Dear aspirant,

The answer to your question is as follows:

Sin A +Sin B=2[{sin(A+B}/2}{cos(A-B)/2}]

proof:

sin(X+Y)+sin(X-Y)=2(sinXcosY)

let X+Y=A

X-Y=B

X=(A+B)/2 & Y=(A-B)/2

by substituting we get:

sin A +sin B=2[{sin(A+B)/2}{cos{A-B)/2}]

Hope this was useful.

There is only one value given by the formula below:

Answer goes like

sinA+ sinB =2[{sin(A+B)/2} { cos(A-B) /2}]

proof is given as follows:

sin(X+Y) + sin(X-Y) = 2(sinXcosY)

let X + Y = A and X - Y = B

X = (A + B)/2 & Y = (A - B)/2

By Substituting, we obtain the following equation

sinA+ sinB =2[{sin(A+B)/2} { cos(A-B) /2}]

Hence the above equation gives the value of your question

Hope you undersood.

Thank You!!