# Simplifying a Trig Expression

• atomicpedals
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.

Mentor
Hint: ##(1+x)(1-x) = 1 - x^2##

Michael Price, neilparker62, YoungPhysicist and 1 other person
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 :-)

chwala, YoungPhysicist and PeroK
Homework Helper
Hint: ##(1+x)(1-x) = 1 - x^2##
Triggered a 'Eureka' moment!

Gold Member
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...