Variation of De Moivre's formula

Question

I heard that in general De Moivre's formula:

$$(\sin a + i \cos a)^n= \sin na + i \cos na $$ can be used when n has a particular (real number? ) condition.

I can't figure out how to narrow the exponent down (specialize it) to an integer.

Help me please.

Original De Moivre formula is (cos(a)+isin(a))n=cos(na)+isin(a) and when we switch sin and cos, n needs more condition than only integer and I don't know what it is. — wormname, Nov 22 '19 at 13:07

score 0 · Answer 1 · answered Nov 22 '19 at 13:07

0

Use How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?

$$\sin A+i\cos A =i(\cos(- A)+i\sin(- A))=e^{i\pi/2}\cdot e^{-iA}=e^{i(\pi/2-A)}$$

Similarly, $$\sin nA+i\cos nA =e^{i(\pi/2-nA)}$$

$$(e^{i(\pi/2-A)})^n$$ will be $$=e^{i(\pi/2-nA)}$$

if $n\dfrac\pi2-nA-\left(\dfrac{\pi}2-nA\right)=2m\pi$ for some integer $m$

$$\dfrac{(n-1)\pi}2=2m\pi\iff n-1=4m$$

answered Nov 22 '19 at 13:07

lab bhattacharjee

274,582

Thank you very much. – wormname Nov 22 '19 at 13:12

score 0 · Answer 2 · answered Nov 22 '19 at 13:09

Note that$$\sin(\alpha)+i\cos(\alpha)=i\bigl(\cos(\alpha)-i\sin(\alpha)\bigr)=i\bigl(\cos(-\alpha)+i\sin(-\alpha)\bigr).$$So,\begin{align}\bigl(\sin(\alpha)+i\cos(\alpha)\bigr)^n&=i^n\bigl(\cos(-n\alpha)+i\sin(-n\alpha)\bigr)\\&=i^n\cos(-n\alpha)+i^{n+1}\sin(-n\alpha)\\&=i^n\cos(n\alpha)-i^{n+1}\sin(n\alpha).\end{align}Can you take it from here?

Variation of De Moivre's formula

2 Answers2