Question

Can you help me?

Answer 1

look at the picture below

Answer 2

@UnkleRhaukus

Answer 3

@UnkleRhaukus come here please if you are not busy...

Answer 4

@waterineyes have you idea?/

Answer 5

If I have some Idea, then I would not have called Uncle Rocks here.. :)

Answer 6

this question is way beyond me

Answer 7

how about this

Answer 8

i have no idea.., :(

Answer 9

for the 1st one i got LHS to be \[\frac{ 1 }{ 2 }(-\cos(nx)+\cot \frac{ x }{ 2 }\sin(nx)+1)\]

Answer 10

@gerryliyana ??

Answer 11

still confused.., i have a question., how to transform
\[\sum_{n=0}^{N-1} (e^{ix})^{n} \]
to
\[\frac{ 1-e^{iNx} }{ 1-e^{ix} }\]

???

Answer 12

let \[\large{z=e^{ix}=\cos x + i\sin x\\z^n=e^{inx}=\cos nx + i\sin nx}\]
\[\large \sum_{n=0}^{N-1}z^n=\sum_{n=0}^{N-1}(\cos nx + i\sin nx)=\sum_{n=0}^{N-1}\cos nx+i\sum_{n=0}^{N-1}\sin nx\]
now
\[\large {\sum_{n=0}^{N-1}z^n=\frac{1-z^N}{1-z}=\frac{1-(\cos Nx+i\sin Nx)}{1-(\cos x+i\sin x)}\\\quad = \frac{(1-\cos Nx)-i\sin Nx}{(1-\cos x)-i\sin x}\cdot \frac{(1-\cos x)+i\sin x}{1-\cos x)+i\sin x}}\]

Answer 13

\[\quad =\frac{(1-\cos Nx)(1-\cos x)+\sin Nx \sin x}{(1-\cos x)^2+\sin^2 x}+i\frac{\sin x(1-\cos Nx)-\sin Nx(1-\cos x)}{(1-\cos x)^2+\sin^2x}\]
therefore,
\[\sum_{n=0}^{N-1}\cos nx=\frac{(1-\cos Nx)(1-\cos x)+\sin Nx \sin x}{(1-\cos x)^2+\sin^2 x}\] and
\[\sum_{n=0}^{N-1}\sin nx =\frac{\sin x(1-\cos Nx)-\sin Nx(1-\cos x)}{(1-\cos x)^2+\sin^2x}\]

just simplify the RHSs

Answer 14

here's the complete solution:

Answer 15

thank you so much @sirm3d

Answer 16

i got it now.., thank you again :)