# Quick Trigonometric Identity Question

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1. May 8, 2016

### Draconifors

Hi! I have an integral to solve (that's not the point, though) and the inside of the integral is almost a trig identity:

1. The problem statement, all variables and given/known data

$sin\frac{(x+y)} {2}*cos\frac{(x-y)} {2}$

2. Relevant equations

I noticed this was very similar to $sinx+siny = 2sin \frac{(x+y)} {2} * cos\frac{(x-y)} {2}$

3. The attempt at a solution
Initially, within the context of the problem (a double integral over a certain region) I had used a change of variables which, while tedious, was doable (I can share that work, if you want). While reviewing my work, I recalled this identity, and just wanted to make sure whether or not I could transform the equation into $\frac {1} {2} (sinx+siny) = sin \frac{(x+y)} {2} * cos\frac{(x-y)} {2}$. I don't see why I couldn't, but I just want a confirmation.

2. May 8, 2016

### SammyS

Staff Emeritus
That looks perfectly fine to me.

What is region of integration?

3. May 8, 2016

### Draconifors

The region is bounded by $x+y=0$, $x+y=2$ and $y=0$.

That's why I had initially defined $u=x-y$ and $v=x+y$. It was a doable but kind of long integral to do, so I wanted to see whether I could shorten it down.

4. May 8, 2016

### SammyS

Staff Emeritus
That region is not bounded. Is there a typo ?

5. May 8, 2016

### Draconifors

Yes, I'm sorry!

It should read $x-y=0$ for the first equation.

And I'm redoing my problem using the trigonometric identity, and I notice that it's not actually shorter because of my upper bound being 2-x for y.

6. May 8, 2016

### SammyS

Staff Emeritus
Well, it was a good idea anyway. It just didn't work out in this case.