Question

If A + B + C = $180{}^\circ$, then prove that sin 2A + sin 2B + sin 2C = 4 sinA sinB sinC.

Answer 1

Hint: Start with the LHS. Use sin (X + Y) + sin (X – Y) = 2 sin X cos Y, and sin 2C = 2 sinC cosC to get 2sin (A + B) cos (A – B) + 2 sinC cosC. Then simplify sin (A + B) and cosC. Then use cos (A – B) - cos (A + B) = 2 sinA sinB, to get 4 sinA sinB sinC, which is equal to the RHS.

Complete step-by-step answer:
In this question, we are given that A + B + C = $180{}^\circ$.
Using this information, we need to prove that sin 2A + sin 2B + sin 2C = 4 sinA sinB sinC.
Let us start with the LHS.
LHS = sin2A + sin2B + sin2C
Now, we know the property that:
sin (X + Y) + sin(X – Y) = 2 sin X cos Y
Using this property with X = A + B and Y = A – B, we will get the following:
2sin (A + B) cos (A – B) + sin 2C
Now, we know that sin 2C = 2 sinC cosC
Substituting this in the above equation, we will get the following:
2sin (A + B) cos (A – B) + 2 sinC cosC
Now, A + B = $180{}^\circ$ - C
So, sin (A + B) = sin ($180{}^\circ$ - C) = sinC
Substituting this, we will get the following:
2sinC cos (A – B) + 2 sinC cosC
Taking 2sinC common, we will get the following:
2sinC [cos (A – B) + cosC]
Now, cosC = cos[$180{}^\circ$ - (A + B)] = - cos (A + B)
So, we will get the following:
2sinC [cos (A – B) - cos (A + B)]
We know that cos (A – B) - cos (A + B) = 2 sinA sinB
Using this, we will get the following:
2sinC(2 sinA sinB )
4 sinA sinB sinC = RHS
So, LHS = RHS
sin 2A + sin 2B + sin 2C = 4 sinA sinB sinC
Hence proved.

Note: In this question, it is very important to know various trigonometric properties like sin (X + Y) + sin (X – Y) = 2 sin X cos Y, sin 2C = 2 sinC cosC, and cos (A – B) - cos (A + B) = 2 sinA sinB. Students should remember important trigonometric formulas and identities for solving these types of questions.