How to deduce trigonometric formulae like $2 \cos(\theta)^{2}=\cos(2\theta) +1$? Very important in integrating things like $\int \cos^{2}(\theta) d\theta$ but it is hard for me to remember them. So how do you deduce this type of formulae? If I can remember right, there was some $e^{\theta i}=\cos(\theta)+i \sin(\theta)$ trick where you took $e^{2 i \theta}$ and $e^{-2 i \theta}$. While I am drafting, I want your ways to remember/deduce things (hopefully fast).
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*About TPV's suggestion, how do you attack it geometrically??


$\cos^{2}(x) - \sin^{2}(x)=\cos(2x)$
plus $2\sin^{2}(x)$, then
$\cos^{2}(x)+\sin^{2}(x)=\cos(2x)+2\sin^{2}(x)$
and now solve homogenous equations such that LHS=A and RHS=B, where $A\in\mathbb{R}$ and $B\in\mathbb{R}$. What can we deduce from their solutions?

In my experience, almost all trigonometric identities can be obtained by knowing a few values of $\sin x$ and $\cos x$, that $\sin x$ is odd and $\cos x$ is even, and the addition formulas:
\begin{align*}
\sin(\alpha+\beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta,\\
\cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta.
\end{align*}
For example, to obtain the classic $\sin^2x + \cos^2x = 1$, simply set $\beta=-\alpha$ in the formula for the cosine, and use the facts that $\cos(0)=1$ and  that $\sin(-a)=-\sin(a)$ for all $a$.
For the one you have, we use the formula for the cosine with $\alpha=\beta=\theta$ to get
$$\cos(2\theta) = \cos^2\theta - \sin^2\theta.$$
Then using $\sin^2 \theta+\cos^2\theta=1$, so $-\sin^2\theta = \cos^2\theta-1$ gives
$$\cos(2\theta) = \cos^2\theta +\cos^2\theta - 1 = 2\cos^2\theta - 1.$$
If you know the basic values (at $\theta=0$, $\frac{\pi}{6}$, $\frac{\pi}{4}$, $\frac{\pi}{3}$, $\frac{\pi}{2}$, $\pi$, $\frac{3\pi}{2}$), parity, and the addition formulas, you can get almost any of the formulas.
