# Math Help cos^6 x

Rewrite the expression in terms of the first power of the cosine:

cos^6 x

My guess would be that it would be cos(x)*cos(x)*cos(x)*cos(x)*cos(x)*cos(x)

I'm not sure if that is it, or any way to start the problem even

edit :

cos^6(x)=cos^2(x)*cos^2(x)*cos^2(x)
=.5(1+cos2x)*.5(1+cos2x)*.5(1+cos2x)
then ???
?

Last edited:

## Answers and Replies

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Maybe $$\cos^2{x}=\frac{1+\cos{2x}}{2}$$ will help?

nolachrymose said:
Maybe $$\cos^2{x}=\frac{1+\cos{2x}}{2}$$ will help?

Isn't that what i have?

cos^6(x)=cos^2(x)*cos^2(x)*cos^2(x)
=.5(1+cos2x)*.5(1+cos2x)*.5(1+cos2x)

i am lost again

Last edited:
Hurkyl
Staff Emeritus
Gold Member
Do it again.

Hurkyl said:
Do it again.
what do you mean do it again

Logitech said:
Isn't that what i have?

cos^6(x)=cos^2(x)*cos^2(x)*cos^2(x)
=.5(1+cos2x)*.5(1+cos2x)*.5(1+cos2x)

i am lost again
Sorry, I must have posted right before you edited your original post.

nolachrymose said:
Sorry, I must have posted right before you edited your original post.
what would i need for the next step

I really really need help someone

futb0l
Hmmm.. maybe it's

$$\cos^6 x = (\cos^2 x)^3$$

so then ...

$$\cos^6 x = ( \frac { 1 + cos 2x } { 2 } )^3$$

how is cos2x = 1 + cos2x / 2 ???

isnt cos2x = 1-sin2x ??????

jai6638 said:
how is cos2x = 1 + cos2x / 2 ???

isnt cos2x = 1-sin2x ??????
$$cos(2x)=cos^2(x)-sin^2(x)$$
$$cos(2x)=cos^2(x)-(1-cos^2(x))$$
$$cos(2x)=2cos^2(x)-1$$
$$cos^2(x)= (\frac { 1 + cos 2x } { 2 } )$$

Since we can express cos(2x) and cos(3x) in terms of cos(x) it is a good idea to do so right away when you are asked such a question because you cannot break it up like (5x+x) because that would require a polynomial for cos(5x) in terms of cos(x) which you are not likely to remember or even derive easily. So the sensible thing to do would be to break it up as cos(3x+3x) or cos(2(3x)) and proceed further.