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.

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

\( \begin{aligned}
\text{Let } A &= x + y \text{ and } B = x – y \\
A + B &= 2x \\
x &= \frac{A + B}{2} \\
A – B &= 2y \\
y &= \frac{A – B}{2} \\
\sin (x + y) + \sin (x – y) &= 2\sin x \cos y &\color{green} (1) + (2) \\
\sin A + \sin B &= 2 \sin \frac{A + B}{2} \cos \frac{A – B}{2} &\color{green} (3) \\
\sin (x + y) – \sin (x – y) &= 2\sin y \cos x &\color{green} (1) – (2) \\
\sin A – \sin B &= 2 \sin \frac{A – B}{2} \cos \frac{A + B}{2} &\color{green} (4) \\
\end{aligned} \\ \)
The following Example Question covers one of 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 \(A + B + C = \pi\).

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

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