# Trigonometric Proof using Compound Angle Formula

There are many areas to apply the compound angle formulas, and trigonometric proof using the compound angle formula is one of them.

\begin{aligned} \require{AMSsymbols} \require{color} \sin (x + y) &= \sin x \cos y + \sin y \cos x &\color{green} (1) \\ \sin (x-y) &= \sin x \cos y-\sin y \cos x &\color{green} (2) \end{aligned}
Using Compound Angle Formula, we can abstract two similar formulas using these identities for Trigonometric Proof.

\begin{aligned} \text{Let } A &= x + y \text{ and } B = x-y \\ A + B &= 2x \\ x &= \frac{A + B}{2} \\ A-B &= 2y \\ y &= \frac{A-B}{2} \\ \sin (x + y) + \sin (x-y) &= 2\sin x \cos y &\color{green} (1) + (2) \\ \sin A + \sin B &= 2 \sin \frac{A + B}{2} \cos \frac{A-B}{2} &\color{green} (3) \\ \sin (x + y)-\sin (x-y) &= 2\sin y \cos x &\color{green} (1)-(2) \\ \sin A-\sin B &= 2 \sin \frac{A-B}{2} \cos \frac{A + B}{2} &\color{green} (4) \end{aligned}
The following Example Question covers one of the popular ways to prove trigonometric identities.
Let’s have a look at it now!

## Example

Prove $\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C$, if $A + B + C = \pi$.

\begin{aligned} \displaystyle \require{AMSsymbols} \require{color} \text{LHS} &= \sin 2A + \sin 2B + \sin 2C \\ &= 2 \sin (A + B) \cos (A-B) + \sin 2C &\color{green} \text{apply (3)} \\ &= 2 \sin (A + B) \cos (A-B) + 2 \sin C \cos C &\color{green} \text{double angle formula} \\ &= 2 \sin (\pi-C) \cos(A-B) + 2 \sin C \cos (\pi-A-B) &\color{green} A + B + C = \pi \\ &= 2 \sin C \cos(A-B) + 2 \sin C \cos (\pi-A-B) &\color{green} \sin (\pi-\theta) = \sin \theta \\ &= 2 \sin C \cos(A-B)-2 \sin C \cos (A + B) &\color{green} \cos (\pi-\theta) = -\cos \theta \\ &= 2 \sin C \big[\cos (A-B)-\cos(A + B)\big] &\color{green} \text{common factor of } 2 \sin C \\ &= 2 \sin C \big[(\cos A \cos B + \sin A \sin B)-(\cos A \cos B-\sin A \sin B)\big] &\color{green} \text{compound angle formulas} \\ &= 2 \sin C (\cos A \cos B + \sin A \sin B-\cos A \cos B + \sin A \sin B) \\ &= 2 \sin C (2 \sin A \sin B) \\ &= 4 \sin C \sin A \sin B \\ &= \text{RHS} \end{aligned}