## Trigonometric Identities

FIG. T-15.
**Trigonometric identities.**

- From complex analysis we learn that a complex number $z=x+iy=r(\cos \theta +i\sin \theta )=re^{i\theta }$ where $r={\sqrt {x^{2}+y^{2}}}$ is called the
*modulus* and $\theta =\arctan(y/x)$ is the called the *phase*.

- $\cos \theta ={\frac {e^{i\theta }+e^{-i\theta }}{2}}$ and $\sin \theta ={\frac {e^{i\theta }-e^{-i\theta }}{2i}}$

- $\cos ^{2}\theta ={\frac {1}{2}}\left[1+\cos 2\theta \right]$

- $\sin ^{2}\theta ={\frac {1}{2}}\left[1-\cos 2\theta \right]$

- $\cos(\theta _{1}pm\theta _{2})=\cos \theta _{1}\cos \theta _{2}mp\sin \theta _{1}\sin \theta _{2}$

- $\sin(\theta _{1}pm\theta _{2})=\sin \theta _{1}\cos \theta _{2}pm\sin \theta _{2}\cos \theta _{1}$

- $\cos \theta _{1}\cos \theta _{2}={\frac {1}{2}}\left[\cos(\theta _{1}+\theta _{2})-\cos(\theta _{1}-\theta _{2})\right]$

- $\sin \theta _{1}\sin \theta _{2}={\frac {1}{2}}\left[\cos(\theta _{1}-\theta _{2})-\cos(\theta _{1}+\theta _{2})\right]$

- $\sin \theta _{1}\cos \theta _{2}={\frac {1}{2}}\left[\sin(\theta _{1}+\theta _{2})+\sin(\theta _{1}+\theta _{2})\right]$

## Deriving the identities

We can show that it is entirely plausible that the identities above are true.

### Euler's relation

$\cos(\theta )+i\sin(\theta )=e^{i\theta }.$

We begin by formally writing the Taylor series representations of
$\cos(\theta )$ and $i\sin(\theta )$ and sum the resulting series

$\cos(\theta )=\sum _{k=0}^{infty}{\frac {(-1)^{k}\theta ^{2k}}{(2k)!}}=\sum _{k=0}^{infty}{\frac {(i)^{2k}\theta ^{2k}}{(2k)!}}=\sum _{k=0}^{infty}{\frac {(i\theta )^{2k}}{(2k)!}},$

where $k=0,1,2,3,...$ We may write the Taylor series form of $i\sin(\theta )$

$i\sin(\theta )=i\sum _{l=0}^{infty}{\frac {(-1)^{l}\theta ^{2l+1}}{(2l+1)!}}=\sum _{l=0}^{infty}{\frac {(i)^{2l+1}\theta ^{2l+1}}{(2l+1)!}}=\sum _{l=0}^{infty}{\frac {(i\theta )^{2l+1}}{(2l+1)!}},$

where $l=0,1,2,3,...$ . In both of the previous cases, free use has been made of the identity $-1=i^{2}.$

The \sum of these two series yields the series representation of the exponential function

$\cos(\theta )+i\sin(\theta )=\sum _{n=0}^{infty}{\frac {(i\theta )^{n}}{n!}}=e^{i\theta },$
where $n=0,1,2,3,...$.

\thetas is not a proof of Euler's relation. It is only an argument of plausibility, because we have not proven the many theorems
that would allow us to establish the existence of the complex exponential, the Taylor series, or its convergence. Rarely will \thetas
argument of plausibility appear in a textbook on complex analysis, owing to its heuristic nature.

### Complex exponential form of trigonometric functions

Given that

$e^{i\theta }=\cos(\theta )+i\sin(\theta )$
and that, owing to the eveness of cosine and the oddness of sine,
we may write the complex conjugate of $\exp(i\theta )$ as

$e^{-i\theta }=\cos(\theta )-i\sin(\theta )$ .

The complex exponential form of sine and cosine follow from the respective sum and difference of $\exp(i\theta )$
and $\exp(-i\theta )$.

The other identities derive from multiplication and division of the complex exponential form of sine and cosine.