# Yet another trig identity .

1. Sep 11, 2008

### banfill_89

yet another trig identity.....

1. The problem statement, all variables and given/known data

prove: cos^(x)= 5/16+15/32(cos2x)+3/16(cos4x)+1/32(cos6x)

2. Relevant equations

3. The attempt at a solution

i attempted to use the formula cos^2(x)=(1+cos2x)/(2), and square both sides, then use it again for the square roots, then multiply the answer by (1+cos2x)/(2) again thus making the left side cos^6.........................not getting the right answer

2. Sep 11, 2008

### banfill_89

Re: yet another trig identity.....

i think im on the right trrack?

3. Sep 11, 2008

### Defennder

Re: yet another trig identity.....

You mean cos^2 (x) on the left?

4. Sep 11, 2008

### banfill_89

Re: yet another trig identity.....

hey....the original proof is cos^6(x) on the left side

5. Sep 11, 2008

### banfill_89

Re: yet another trig identity.....

i get to a certain point where i get like....cos^2(4x) times cos (2x) wth a bunch of other stuff on the left....but i dont knwo what to do with it

6. Sep 11, 2008

### Defennder

Re: yet another trig identity.....

Oh yeah, should have read it more closely. I haven't tried it out yet, but if you know the left is function of cos x only, you should repeatedly apply the double angle cos and sin formula to change everything to functions of sin x and cos x. Then express those functions of sin x as functions of cos x. It should all cancel out.

7. Sep 11, 2008

### banfill_89

Re: yet another trig identity.....

lol its cool.....so wadda u think? do you get what i tried to do? like is it clear when i explained it?

8. Sep 12, 2008

### banfill_89

Re: yet another trig identity.....

thats what i keep doing...

9. Sep 12, 2008

### Defennder

Re: yet another trig identity.....

I just proved it using the approach suggested earlier. Just convert everything on the right to a function of cos x. A lot of terms will cancel out to give cos^6 x.