# Thread: Proving Trig Identities

1. ## Proving Trig Identities

I'm having trouble making sense of addition and subtraction in trig identities.

cos^6x + sin^6x = 1 - 3 sin^2x + 3 sin^4x

2. Originally Posted by casey_k
I'm having trouble making sense of addition and subtraction in trig identities.

cos^6x + sin^6x = 1 - 3 sin^2x + 3 sin^4x
Notice that $\cos^6{x} = (\cos^2{x})^3$

$= (1 - \sin^2{x})^3$

$= \sum_{k = 0}^{3}\left(_k^n\right)1^{n - k}+(-\sin^2{x})^k$

$= 1 - 3\sin^2{x} + 3\sin^4{x} - \sin^6{x}$.

So we can rewrite the original function as

$1 - 3\sin^2{x} + 3\sin^4{x} - \sin^6{x} + \sin^6{x}$.

What does this equal?

3. Originally Posted by casey_k
I'm having trouble making sense of addition and subtraction in trig identities.

cos^6x + sin^6x = 1 - 3 sin^2x + 3 sin^4x
two hints ...

$(a^2)^3 + (b^2)^3 = (a^2 + b^2)(a^4 - a^2b^2 + b^4)$

$\cos^2{x} = 1 - \sin^2{x}$

4. I didn't get it at first because I expanded the right side both (cos^2x)^3 + (sin^2x)^3 but then I realized I only had to expand cos.

Thank you!