# Is $\sin^4 x-\cos^4 x = \cos2x$ or is it $-\cos2x=\cos2x$?

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.

• "tragicomic"? Really? – Blue May 5 '15 at 6:06
• $-\cos ax \ne \cos ax$ as $-1 \ne 1$. Your answer looks right, or the professor or problem missed a sign. – suneater May 5 '15 at 6:22

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: $\cos^2x-\sin^2x = \cos2x$