How to simplify $\cos(x)\cos(x)-\sin(x)\sin(x)$ to $\cos(2x)$ I am trying to work on calculus, finding extrema, convacity, critical numbers and so forth but when It comes to trig problems I am struggling with problems where it is not as simple as trig$(x) = n$.
For example, how would I simpify $\cos(x)\cos(x)-\sin(x)\sin(x)$ to $\cos(2x)$ so I can solve for it? I've been told several times how it simplifies but I don't understand how.
 A: we have $\cos(2x)=\cos(x+x)=\cos(x)\cos(x)-\sin(x)\sin(x)=\cos(x)^2-\sin(x)^2$
A: It comes from the identity $\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)$.
Using complex numbers we know that $e^{ix} = \cos(x) + i\sin(x)$ therefore;
\begin{equation}
e^{i(x+y)} = e^{ix}e^{iy} = (\cos(x) + i\sin(x))(\cos(y) + i\sin(y)) \\
= \cos(x)\cos(y) -\sin(x)\sin(y) + i\cos(x)\sin(y) + i\cos(y)\sin(x)
\end{equation}
Now we have $Re\{e^{i(x+y)}\} = \cos(x+y)$ so we take the real part of the above to get:
\begin{equation}
\cos(x+y) = \cos(x)\cos(y) -\sin(x)\sin(y)
\end{equation}
Now substituting $y=x$ into that we get the desired identity.
A: It depends on what you are aiming for.  $\cos(x) \cos(x)- \sin(x) \sin(x)= \cos^2(x) - \sin^2(x)\\ = \cos^2(x) - (1-\cos^2(x))= 2cos^2(x)-1\\ $ 
This last part is true because $\sin^2(x)+\cos^2(x)=1$.  This also implies the following $\cos^2(x) - \sin^2(x)\\ = 1-\sin^2(x) - \sin^2(x) = 1 - 2\sin^2(x) $
A: 
Apply the Law of Cosines to the diagram of a reflected right triangle (above) to derive
$$\cos2\theta = 1-2\sin^2\theta$$
and note that
$$1-2\sin^2\theta = (1-\sin^2\theta)-\sin^2\theta=\cos^2\theta-\sin^2\theta$$
