Difference between revisions of "Angle addition identities"

Revision as of 19:20, 7 September 2023

The trigonometric angle addition identities state the following identities:

$\sin(x + y) = \sin (x) \cos (y) + \cos (x) \sin (y)$ $\cos(x + y) = \cos (x) \cos (y) - \sin (x) \sin (y)$ $\tan(x + y) = \frac{\tan (x) + \tan (y)}{1 - \tan (x) \tan (y)}$

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 <math>\cos(x + y) = \cos (x) \cos (y) - \sin (x) \sin (y)</math>
 <math>\tan(x + y) = \frac{\tan (x) + \tan (y)}{1 - \tan (x) \tan (y)}</math>
-There are many proofs of these identities. For the sake of brevity, we list only one here.
-Euler's identity states that <math>e^{ix} = \cos (x) + i \sin(x)</math>. 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 <math>\cos (x) \cos (y)</math>, 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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Difference between revisions of "Angle addition identities"

Revision as of 19:20, 7 September 2023

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