# Thread: [SOLVED] Trig identities - need suggestions

1. ## [SOLVED] Trig identities - need suggestions

I understand what to do (for the most part), but need a suggestion or two in how to approach these problems:

Simplify and reduce:

$\displaystyle (sinx + cscx)(sin^2x + csc^2x-1)$

Factor and reduce:

$\displaystyle sin^4 - cos^4$

The answer to the first one is $\displaystyle sin^3 . cos^3$ but i have no idea how to solve it.. ive tried it a hundred different ways.

for the second one, how would I start?

2. $\displaystyle (\sin{x} + \csc{x})(\sin^2{x} + \csc^2{x} - 1)$

multiply it out ...

$\displaystyle \sin^3{x} + \csc{x} - \sin{x} + \sin{x} + \csc^3{x} - \csc{x}$

combine like terms.

$\displaystyle \sin^4{x} - \cos^4{x}$

$\displaystyle (\sin^2{x} - \cos^2{x})(\sin^2{x} + \cos^2{x})$

can you finish up from here?

3. For the first problem:

$\displaystyle (\sin x + \csc x)(\sin^2 x + \csc^2 x - 1)$

$\displaystyle \sin^3 x + \sin x \csc^2 x - \sin x + \csc x \sin^2 x + \csc^3 x - \csc x$

$\displaystyle \sin^3 x + \csc x - \sin x + \sin x + \csc^3 x - \csc x$

$\displaystyle \sin^3 x + \csc^3 x$

Not sure where you would go from there.

For the second problem:

$\displaystyle \sin^4 x - \cos^4 x$

$\displaystyle (\sin^2 x + \cos^2 x)(\sin^2 x - \cos^2 x)$

$\displaystyle \sin^2 x - \cos^2 x$

$\displaystyle 1 - 2 \cos^2 x$

$\displaystyle -\cos 2x$

4. Thanks, this helped a lot