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