Question

De Moivre's Theorem help. Will give medal and fan

Answer 1

Can I have a step by step explanation as to how to prove this problem? *attached below*

Answer 2

lol i was all set to go, until i saw the problem
not sure why it is demoivre, are you?

Answer 3

Haha, I'm not sure..the teacher said to use De Moivre to show it :(

Answer 4

complex numbers ?

Answer 5

Yes!

Answer 6

usually the gimmick is to get an expression and show that the real part is equal to whatever you know it is equal to
my first guess was to start with
\[\sin^2(x)+\cos^2(x)=1\] square it and get
\[\sin^4(x)+\cos^4(x)+2\sin^2(x)\cos^2(x)=1\] or
\[\sin^4(x)+\cos^4(x)=1-2\sin^2(x)\cos^2(x)\]
let me think of how we could finish it from there

Answer 7

or maybe if you have the stomach for it, you could actually compute \((\cos(x)+\sin(x))^4\) by hand
by which i mean multiply it out

Answer 8

oops i meant
\[(\cos(x)+i\sin(x))^4\]

Answer 9

we might could try that
one answer is
\[\cos(4x)+i\sin(4x)\] by demoivre, which is what gave me the hint

Answer 10

Okay! Thank you! Let me just process everything

Answer 11

i am getting there

Answer 12

it is also going to require some trig identity that i do not know, but i can tell you the idea of the proof using demoivre if you like

Answer 13

Okay..take your time :)
Yes, please tell me the idea of the proof

Answer 14

start with
\[\cos(x)+i\sin(x)\] and raise it to the power of 4
do it two ways, one way by hand, i.e. multiply it out
the other way (the easy way) is by demoivre

Answer 15

once you do it by hand, you will get a real part, and an imaginary part
i believe it will be
\[\cos^4(x)+\sin^4(x)-6\sin^2(x)\cos^2(x)+i(4\sin^3(x)\cos(x)-4\cos^3(x)\sin^3(x))\]
you can check that, i did it with wolfram sort of
ignore the imaginary part, and focus on the real part
\[\cos^4(x)+\sin^4(x)-6\sin^2(x)\cos^2(x)\]

Answer 16

when you do it using demoivre, it is much much easier
\[(\cos(x)+i\sin(x))^4=\cos(4x)+i\sin(4x)\]

Answer 17

but that tells you that both are the same, and the only way that is true is if the real part is equal the real part and the imaginary part is equal the imaginary part

Answer 18

that means
\[\cos^4(x)+\sin^4(x)-6\sin^2(x)\cos^2(x)=\cos(4x)\]

Answer 19

go from there to
\[\cos^4(x)+\sin^4(x)=\cos(4x)+6\sin^2(x)\cos^2(x)\]

Answer 20

and then some trig identity, i think a double angle formula or something , will finish it

Answer 21

Alrighty! Wow..good explanation! Thank you sooooo much!

Answer 22

yw
byw first method without demoivre works just as well

Answer 23

using the first method to get
\[\sin^4(x)+\cos^4(x)=1-2\sin^2(x)\cos^2(x)\] it is some trig identity to finish
i checked it with wolfram and it works here too

Answer 24

hence "tickonometry"

Answer 25

Ahhh! I get it (Y) I love how there's more than one day to do it. Thank you so much!