Question

prove this

Answer 1

what to prove?

Answer 2

\[\cos n \theta = \cos^{n} \theta -\left(\begin{matrix}n \\ 2\end{matrix}\right) \cos^{n-2}\theta \sin^{2} \theta + \left(\begin{matrix}n \\ 4\end{matrix}\right) \cos^{n-4} \theta \sin^{4}\theta - ......\]

Answer 3

can you help me guys??

Answer 4

note:
\[\left(\begin{matrix}n \\ m\end{matrix}\right) = \frac{ n! }{ (n-m) ! m!}\]

Answer 5

are you alowed to use complex analysis?

Answer 6

then it's easy

Answer 7

how to prove it??

Answer 8

1
11
121
1441
_what do they call this pyramid again??

that statement is somewhat related to it if im not mistaken

surely i already forgot that topic but i hope i could still remember it

Answer 9

i mean 1331 not 1441

Answer 10

use de Moivre's formula for:
\((cos\theta + i sin\theta)^n=\cos n\theta+i\sin n\theta\)
so it's a binomial expantion. To find \(\cos n\theta\) make equal real parts of bouth sides of the equality

Answer 11

got it?

Answer 12

using
http://hyperphysics.phy-astr.gsu.edu/%E2%80%8Chbase/imgmth/algb1.gif

yes. , i got it! thank you

Answer 13

yes, rigth that's the binomial expantion theorem