Does $\sin^2 x - \cos^2 x = 1-2\cos^2 x$? I am finishing a proof.  It seems like I can use $\cos^2 + \sin^2 = 1$ to figure this out, but I just can't see how it works.  So I've got two questions.  
Does $\sin^2 x - \cos^2 x = 1-2\cos^2 x$? 
And if it does, then how?
 A: Not really different but just another way of looking at it. Start with $\sin^2x+\cos^2x=1$ and subtract $2\cos^2x$ from both sides. Done!
A: Here is my favorite way to verify trigonometric identities:
First note that the equation of a circle gives us the rational parameterizations
$$\sin\theta=\frac{2t}{1+t^2}\qquad\cos\theta=\frac{1-t^2}{1+t^2}.$$
Substitute these expressions in. Now the equation we want to verify is
$$\left(\frac{2t}{1+t^2}\right)^2-\left(\frac{1-t^2}{1+t^2}\right)^2\overset{?}{=}1-2\left(\frac{1-t^2}{1+t^2}\right)^2.$$
Now just find a common denominator and compare numerators, so we want to know
$$(2t)^2-(1-t^2)^2\overset{?}{=}(1+t^2)^2-2(1-t^2)^2.$$
But this is true because
$$(1+t^2)^2-(1-t^2)^2=(1+2t^2+t^4)-(1-2t^2+t^4)=4t^2=(2t)^2,$$
thus the identity is true.
A: Observe that
$$
\begin{align*}
\sin^2(x)-\cos^2(x)&=\sin^2(x)+\bigl(\cos^2(x)-\cos^2(x)\bigr)-\cos^2(x)\\
&= (\sin^2(x)+\cos^2(x))-2\cos^2(x)\\
&= 1-2\cos^2(x).
\end{align*}
$$
More easily, just subtract $2\cos^2(x)$ from both sides of $\sin^2(x)+\cos^2(x)=1$ to get the result.
A: To go with your idea, you could try solving $\cos^2 x + \sin^2 x = 1$ for $\sin^2 x - \cos^2 x$. i.e. do algebraic manipulations to make the left hand side of the equation $\sin^2 x - \cos^2 x$.
