# Homework Help: Proving Trigonometric Identities

1. Aug 28, 2012

### acen_gr

1. The problem statement, all variables and given/known data
Prove that sin6 + cos6 = 1 - 3sin2cos2

2. Relevant equations
(1)

3. The attempt at a solution
I tried to convert those all in terms of sine and cosine only but it didn't work.

2. Aug 28, 2012

### Mentallic

Notice that $\sin^6x=\left(\sin^2x\right)^3$ and that $\sin^2x=1-\cos^2x$

Start by trying to apply these two ideas to the LHS of the equation.

3. Aug 28, 2012

### acen_gr

This is my work. Is this right?

sin6x + cos6x = 1 - 3sin2xcos2x
(sin2x)3 + cos6x = RHS
(1 - cos2x)3 + cos6 = RHS
1 - 3cos2x + 3cos4x - cos6x + cos6x = RHS
1 - 3cos2x + 3(cos2x)(cos2x) = RHS
1 - 3cos2x + 3(cos2x)(1 - sin2x) = RHS
1 - 3cos2x[1-(1 -(sin2x)] = RHS
1 - 3cos2x(sin2x) = RHS
1 - 3cos2xsin2x = 1 - 3sin2xcos2x

4. Aug 28, 2012

### Mentallic

Every step is mathematically equivalent to its preceding step (so you haven't broken any rules) so yes, it's right

If I'm to be a bit picky however, I'll mention that it's not necessary to write ... = RHS on every line. You should instead set out your proof as so:

LHS = ...
= ...
= ...
= RHS

You don't need to keep writing LHS either, unless you make any manipulations to both sides of the equality such as:

LHS = ...
LHS + 1 = ... + 1

Although you shouldn't do this when trying to prove LHS = RHS. If you needed to add 1 to the RHS then you can minus one to keep it balanced as so:

LHS = ...
= ... + 1 - 1

I'd also skip this line altogether (unless it's to help you to personally keep track of things and not get confused), but if not, you should just go ahead and factorize straight away:

$$=1-3\cos^2x+3cos^4x$$
$$=1-3\cos^2x(1-cos^2x)$$

Also this last line is unnecessary, but there's no harm done if you feel like doing it.

5. Aug 28, 2012

### acen_gr

Thank you!:) Now I know why my teachers hate my solutions :P I'll take note of everything you've said. God bless! Hope you could help more people like me:) I'm starting to love this forum! :D

6. Aug 28, 2012

### Mentallic

You're welcome and that's glad to hear!