Question

Quick Trig question: http://prntscr.com/47dg11

Answer 1

\[e^{ix}=cosx+isinx \\
e^{-ix}=cosx-isinx\]
\[cosx=\frac{e^{ix}+e^{-ix}}{2}\]
\[sinx=\frac{e^{ix}-e^{-ix}}{2i}\]

\[sin(3x)+cos(x)=\frac{e^{i3x}-e^{-i3x}}{2i}+\frac{e^{ix}+e^{-ix}}{2}\]

Answer 2

now lets break that e^(i3x) stuff down

Answer 3

I don't remember seeing this anywhere in the entire chapter on this stuff!

Answer 4

\[sin(3x)+cos(x)=\frac{e^{i3x}-e^{-i3x}}{2i}+\frac{e^{ix}+e^{-ix}}{2}\\
=\frac {(e^{ix})^{3}-(e^{-ix})^{3}}{2i}+\frac{e^{ix}+e^{-ix}}{2}\\
=\frac{(cos(x)+isin(x))^3-(cos(x)-isin(x))^3}{2i}+cosx\]
we can also use the fact that 1/i=-i to simplify further, but lets see how this simplfies right now

Answer 5

ok. I'm following

Answer 6

okay so umm lets just use some quick binomial expansion

Answer 7

ok

Answer 8

Wait, I have a quick off topic question...
Does 2sin2x = sin4x?

Answer 9

@Kainui

Answer 10

\[
=\frac{(cos(x)+isin(x))^3-(cos(x)-isin(x))^3}{2i}+cosx\\
\]
let a = cos x
let b = isin x
\[\frac{(a+b)^3-(a-b)^3}{2i}+cosx\]
\[\frac{a^3+3a^2b+3ab^2+b^3-(a^3-3a^2b+3ab^2-b^3)}{2i}+cosx\]
\[\frac{6a^2b+2b^3}{2i} +cosx\]
\[\frac{3cos^2x*isinx+i^3sin^3x}{i}+cosx\]
\[3cos^2x*sinx-sin^3x+cosx\]

Answer 11

Nope, 2sin(2x) is twice as high as sin(x) and makes waves twice as fast.

sin(4x) is the same height has sin(x) gets but waves four times as fast.

Answer 12

ok, Kainui

Answer 13

Thanks, @RaphaelFilgueiras

Answer 14

oh question had -cos x...

Answer 15

I actually have a few more questions...

Answer 16

you are welcome

Answer 17

\[3cos^2x*sinx-sin^3x-cosx\]

Answer 18

Posting a few more question now.

Answer 19

ok :)