Trigonometric Proof using Compound Angle Formula

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 \(A + B + C = \pi\).

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

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