Question

Use mathematical induction to prove De Moivre's theorem

[ R (cos t + i sin t) ] n = R n(cos nt + i sin nt)


for n a positive integer.

Answer 1

@Preetha
@pooja195
@sarahnichols3
@BAdhi
@iambatman

Answer 2

@Marki

Answer 3

`this is the theorem `
\([ R (\cos ~t + i ~\sin ~t) ]^n = R^n(\cos~ nt + i~ \sin~ nt) \)
make sure next time to type it fully

Answer 4

maybe @mathmath333 free to solve it for u

Answer 5

note that
\([ R (\cos ~t + i ~\sin ~t) ]^n = R^n(\cos~ nt + i~ \sin~ nt)=e^{nR\theta}\)

Answer 6

[ R (cos t + i sin t) ] k R (cos t + i sin t) = R k(cos kt + i sin kt) R (cos t + i sin t) ?

Answer 7

(cos kt cos t - sin kt sin t) = cos(kt + t) = cos(k + 1)t

(sin kt cos t + cos kt sin t) = sin(kt + t) = sin(k + 1)t
????

Answer 8

\[(\cos kt + i \sin kt )(\cos t + i \sin t) =\\
[ \cos (kt)\cos t - \sin(kt)\sin t]+i[\sin(kt)\cos t+ \cos(kt)\sin t]\]

Use above identity of yours

Answer 9

[ R (cos t + i sin t) ] k + 1 = R k + 1 [ cos (k + 1)t + sin(k + 1)t ]

Answer 10

Please use brackets where needed tell us what else you need to know and where you are stuck :)

Answer 11

\(\large \begin{align} \color{black}{ (\cos x + i \sin x)^n = (\cos nx + i \sin nx) \hspace{.33em}\\~\\
\normalsize \text{for n=1} \hspace{.33em}\\~\\
(\cos x + i \sin x)^n \hspace{.33em}\\~\\
= (\cos x + i \sin x)^1 \hspace{.33em}\\~\\
= (\cos x + i \sin x)\hspace{.33em}\\~\\
\normalsize \text{True} \hspace{.33em}\\~\\
\normalsize \text{assume it is true for n=k, so} \hspace{.33em}\\~\\
(\cos x + i \sin x)^k = (\cos kx + i \sin kx)------(1) \hspace{.33em}\\~\\
\normalsize \text{prove that it is also true for (n=k+1) such that } \hspace{.33em}\\~\\
(\cos x + i \sin x)^{k+1} = (\cos (k+1)x + i \sin (k+1)x) \hspace{.33em}\\~\\~\\
\underline {~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ }\hspace{.33em}\\~\\
(\cos x + i \sin x)^{k+1} \hspace{.33em}\\~\\
=(\cos x + i \sin x)^{k}(\cos x + i \sin x) \hspace{.33em}\\~\\
=(\cos x + i \sin x)^{k}(\cos x + i \sin x) \hspace{.33em}\\~\\
=(\cos kx + i \sin kx)(\cos x + i \sin x) -------(from~~1)\hspace{.33em}\\~\\
=\cos kx\cdot \cos x + \cos kx\cdot i\cdot \sin x +i\cdot \sin kx\cdot \cos x+i\cdot \sin kx\cdot i \sin x\hspace{.33em}\\~\\
=\cos kx\cdot \cos x -\cdot \sin kx\cdot \sin x + i(\cos kx\cdot \cdot \sin x +\cdot \sin kx\cdot \cos x)\hspace{.33em}\\~\\
=\cos (kx+x) + i(\sin kx+x)\hspace{.33em}\\~\\
=\cos x(k+1) + i(\sin x(k+1))\hspace{.33em}\\~\\


}\end{align}\)