# 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.