# prove that sin^4(theta)-cos^4(theta)=sin^2(theta)-cos^2(theta)trigonometry

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### 2 Answers

We have to prove: sin^4(theta) - cos^4(theta) = sin^2(theta) - cos^2(theta)

First let's write the terms in a standard form and use x instead of theta.

So we have to prove (sin x)^4 - (cos x)^4 = (sin x)^2 - (cos x)^2

Start with the left hand side:

(sin x)^4 - (cos x)^4

we use the relation x^2 - y^2 = (x - y)(x + y)

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

we know that [(sin x)^2 + (cos x)^2] = 1

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

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

which is the right hand side.

**This proves that sin^4(theta) - cos^4(theta) = sin^2(theta) - cos^2(theta)**

We'll consider the difference of the terms from the left side as a difference of squares:

[(sin theta)^2]^2 -[(cos theta)^2]^2 = [(sin theta)^2 - (cos theta)^2][(sin theta)^2 +(cos theta)^2]

But, from the fundamental formula of trigonometry, we could substitute [(sin theta)^2 +(cos theta)^2 = 1

**(sin theta)^4 -(cos theta)^4 = [(sin theta)^2 - (cos theta)^2] q.e.d.**