# Sin^4Ө =3/8-3/8cos(2Ө) Prove the following trigonometric identity

## Homework Statement

Prove the following trigonometric identity. The question is sin^4Ө =3/8-3/8cos(2Ө)

## Homework Equations

I think I'm supposed to use the power reducing formulas for trigonometric identities which are
sin^2(u)= (1- cos(2u))/2
cos^2(u)=(1+cos(2u))/2
*Let u represent any integer/value*

## The Attempt at a Solution

To expand the equation I separated the equation into (1- cos^2Ө)(1- cos^2Ө) = 1- 2cos^2Ө + (cos^2Ө)^2. I reduced it because sin^2Ө = 1 – cos^2Ө and after this I'm confused on what the next steps are.

BvU
Homework Helper
So, a much better post than the first try. Good !

Your other entry is from the righthand side: what have you got for ##\cos(2\theta)## that might be useful here ?

 ah: your second relevant equation !

jedishrfu
Mentor
So you applied the sin^2(u) to the lefthand side of the equation and got (1/4) * (1 - cos(2u) )^2 right?

Next expand the

(1/4)*(1 - cos(2u) )^2 = (1/4) * ( 1 - 2*cos(2u) - cos(2u)^2 )

next apply the cos^2(u) rule to the last term and see what you get.

bubbly616
As well as the power reducing formulas, you'll want to glance at the double angle formulas.

bubbly616
BvU
Homework Helper

## Homework Statement

Prove the following trigonometric identity. The question is sin^4Ө =3/8-3/8cos(2Ө)

## Homework Equations

I think I'm supposed to use the power reducing formulas for trigonometric identities which are
sin^2(u)= (1- cos(2u))/2
cos^2(u)=(1+cos(2u))/2
*Let u represent any integer/value*

## The Attempt at a Solution

To expand the equation I separated the equation into (1- cos^2Ө)(1- cos^2Ө) = 1- 2cos^2Ө + (cos^2Ө)^2. I reduced it because sin^2Ө = 1 – cos^2Ө and after this I'm confused on what the next steps are.
But now I see a problem coming up:
let ##\theta = \pi/2\ \ ## then ##(sin\theta)^4 = 1\ \ ## and ##\ \ 3/8-3/8\cos(2\theta)=3/4## !?
So no identity at all !
Or did I read the original thingy in the wrong way ?

bubbly616
haruspex
Homework Helper
Gold Member
2020 Award
But now I see a problem coming up:
let ##\theta = \pi/2\ \ ## then ##(sin\theta)^4 = 1\ \ ## and ##\ \ 3/8-3/8\cos(2\theta)=3/4## !?
So no identity at all !
Or did I read the original thingy in the wrong way ?
I believe the identity ought to read ##\sin^4(\theta)=\frac 38-\frac 12\cos(2\theta)+\frac 18\cos(4\theta)##.
Looks like someone turned that ##4\theta## into another ##2\theta##.

bubbly616, Qwertywerty and jedishrfu
Why would u have to be an integer?