154399

Equation or Identity problem

7) IDENTITY:

sin(2A)+sin(2B)+sin(2C)=4sinAsinBsinC
For starters A+B+C= 180° as they are the sum of the three interior angles of the triangle. Then A+B+C=π and thus A+B= π-C. The first two terms of LHS are sin(2A)+sin(2B) which can be written in the form of sinC + sinD. The formula to denote this is 2 sin⁡((2A+2B)/2)+cos⁡((2A-2B)/2)+2sinC cosC. When simplified this gives: sin2A+sin2B+sin2C = 2sin(A+B)cos(A-B)+2sinC cos C
=2sin C(cos(A-B)+cos C). This is due to the fact that sin(A+B) = sin[180-(A+B)]=sin C. Then I can further simplify the RHS to give
=4sin C cos1/2(A+C-B) cos1/2(A-B-C)

=4sin C cos1/2([180-B]-B) cos1/2(A-[180-A])

=4sin C cos(90-B) cos(A-90)

=4sin C cos(90-B) cos(90-A)

sin(2A)+sin(2B)+sin(2C)=4sinAsinBsinC . This therefore shows that the LHS=RHS thus proving that this is an identity.