# Proof that sin a + sin b= 2sin [(a+b)/2]cos [(a-b)/2]i need it very urgent please

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the formula sinC+sinD=2sin[(c+d)/2]cos[(c-d)/2] applies

in the case of sina +sinb:2sin[(a+b)/2]cos[(a-b)/2]

Let a = x+y....................................(1)and

b = x-y. ........................................(2)

Therefore, a+b=2x and a-b=2y or

x = (a+b)/2 ....................................(3)and

y= (a-b)/2.......................................(4)

Sin(x+y) = sinx*cosy + cosx*siny .......(5)

sin(x-y) = sinx*cosy - cosx*siny.........(6)

Adding (4) and (5), we get:

sin(x+y) + sin(x-y) = 2sinx*cosy................(7)

Replace x+y=a from (1) and x-y = b from (2), x=(a+b)/2 from (3) and y = (a-b)/2 from (4) in and get:

sina+sinb = 2*sin[(a+b)/2]*cos[(a-b)/2]

Imagine that a is the sum of 2 angles, u and v, so u+v=a, and b is the difference of the same 2 angle, so b=u-v

If we'll add a+b=(u+v)+(u-v), we'll reduce the similar terms v and (-v) and the result will be a+b=2u, so **u=(a+b)/2.**

If we'll make the difference between the 2 terms

a-b=(u+v)-(u-v) and we'll reduce the similar terms, the result will be a-b=2v, so **v=(a-b)/2**

**Let's substitute what we have found:**

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

In the right side of the equal, we'll open the paranthesis in this way:

sin (u+v)=sin u*cos v+ sin v*cos u

sin (u-v)=sin u*cos v- sin v*cos u

If we'll substitute the sin (u+v) + sin (u-v) with what we've found out, the result will be:

sinu*cosv+sin v*cosu+sin u*cosv-sinv*cosu=2 sin u*cos v

We'll reduce the similar terms (-sinv*cosu) with (sinv*cosu) and the result will be:

**sinu*cosv+sinu*cosv=2 sin u*cos v**

**2 sin u*cos v=2 sin u*cos v **

**q.e.d!**