# Answer to Question #14863 in Trigonometry for geethanjali

Question #14863

Prove that: sin4A-cos4A=sin2A-cos2A

Expert's answer

This statement is wrong. Let's take A = π/12, for example:

sin4A-cos4A = sin4(π/12)-cos4(π/12) = sin(π/3) - cos(π/3) = √3/2 - 1/2.

But

sin2A-cos2A = sin2(π/12)-cos2(π/12) = sin(π/6)-cos(π/6) = 0.5 - √3/2

and

& radic;3/2 - 1/2 ≠ 0.5 - √3/2.

sin4A-cos4A = sin4(π/12)-cos4(π/12) = sin(π/3) - cos(π/3) = √3/2 - 1/2.

But

sin2A-cos2A = sin2(π/12)-cos2(π/12) = sin(π/6)-cos(π/6) = 0.5 - √3/2

and

& radic;3/2 - 1/2 ≠ 0.5 - √3/2.

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