# Trig Identity

1. Jan 8, 2008

### tornzaer

1. The problem statement, all variables and given/known data
1. (sinx - cosx)(sinx + cosx) = 2sin^2x-1
2. (2sinx + 3cos)^2 + (3sinx - 2cosx)^2 = 13

2. Relevant equations
N/A

3. The attempt at a solution

For 1. L.S. = sinx^2+sinxcosx-sinxcosx-cosx^2, the sinxcosx cancels and I'm lost.

I haven't a clue how to do the second one.

I missed a couple of day of school because of the flu before christmas break, so I need to understand this. Some please help.

2. Jan 8, 2008

### blochwave

For 1. L.S. = sinx^2+sinxcosx-sinxcosx-cosx^2, the sinxcosx cancels and I'm lost.

that part is right. Now you just need sin(x)^2+cos(x)^2=1

"But what can I do with sin(x)^2-cos(x)^2?"

Nothing, but look harder. Rearrange the identity. Substitute.

Same thing with the second one, looks like you just multiply it out and substitute with that idenitity

3. Jan 8, 2008

### rock.freak667

For the first one: the sinxcosx cancels out and leaves you with
$sin^2 x - cos^2 x$
Then recall that $sin^2 x + cos^2 x=1$ from that find cos$^2$x in terms of sin$^2$x and sunstitute.

For the second one expand out the LHS and use the identity $sin^2 x + cos^2 x=1$

4. Jan 8, 2008

### rocomath

1. Left side is simply difference of squares, put it in (x^2-y^2) form and then use another trig identity to make it look like the right.

2. Square the terms and the middle terms with cancel the middel term of the other. From there use the fact that sin^(2)x + cos^(2)x = 1.

5. Jan 8, 2008

### tornzaer

Ok, thank you all. I understand now.