^2 * (cos y)^2 + (sin x)^2 * (sin y) ^2 + (sin x)^2 * (cos y)^2 + (sin y)^2 * (cos x)^2 = 1**

L:H:S = cos²x cos²y + sin²x sin²y + sin²x cos²y + sin²y cos²x

= cos²x (cos²y+ sin²y) + sin²x (sin²y + cos²y)

we know that sin²θ+cos²θ=1

= cos²x+sin²x

= 1

= R:H:S