Why is the e in the LHS equals to cos + isin in the RHS?

Question

In the Method 1 and 2 how is the $e$ in the LHS equal to the RHS?

In the junior classes I was just given to learn that $cos(\frac{π}{2}±\theta)=\mp sin\theta$ but we never proved it.

I have no idea how the e got there in the first statement of the proof.

Can anyone prove the first statement of the proof of Method 1 and 2?

Any help would be appreciated.

how the $e$ got there: $e^{ix}=\cos{x}+i\sin x$ is Euler's formula — J. W. Tanner, Nov 19 '19 at 01:38

score 1 · Accepted Answer · answered Nov 19 '19 at 01:46

1

$e^{ix}=\cos{x}+i\sin x$ is Euler's formula.

Also, $\cos\left(\dfrac\pi2\pm\theta\right)=\mp \sin\theta$ follows from the angle addition formula

$\cos\left(\dfrac\pi2\pm\theta\right)=\cos\left(\dfrac\pi2\right)\cos(\theta)\mp\sin\left(\dfrac\pi2\right)\sin(\theta)$.

answered Nov 19 '19 at 01:46

J. W. Tanner

Euler's formula could be proved using power series – J. W. Tanner Nov 19 '19 at 02:05

1 Answers1