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?
 A: 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.
A: It might be useful to remember that $\cos^2 x$ oscillates twice as fast as $\cos x$. This is something that people who work with alternating current know very well; the effect (which is proportional to the square of the current) has twice the frequency. For example, a light bulb flickers at 100 Hz if the AC frequency is 50 Hz. This means that $\cos^2 x$ should be "something with $\cos 2x$".
Next, since $\cos x$ oscillates between -1 and 1, $\cos^2 x$ will oscillate between 0 and 1. The average around which the curve oscillates will be 1/2, and the amplitude will also be 1/2 (so that you reach down to 0 and up to 1 from the central level 1/2). Thus $\cos^2 x$ should be "1/2 + (1/2)*oscillating term".
Combining these two facts, it's not too hard to remember that
$$ \cos^2 x = \frac12 + \frac12 \cos 2x.$$
One has to be a little bit careful with the sign before the second term, but it must be plus if the formula is to hold when $x=0$.
If you choose the minus sign, you get the related formula
$$ \sin^2 x = \frac12 - \frac12 \cos 2x.$$
(So $\sin^2 x$ also oscillates around 1/2 with amplitude 1/2 and twice the frequency. Note that when you add the two formulas up, the oscillations cancel, and you get $\cos^2 x + \sin^2 x = 1/2+1/2 = 1$.)
A: 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:


*

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

*$\displaystyle \tan{3\alpha} = \frac{3\tan\alpha - \tan^{2}\alpha}{1-3\tan^{2}\alpha}$. Look here
A: By far, the best approach to trigonometric formulae is to use Euler's identity and its corollary, de Moivre's formula, recalling that $\sin^2 + \cos^2 = 1$.
A: For this you need only two very basic trigonometric identities:
$$ \cos^2(x) + \sin^2(x) = 1 $$
and
$$ \cos^2(x) - \sin^2(x) = \cos(2x) $$
Add or subtract them, and you have the identities for $\cos^2(x)$ and $\sin^2(x)$.
A: All of trig identities can be derived from $\exp{(\hat{i}\theta)}=\cos(\theta)+\hat{i}\sin\theta$. How?
Example
$$\cos(2\,x) = \mathrm{Re}\left(\exp{(2\hat{i}\,x)}\right)$$ 
$$ = \mathrm{Re}\left(\exp{(\hat{i}x)}\,\exp{(\hat{i}x)}\right) $$
$$ = \mathrm{Re}\left((\cos x+\hat{i}\sin x)\,(\cos x+\hat{i}\sin x)\right) $$
$$ = \mathrm{Re}\left( \cos^2 x - \sin^2 x +2\hat{i}\cos x \sin x \right) $$
$$ = \cos^2 x - \sin^2 x $$
A: Here's yet another method for your amusement.
Let $y = \cos^2(t)$.  Then note that $y'(t) = 2\cos(t)\sin(t)$ and therefore 
$$
y''(t) = 2\cos^2(t) - 2\sin^2(t) = 2\cos^2(t) - 2(1 - \cos^2(t)) = 4 y - 2.
$$
The general solution to this nonhomogeneous ode is
$$
y = C_1 \sin(2t) + C_2 \cos(2t) + 1/2
$$
and since we know that $y$ satisfies the initial conditions $y(0) = 1$ and $y'(0) = 0$, we can solve for the coefficients and get $C_1 = 0$ and $C_2 = 1/2$.
