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?

For Proving $\sin(\alpha+\beta)=\sin\alpha\cdot \cos\beta + \cos\alpha \cdot \sin\beta$ you can see this link:


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*http://www.math.wisc.edu/~leili/teaching/math222s11/problems/quizzes/trig.pdf
By the above by substituting $\beta=\alpha$ you have 


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*$\sin{2\alpha} = \sin\alpha \cdot \cos\alpha + \cos\alpha \cdot \sin\alpha =2 \sin\alpha \cos\alpha$

*$\sin{2\alpha} = \frac{2 \tan\alpha}{1+\tan^{2}\alpha} = \displaystyle 2\tan{\alpha}\cdot\cos^{2}\alpha=2\sin\alpha \cdot \cos\alpha$
We have $\cos(\alpha + \beta) = \cos\alpha\cdot \cos\beta - \sin\alpha \cdot \sin\beta$. Using this you can obtain the following:


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*$\cos{2 \alpha} = \cos\alpha\cdot \cos\alpha - \sin\alpha\cdot \sin \alpha =\cos^{2} \alpha - \sin^{2} \alpha$

*$\cos{2 \alpha}= 1 - 2 \sin^{2}\alpha \qquad \Bigl[ \because \ \sin^{2}\theta = 1 - \cos^{2}\theta \Bigr]$

*$ \cos{2 \alpha}=2 \cos^{2}\alpha - 1$

*$\cos{2 \alpha}=\displaystyle \cos^{2}\alpha \cdot \Bigl[1-\frac{\sin^{2}\alpha}{\cos^{2}\alpha}\Bigr]=\displaystyle \frac{1-\tan^{2}\alpha}{1+\tan^{2}\alpha} \  \Bigl[\because \ \cos^{2}\alpha = \frac{1}{\sec^{2}\alpha}\Bigr]$.
Using the above formulas for $\sin(\alpha + \beta)$ and $\cos(\alpha + \beta)$ we have $$\tan(\alpha+\beta) = \frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)} = \frac{\sin\alpha\cdot \cos\beta + \cos\alpha \cdot \sin\beta}{\cos\alpha\cdot \cos\beta - \sin\alpha\cdot \sin\beta}$$
Dividing the last term by $\cos\alpha \cdot \cos\beta$ gives, $$\tan(\alpha + \beta) = \frac{\tan\alpha + \tan \beta}{1 - \tan\alpha \cdot \tan\beta}$$


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*From this if you put $\beta = \alpha$ you get $$\tan{2\alpha} = \frac{\tan\alpha + \tan\alpha}{1 - \tan\alpha \cdot \tan\alpha} = \frac{2 \tan \alpha}{1 - \tan^{2}\alpha}$$


Now moving on to formulas for $3\alpha$.
\begin{align*}
\sin{3\alpha} &= \sin(2\alpha + \alpha) = \sin(2\alpha)\cdot\cos{\alpha} + \cos(2\alpha)\cdot\sin\alpha \\ &= 2\sin\alpha\cdot \cos^{2}\alpha + (\cos^{2}\alpha - \sin^{2}\alpha) \cdot \sin\alpha \\ &= 2\sin\alpha \cdot (1-\sin^{2}\alpha) + (1-\sin^{2}\alpha)\sin\alpha - \sin^{3}\alpha \\ &= 3\sin{\alpha} - 4 \sin^{3}\alpha
\end{align*}
Again proceed similarly for $\cos{3\alpha}$ and convert all the $\sin^{2}\alpha$'s which appear in the expression to $(1-\sin^{2}\alpha)$. Then you get the value as:


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*$\cos(3\alpha) = 4\cos^{3}\alpha - 3 \cos\alpha$.

*$\displaystyle \tan{3\alpha} = \frac{3\tan\alpha - \tan^{2}\alpha}{1-3\tan^{2}\alpha}$. Look here
