Math Help cos^6 x

1. Dec 5, 2004

Logitech

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: Dec 5, 2004
2. Dec 5, 2004

nolachrymose

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

3. Dec 5, 2004

Logitech

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: Dec 5, 2004
4. Dec 5, 2004

Hurkyl

Staff Emeritus
Do it again.

5. Dec 5, 2004

Logitech

what do you mean do it again

6. Dec 5, 2004

nolachrymose

Sorry, I must have posted right before you edited your original post.

7. Dec 5, 2004

Logitech

what would i need for the next step

8. Dec 5, 2004

Logitech

I really really need help someone

9. Dec 5, 2004

futb0l

Hmmm.. maybe it's

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

so then ...

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

10. Dec 5, 2004

jai6638

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

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

11. Dec 5, 2004

ascky

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

12. Dec 5, 2004

maverick280857

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.