Answer
Question of proof
Work Step by Step
To Prove That:-
$\cos(bx)=\frac{e^{ibx}+e^{-ibx}}{2}$
and
$\sin(bx)=\frac{e^{ibx}-e^{-ibx}}{2i}$
We will use the Euler's Formula
\[e^{i\theta}=\cos\theta+i\sin\theta\]
Consider $\;e^{ibx}$
Using Euler's formula
$e^{ibx}=\cos (bx)+i\sin (bx)$ ___(1)
Consider $\;\;e^{-ibx}$
$e^{-ibx}=\cos (-bx)+i\sin (-bx)$
$\Rightarrow e^{-ibx}=\cos (bx)-i\sin (bx)$ ___(2)
Adding equation (1) and (2)
$2\cos(bx)=e^{ibx}+e^{-ibx}$
$\Rightarrow \cos(bx)=\frac{e^{ibx}+e^{-ibx}}{2}$
Subtracting equations (1) and (2)
$2i\sin (bx)=e^{ibx}-e^{-ibx}$
$\Rightarrow \sin(bx)=\frac{e^{ibx}-e^{-ibx}}{2i}$
Hence,
$ \cos(bx)=\frac{e^{ibx}+e^{-ibx}}{2}$
and
$\sin(bx)=\frac{e^{ibx}-e^{-ibx}}{2i}$
Hence prove.
