# Prove the Sum to Product Trigonometry Identities

Prove and Calculate the values for sin(u)+sin(v), sin(u)-sin(v), cos(u)+cos(v) and cos(u)-cos(v) against selected values of u and v angle.

## Formulas:

sin(u)+sin(v) = 2(sin( u +2v) ( cos( u -2v))

sin(u)-sin(v) = 2(cos( u +2v) ( sin( u -2v))

cos(u)+cos(v) = 2(cos( u +2v) ( cos( u -2v))

cos(u)-cos(v) = -2(sin( u +2v) ( sin( u -2v))

## Proof of sum to product Identities

the product to sum and sum to product identities can be easily derived from the sum and difference identities as follows:
sin(u)+sin(v) = 2(sin( u +2v) ( cos( u -2v)) Lets start :-
$$\sin u . \cos v = \dfrac{1}{2}{[\sin(u+v) + \sin(u-v)]}$$ we will muliply 2 in both side .. $$2 \sin u \cos v = 2(\dfrac{1}{2}{[\sin(u+v) + \sin(u-v)])}$$ $$2 \sin u \cos v = {[\sin(u+v) + \sin(u-v)]}$$ now we can rewite equation as $${\sin(u+v) + \sin(u-v)} = 2 \sin u \cos v$$ lets apply in both side of equation $$u = \dfrac{u + v}{2} and v = \dfrac{u - v}{2}$$ so now we will get as $${ \sin({ \dfrac{(u + v)}{2}} + { \dfrac{(u - v)}{2})} + \sin({ \dfrac{(u + v)}{2}} - { \dfrac{(u - v)}{2})} = 2 \sin { \dfrac{(u + v)}{2}} \cos { \dfrac{(u - v)}{2}}}$$ $${ \sin{ \dfrac{2(u)}{2}} + \sin{ \dfrac{2(v)}{2}} = 2 \sin { \dfrac{(u + v)}{2}} \cos { \dfrac{(u - v)}{2}}}$$ $$\sin u + \sin v = 2 \sin { \dfrac{(u + v)}{2}} \cos { \dfrac{(u - v)}{2}}$$ simlarly we can prove other Identities.