De Moivre's Theorem

[tall lichen]

Hey guys! Can someone verify if this is even true??😭

I've tried it a few times, and the math just isn't mathing. Did anyone actually manage to prove this, or is it just a typo in the question? I don't want to keep bashing my head against a wall if the question is broken T_T T_T

[simple urchin]

cos(6(0)) = 1

RHS: cos^6(0) = 1
-3cos^4 (0) = -3
3cos^2(0) = +3
-1

adding them up: 1 - 3 + 3 - 1 = 0
then its 1 = 0, which is a clear contradiction
hence this is NOT an identity in general

[tall lichen]

Thanks! If it's not a general identity, do you think the question itself is wrong? Or is there another way to approach this that I’m just totally missing? T_T

[atomic pecan]

its quicker to approach like this...but you can also do it the harder way by breaking cos6a in 2(cos3a)²-1
after this whatever way you try to expand in cosa terms the term cos⁶a will always end up having a coefficient cuz it'll be multiplied by 2 in the end
¯_(ツ)_/¯

[worthy fiber]

if you provide a counter-example then it's wrong

for example sps something = another thing

this will create an identity like in the above

BUT if you find even one counter-example to this

it'll be considered false

[wraith escarp]

we can make it a quadratic by substituting cos^2(theta) = t

[worthy fiber]

yeah that also works

[wraith escarp]

ye, then we get the roots

i'll find tht

t = 1

[floral dragon]

using de moivres i got $32\cos^6(\theta) -48\cos^4(\theta) + 18\cos^2(\theta) - 1$

[iron vector]

spaghetti

[wraith escarp]

hm..it can be turned to a cubic

[wraith escarp]

well.. the question isint always true..

true for cos = 2n(pie) or 2n+1(pie)

damn i forgot it, i have to revose

[simple urchin]

maybe its an equation rather than identity

[worthy fiber]

"show that" usually implies that you need to prove the identity is true

if it was an equation it would explicitly say "solve for t"

[simple urchin]

fair point

[wraith escarp]

ye, but it isint an identity