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

[bubbly616]

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]

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 ?

[edit] ah: your second relevant equation !

 

[jedishrfu]

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.

 

[laplacean]

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

 

[BvU]

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 ?

 

[haruspex]

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 $2\theta$.

 

[HOI]

Why would u have to be an integer?