Quick Trigonometric Identity Question

[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.

Thank you for your time!

 

[SammyS]

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.

Thank you for your time!

That looks perfectly fine to me.

What is region of integration?

 

[Draconifors]

That looks perfectly fine to me.

What is region of integration?

Thank you for your answer!

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.

 

[SammyS]

Thank you for your answer!

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 ?

 

[Draconifors]

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.

 

[SammyS]

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.