# Verify the identity $\cos^2x-\sin^2x = 2\cos^2x-1$

I am having problems understanding how to verify this identity. I am quite sure that it is to be solved using the Pythagorean identities but, alas, I'm not seeing what might otherwise be obvious.

I need to verify the identity

$$\cos^2x-\sin^2x = 2\cos^2x-1$$

• You know that $\cos^2x+\sin^2x=1$, then write $\sin^2x$ in terms of $\cos^2x$, then substitute your expression in $\cos^2x-\sin^2x$ and simplify. – Hakim Nov 17 '14 at 13:02
• find some way of getting rid $\sin^2x$ on the LHS. – John Joy Nov 17 '14 at 15:00

$$\cos^2 x + \sin^2 x = 1 \iff \color{blue}{\sin^2 x = 1-\cos^2 x}$$
\begin{align}\cos^2x-\color{blue}{\sin^2x} & = \cos^2 x - \color{blue}{(1-\cos^2 x)}\\\\ &= \cos^2 x - 1 + \cos^2 x \\ \\ & = 2\cos^2 x - 1\end{align}
Remember that $sin^2(x) = 1 - cos^2(x)$.
Then $cos^2x - sin^2x = ?$
we have $\cos(x)^2-\sin(x)^2=\cos(x)^2-(1-\cos(x)^2)=2\cos(x)^2-1$ since $\sin(x)^2+\cos(x)^2=1$