Answer
Proof
Work Step by Step
We have to prove that :- $\;\sin ^2x+\cos ^2x=1$
According to question we will use the result (1) and (2)
\[\cos bx=\frac{1}{2}\left(e^{ibx}+e^{-ibx}\right)\;\;\;...(1)\]
\[\sin bx=\frac{1}{2i}\left(e^{ibx}-e^{-ibx}\right)\;\;\;...(2)\]
Using (1) and (2) with $b=1$
\[\cos x=\frac{e^{ix}+e^{-ix}}{2}\;\;\;...(3)\]
\[\sin x=\frac{e^{ix}-e^{-ix}}{2i}\;\;\;\]
\[\Rightarrow \sin x=\frac{(e^{-ix}-e^{ix})i}{2}\;\;\;...(4)\]
Using (3)
\[\cos ^2x=\frac{e^{2ix}+e^{-2ix}+2}{4}\;\;\;...(5)\]
Using (4)
\[\sin ^2x=\frac{-(e^{-2ix}+e^{2ix}-2)}{4}\;\;\;...(6)\]
Adding (5) and (6)
\[\cos ^2x+\sin ^2x=\frac{e^{2ix}+e^{-2ix}+2}{4}+
\frac{-(e^{-2ix}+e^{2ix}-2)}{4}\]
\[ \cos ^2x+\sin ^2x =\frac{4}{4}=1\]
Hence $\;\;\cos ^2x+\sin ^2x=1$.
