### Product to Sum Trigonometry Identities Calculation

Calculate the values for sin(u)sin(v), sin(u)cos(v) and cos(u)sin(v) against selected values of u and v angle, using following formulas:
sin(u)sin(v) = ½(cos(u-v) – cos(u+v))
sin(u)cos(v) = ½(sin(u+v) + sin(u-v))
cos(u)sin(v) = ½(sin(u+v) – sin(u-v))
cos(u)cos(v) = ½(cos(u+v) + cos(u-v))
The group of trigonometry identities are known as the product-to-sum identities. These identities are the true trigonometry statement that shows how to go from the product of two trigonometry functions to the sum of two trigonometry functions as shown above. We can use the product-form or sum-form to describe the same things because these are definition and interchangeable.
 Please enter u angle in degree: Please enter v angle in degree:
 Result

Proof of product to sum 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) = ½(cos(u-v) – cos(u+v)) $$\sin u . \sin v = \dfrac{1}{2}{[\cos(u-v) – \cos(u+v)]}$$ we know that $$\cos (u - v) = \cos u \cos v + \sin u \sin v$$ $$\cos (u + v) = \cos u \cos v - \sin u \sin v$$ by subtracting both side of equation we will get - $$\cos (u - v) - \cos (u + v) = \cos u \cos v + \sin u \sin v - ( \cos u \cos v - \sin u \sin v)$$ $$\cos (u - v) - \cos (u + v) = \cos u \cos v + \sin u \sin v - \cos u \cos v + \sin u \sin v)$$ $$\cos (u - v) - \cos (u + v) = 2 \sin u \sin v$$ now divide both side by 2 $$\dfrac{\cos (u - v) - \cos (u + v)}{2} = \dfrac{2 \sin u \sin v}{2}$$ $$\dfrac{1}{2} [\cos (u - v) - \cos (u + v)] = \sin u \sin v$$
Proof of sin(u)cos(v) = ½(sin(u+v) + sin(u-v)) $$\sin u . \cos v = \dfrac{1}{2}{[\sin(u+v) + \sin(u-v)]}$$ we know that $$\sin (u + v) = \sin u \cos v + \cos u \sin v$$ $$\sin (u - v) = \sin u \cos v - \cos u \sin v$$ by adding both side of equation we will get - $$\sin (u + v) + \sin (u + v) = \sin u \cos v + \cos u \sin v + \sin u \cos v - \cos u \sin v$$ $$\sin (u + v)+ \sin (u- v) = 2 \sin u \cos v$$ now divide both side by 2 $$\dfrac{\sin (u + v) - \sin (u - v)}{2} = \dfrac{2 \sin u \cos v}{2}$$ $$\dfrac{1}{2} [\sin (u+ v) + \sin (u + v)] = \sin u \cos v$$
Proof of cos(u)cos(v) = ½(cos(u-v) + cos(u+v)) $$\cos u . \cos v = \dfrac{1}{2}{[\cos(u-v) + \cos(u+v)]}$$ we know that $$\cos (u - v) = \cos u \cos v + \sin u \sin v$$ $$\cos (u + v) = \cos u \cos v - \sin u \sin v$$ by adding both side of equation we will get - $$\cos (u - v) + \cos (u + v) = \cos u \cos v + \sin u \sin v + \cos u \cos v - \sin u \sin v$$ $$\cos (u - v) + \cos (u + v) = 2 \cos u \cos v$$ now divide both side by 2 $$\dfrac{\cos (u - v) + \cos (u + v) }{2} = \dfrac{2 \cos u \cos v}{2}$$ $$\dfrac{1}{2} [\cos (u - v) + \cos (u + v) ] = \cos u \cos v$$ similarly other equation can be proved.