A test question I received and got wrong stated that $$\sin^4x-\cos^4x = \cos2x$$ After solving the equation from lower powers of tragicomic functions it came out $$\frac{-1}{2}(\cos2x)-\frac{1}{2}(\cos2x)=-cos2x$$ which ≠ RHS. Our professor showed that it is correct by solving it as $$(\sin^2x-\cos^2x)(\sin^2x+\cos^2x) =(\sin^2x-\cos^2x)(1) = \cos2x$$ which works out correctly. Who is right? Here is a Wolfram Alpha solution that shows the solution I came up with to be correct.

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    $\begingroup$ "tragicomic"? Really? $\endgroup$
    – Blue
    May 5, 2015 at 6:06
  • $\begingroup$ $-\cos ax \ne \cos ax$ as $-1 \ne 1$. Your answer looks right, or the professor or problem missed a sign. $\endgroup$
    – zahbaz
    May 5, 2015 at 6:22

2 Answers 2


Note that $\cos^2x-\sin^2x=\cos2x$

So your equation gives $$(\sin^2x-\cos^2x)(\sin^2x+\cos^2x)$$ $$=(-\cos2x)(1)$$


So the mistake was a confusion of $\cos^2x-\sin^2x \neq \sin^2x-\cos^2x$.

The following is true: $\cos^2x-\sin^2x = \cos2x$


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