Solving a Sin Equation: Find Value of cos6x-4cos4x+8cos2x

[Saitama]

Homework Statement

If sin x + sin2x + sin3x= 1, then find out the value of cos6x-4cos4x+8cos2x.

Homework Equations

The Attempt at a Solution

How should i start?
I don't find any way to convert them to cos 6x or cos 4x.

 

[SammyS]

Try using \displaystyle \sin \theta + \sin \varphi = 2 \sin\left( \frac{\theta + \varphi}{2} \right) \cos\left( \frac{\theta - \varphi}{2} \right) to combine sin(x) + sin(3x) .

-- Just a possibility.

 

[Saitama]

Try using \displaystyle \sin \theta + \sin \varphi = 2 \sin\left( \frac{\theta + \varphi}{2} \right) \cos\left( \frac{\theta - \varphi}{2} \right) to combine sin(x) + sin(3x) .

-- Just a possibility.

Using this identity i get:-
2sin2xcosx+sin2x=1

But what next?

 

[eumyang]

I don't think the problem was copied correctly. I've seen this problem before. The numbers in front of the x's are supposed to be exponents, not multiples of angles.

The problem should be as follows:
If
\sin x + \sin^2 x + \sin^3 x = 1,
then find out the value of
\cos^6 x - 4\cos^4 x + 8\cos^2 x.

Here's a hint, and hopefully, it's not a big one:

Spoiler

Rewrite as
\sin x + \sin^3 x = \cos^2 x.
Then square both sides and use the identity
\sin^2 x =1 - \cos^2 x.
You should eventually get the answer.

Mods: if this is too big of a hint, then please delete.

 

[Saitama]

I don't think the problem was copied correctly. I've seen this problem before. The numbers in front of the x's are supposed to be exponents, not multiples of angles.

The problem should be as follows:
If
\sin x + \sin^2 x + \sin^3 x = 1,
then find out the value of
\cos^6 x - 4\cos^4 x + 8\cos^2 x.

Here's a hint, and hopefully, it's not a big one:

Spoiler

Rewrite as
\sin x + \sin^3 x = \cos^2 x.
Then square both sides and use the identity
\sin^2 x =1 - \cos^2 x.
You should eventually get the answer.

Mods: if this is too big of a hint, then please delete.

You're right. I am very sorry for my foolishness. Please pardon me.