Question

asinA-bsinB=csin(A-B)

Answer 1

Answer:

The proof is explained below :

Step-by-step explanation:

Using Sine rule in a triangle ,

$\frac{a}{sinA} =\frac{b}{sin B} =\frac{c}{sin C}=k\\\\\implies a=k\cdot sin A , \\b=k\cdot sin B \\c=k\cdot sin C$

Now, LHS = a·sinA - b·sinB

                 = k·sin²A - k·sin²B    { Using Sine rule equations }

                 = k·[sin(A + B)·sin(A - B)]

                 = k·sin(π - C)·sin(A - B)  { By angle sum property of a triangle sum of all angles is 180 ⇒ A + B + C = π ⇒ A + B = π - C }

                 = k·sinC·sin(A - B)

                = c·sin(A - B) = R.H.S.

So, LHS = RHS

Hence, a·sinA - b·sinB = c·sin(A - B)