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.