Mastering Trig Identities: Simplifying a Trig Expression

[atomicpedals]

The other day in a fit of boredom I decided to dust off my old math books (high school and undergrad) and see if I can still crank through the basics.

1. Homework Statement
Show the following: $$ \frac { 3 cos^2(x) } { 2 - 2 sin(x) } = \frac { 3 }{ 2 } ( sin(x) + 1 ) $$

Homework Equations

$$ sin^2(x) + cos^2(x) = 1 $$ $$ cos^2(x) = 1 - sin^2(x) $$

The Attempt at a Solution

I make it about halfway to a solution and then draw a blank.
$$ \frac { 3 cos^2(x) } { 2 - 2 sin(x) } = \frac { 3 (1 - sin^2(x)) } { 2 - 2 sin(x) } $$ $$ = \frac { 3 - 3 sin^2(x) } { 2 - 2 sin(x) } $$ I get the feeling I've overlooked something fairly obvious or swapped a sign somewhere, but it's not jumping off the page at me. Any nudges in the right direction are greatly appreciated.

 

[DrClaude]

Hint: $(1+x)(1-x) = 1 - x^2$

 

[atomicpedals]

I'm guessing I'd apply that before having multiplied through by 3, giving $$ \frac { 3 (1 + sin(x)) (1 - sin(x)) } { 2 - 2 sin(x) } = \frac { 3 (1 + sin(x)) (1 - sin(x)) } { 2 (1 - sin(x)) } $$ Hey! It works! The $( 1 - sin(x) )'s$ cancel out!
I should do a review of algebra and trig more than once a decade :-)

 

[neilparker62]

Hint: $(1+x)(1-x) = 1 - x^2$

Triggered a 'Eureka' moment!

 

[chwala]

I'm guessing I'd apply that before having multiplied through by 3, giving $$ \frac { 3 (1 + sin(x)) (1 - sin(x)) } { 2 - 2 sin(x) } = \frac { 3 (1 + sin(x)) (1 - sin(x)) } { 2 (1 - sin(x)) } $$ Hey! It works! The $( 1 - sin(x) )'s$ cancel out!
I should do a review of algebra and trig more than once a decade :-)

this was easy though...for a post graduate bingo...