Angle addition identities

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The trigonometric angle addition identities state the following identities:

$\sin(x + y) = \sin (x) \cos (y) + \cos (x) \sin (y)$ $\cos(x + y) = \cos (x) \cos (y) - \sin (x) \sin (y)$ $\tan(x + y) = \frac{\tan (x) + \tan (y)}{1 - \tan (x) \tan (y)}$ There are many proofs of these identities. For the sake of brevity, we list only one here.

Euler's identity states that $e^{ix} = \cos (x) + i \sin(x)$. We have that\begin{align*} \cos (x+y) + i \sin (x+y) &= e^{i(x+y)} \\ &= e^{ix} \cdot e^{iy} \\ &= (\cos (x) + i \sin (x))(\cos (y) + i \sin (y)) \\ &= (\cos (x) \cos (y) - \sin (x) \sin(y)) + i(\sin (x) \cos(y) + \cos(x) \sin(y)) \end{align*}By looking at the real and imaginary parts, we derive the sine and cosine angle addition formulas.

To derive the tangent addition formula, we reduce the problem to use sine and cosine, divide both numerator and denominator by $\cos (x) \cos (y)$, and simplify.\begin{align*} \tan (x+y) &= \frac{\sin (x+y)}{\cos (x+y)} \\ &= \frac{\sin (x) \cos(y) + \cos(x) \sin(y)}{\cos (x) \cos (y) - \sin (x) \sin(y)} \\ &= \frac{\frac{\sin(x)}{\cos(x)} + \frac{\sin(y)}{\cos(y)}}{1 - \frac{\sin (x) \sin(y)}{\cos (x) \cos(y)}} \\ &= \frac{\tan (x) + \tan (y)}{1 - \tan (x) \tan(y)} \end{align*}as desired.

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