1. ## Solving Identities

Prove.

$\displaystyle sin(2x) + sin(2y) = 2sin(x+y)cos(x-y)$

2. Originally Posted by cdx
Prove.

$\displaystyle sin(2x) + sin(2y) = 2sin(x+y)cos(x-y)$
sneaky, sneaky! this is actually one of the sum to product formulas in disguise!

$\displaystyle \sin (A + B) = \sin A \cos B + \sin B \cos A$

$\displaystyle \cos (A - B) = \cos A \cos B + \sin A \sin B$

and try to simplify to get the left side, which, if you know your double angle formulas, you'd realize is $\displaystyle 2 \sin x \cos x + 2 \sin y \cos y$

3. Originally Posted by cdx
Prove.

$\displaystyle sin(2x) + sin(2y) = 2sin(x+y)cos(x-y)$
Another way is to use Euler's formula

$\displaystyle \cos \theta = \frac{e^{i \theta} + e^{-i \theta}}{2},\;\;\;\sin \theta = \frac{e^{i \theta} - e^{-i \theta}}{2}$

Replace the $\displaystyle \theta$ with $\displaystyle (x+y)\;\;\;(x-y)$ where appropriate on the RHS, expand and use these expressions again, to get the LHS.

4. Another Identity that I don't quite understand.

$\displaystyle \frac{1-sin2x}{cos2x} = \frac{cos2x}{1+sin2x}$

5. Originally Posted by cdx
Another Identity that I don't quite understand.

$\displaystyle \frac{1-sin2x}{cos2x} = \frac{cos2x}{1+sin2x}$
you can start with either side here and get the other pretty easily.

say we start with the left: multiply by $\displaystyle \frac {1 + \sin 2x}{1 + \sin 2x}$

it should be easy for you to clean up from there in a step or two (think "difference of two squares")

P.S. you should post new questions in a new thread