# Proving a Trigonometric Identity

1. Feb 28, 2015

### einstein314

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

Prove that:

$\cos^6{(x)} + \sin^6{(x)} = \frac{5}{8} + \frac{3}{8} \cos{(4x)}$

2. Relevant equations

I am not sure. I used factoring a sum of cubes.

3. The attempt at a solution

I tried $\cos^6{(x)} + \sin^6{(x)} = \cos^4{(x)} - \cos^2{(x)} \sin^2{(x)} + \sin^4{(x)}$. But I can't get anywhere beyond this; I must be missing something obvious.

2. Feb 28, 2015

### HallsofIvy

Staff Emeritus
Sounds good to me! Now, you might try factoring "$cos^2(x)$ out of the first two terms: $cos^2(x)(cos^2(x)- sin^2(x))+ sin^4(x)= cos^2(x)cos(2x)- sin^4(x)$ see where you can go from that.

3. Feb 28, 2015

### Simon Bridge

Id normally just throw the euler formula at these things ... unless I had an already proved identity I could use.

4. Feb 28, 2015

### SammyS

Staff Emeritus
Although, x2 - xy + y2 cannot be factored (over the reals), x4 - x2 y2 + y4 can be factored .

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5. Feb 28, 2015

### SteamKing

Staff Emeritus
Rather than attacking the LHS of the identity, I would prefer to look at the expression cos (4x) instead. The multiple angle formulas for cosine I think would be more helpful here than trying to factor polynomials.

6. Mar 2, 2015

### haruspex

The 4x on the right, and the 8s in the denominators, are strong clues. Do you know how to expand cos(2x) in terms of cos(x) and sin(x)? Just apply that (in reverse) a couple of times.

Edit... SteamKing's (equivalent) post wasn't there when I hit reply, even though it seems to have been made hours earlier. Strange.