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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