Expert: Scott A Wilson - 9/29/2009

Question
LHS
Sin2A+sin2B+sin2C
= 2SinA cosA+2sin(B+C).cos(B-C)
=2sinAcosA+2sinA.cos(B-C)
=2sinA{cosA+cos(B-C)}
=2sinA{cos(B-C)-cos(B+C)}
=2sinA 2sinB.sinC=4sinA sinB sinC-RHS

Sir please describe how this statement(i.e Sin2A+sin2B+sin2C) became  ( 2SinA cosA+2sin(B+C).cos(B-C).
Please explain.  

Answer
The equation is sin(2A) + sin(2B) + sin(2C) = 2•sinA•cosA + 2sin(B+C)•cos(B-C).

First, it is a comon identity that
1) sin(A+B) = sinA•cosB + sinB•cosA, so sin(2A) = 2sinA•cosA; and

2) cos(B-C) = cosB•cosC + sinB•sinC; and

3) sin²(A) + cos²(A) = 1.

I will take the first part of the left side of the equation at the top and
show that it is the same as the first part of the right side.

What that is is sin(2A) = 2•sinA•cosA.
This shows we can drop the first term on both sides.


All that we have to do now is show that sin(2B) + sin(2C) = 2sin(B+C)•cos(B-C).

To do the proof, I found it easier to show the right side is the same as the left.
To do it properly, they can be reversed.

It is known that sin(B+C) =  sinB•cosC + sinC•cosB.
It is also known that cos(B-C) = sinB•cosC + sinC•cosB.

From here, we can say that 2•sin(B+C)•cos(B-C) = 2(sinB•cosC + sinC•cosB)(cosB•cosC + sinB•sinC).

Using (w+x)(y+z) = wy + xz + wy +- xz, we get 2(sinB•cosC + sinC•cosB)(cosB•cosC + sinB•sinC) as
2(sinB•cosB•cos²C + sinB•cosB•sin²C + sinC•cosC•sin²B + sinC•cosC•cos²B)

This can be factored into 2(sinB•cosB(cos²C + sin²C) + sinC•cosC(sin²B + cos²B).
Now it is known that sin²C + cos²C = sin²B + cos²B = 1, so this equation becomes
2•sinB•cosB + 2•sinC•cosC.

That is the same as sin(2B) + sin(2C).

I may have made a mistake on a sign in cos(B+C), but I’ll check it again tonight.