# Thread: Verifying Identities

1. ## Verifying Identities

Alright, So I feel that I'm not doing these the right way, so if you guys would be so kind as to give me some general pointers on top of help specific to the problem, I'd be extremely grateful. This particular question asks to verify that the following is true:

$\displaystyle \frac{\sin{x}+\sin{y}}{\cos{x}+\cos{y}}=\tan{\frac {x+y}{2}}$

Here is what I have so far:

$\displaystyle \frac{\sin{x}+\sin{y}}{\cos{x}+\cos{y}}=\tan{\frac {x+y}{2}}$
$\displaystyle \frac{\sin{x}+\sin{y}}{\cos{x}+\cos{y}}=\tan({\fra c{x}{2}+\frac{y}{2}})$
$\displaystyle \frac{\sin{x}+\sin{y}}{\cos{x}+\cos{y}}=\frac{\tan (\frac{x}{2}+\frac{y}{2})} {1-\tan{x}\tan{y}}$
$\displaystyle \frac{\sin{x}+\sin{y}}{\cos{x}+\cos{y}}=\frac{\fra c{1-\cos{x}}{\sin{x}}+\frac{1-\cos{y}}{\sin{y}}} {1-\tan{x}\tan{y}}$

I have more work after that, but I'm fairly sure I've done something wrong by this point. If I am, somehow, on the right track, I'll happily type up the rest of the work I have done, but after this point, I basically just started throwing random identities at it hoping the answer would just come to me.

I also have two more that I've been stuck on that I'll update this thread with as soon as I get them and all the work into latex.

Thanks in advance for the help, and thanks for all the help you guys have given me before I made an account here.

2. Use:

$\displaystyle \sin x+\sin y=2\sin \dfrac{x+y}{2} \cos \dfrac{x-y}{2}$

$\displaystyle \cos x+\cos y=2\cos \dfrac{x+y}{2} \cos \dfrac{x-y}{2}$

3. $\displaystyle \tan(\frac{x+y}{2})=\frac{\sin(\frac{x+y}{2})}{\co s(\frac{x+y}{2})}}$

together with the equations in the tread before you get the result.

4. Thanks guys, that made this way easier than I had hoped.

I'm sure this is also simple, but:

$\displaystyle 2\cos^2{\frac{x}{2}}=\frac{\sin^2{x}}{1-\cos{x}}$

$\displaystyle 2\cos^2{\frac{x}{2}}=\frac{\frac{1-\cos{2x}}{2}} {1-\cos{x}}$

$\displaystyle 2\cos^2{\frac{x}{2}}=\frac{1-\cos{2x}} {2-2\cos{x}}$

$\displaystyle 2\cos^2{\frac{x}{2}}=\frac{-2\cos^2{x}} {2-2\cos{x}}$

5. using again

$\displaystyle \cos x+\cos y=2\cos \dfrac{x+y}{2} \cos \dfrac{x-y}{2}$

you get

$\displaystyle 2\cos^2(\frac{x}{2})=\cos x+1$

then multiply this with

$\displaystyle 1=\frac{1-\cos x}{1-\cos x}$

and use

$\displaystyle \sin^2x+\cos^2x=1$.