Sum to Product Trigonometry Identities Calculation


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, using following 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))
The sum-to-product identities are the trigonometry statements that tells how to convert the summation or subtraction of 2-trigonometry functions into product of 2-trigonometry functions as shown in above formulas. The sum-to-product identities deal only with sine and cosine functions. The are the true trigonometry statements that tell you how to turn the sum or subtraction of two trig functions into the product of two trig functions.

Please enter u angle in degree:
Please enter v angle in degree:
  
 
Result:

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.