Trigonometric Identities: Conceptual Doubts Cleared (8)

Question

Prove the identity:

$\frac{\sin 3\theta}{\sin\theta} - \frac{\cos 3\theta}{\cos\theta} = 2$

Solution — Step by Step

Why This Works

The triple-angle formulas factor the numerators in a way that pairs perfectly with the denominators. Dividing $\sin 3\theta$ by $\sin\theta$ kills one factor of $\sin\theta$, leaving a clean polynomial in $\sin^2\theta$. The same happens on the cosine side.

The Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ then collapses everything to a constant. This is a recurring pattern in trig proofs — reduce to $\sin^2$ and $\cos^2$, then apply $\sin^2 + \cos^2 = 1$.

Alternative Method

Use the product-to-sum identities:

$\sin 3\theta \cos\theta - \cos 3\theta \sin\theta = \sin(3\theta - \theta) = \sin 2\theta = 2\sin\theta\cos\theta$

Combine the original LHS into a single fraction:

$\text{LHS} = \frac{\sin 3\theta \cos\theta - \cos 3\theta \sin\theta}{\sin\theta \cos\theta} = \frac{2\sin\theta\cos\theta}{\sin\theta\cos\theta} = 2$

Three lines, no triple-angle formulas needed.

Common Mistake