- #1

agentc0re

- 2

- 0

## Homework Statement

sin(2x)-cos(2x)=sqrt(2)sin(2x+Api), then the number 0 < A = ____ < 2

## Homework Equations

http://bit.ly/9njiUW <-- trig reference sheet with formulas and identites

## The Attempt at a Solution

So I start with the left side of the equation and attempt to make it look like the right in order to figure out what A is. Using some identities i started out:

2sin(x)cos(x) - cos^2(x) + sin^2(x)

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

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

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

2( 1/2[ sin(x + x) + sin(x - x) ] ) - 1 + 2sin^2(x)

sin(x + x) + sin(x - x) - 1 + 2sin^2(x)

sin(2x) + sin(0) - 1 + 2sin^2(x)

sin(2x) + 0 - 1 + 2sin^2(x)

sin(2x) -1 + 2sin^2(x)

Then I realized I made a big loop with the "sin(2x)" part and am not sure how to go from there.