Proving Trigonometric Identities

Notice that [itex]\sin^6x=\left(\sin^2x\right)^3[/itex] and that [itex]\sin^2x=1-\cos^2x[/itex]

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

 

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:

[tex]=1-3\cos^2x+3cos^4x[/tex]
[tex]=1-3\cos^2x(1-cos^2x)[/tex]

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

 

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