# Quick Trigonometric Identity Question

• Draconifors

#### 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. Homework Statement

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

## Homework Equations

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

## 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.

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. Homework Statement

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

## Homework Equations

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

## 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.

That looks perfectly fine to me.

What is region of integration?

That looks perfectly fine to me.

What is region of integration?

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.

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

That region is not bounded. Is there a typo ?

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.

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