Trigonometric Identities: Exam-Pattern Drill (2)

Question

Prove that

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

A CBSE 11 / JEE Main classic — short but tests the triple-angle expansions.

Solution — Step by Step

Final answer: LHS = $2$. Proved.

Why This Works

The triple-angle formulas convert $\sin 3\theta$ and $\cos 3\theta$ into polynomials in $\sin\theta$ and $\cos\theta$. Once expressed this way, the Pythagorean identity $\sin^2 + \cos^2 = 1$ collapses the expression to a constant.

The identity is independent of $\theta$ — that’s the punchline. It works for any $\theta$ where $\sin\theta$ and $\cos\theta$ are non-zero.

Alternative Method

Combine to a single fraction over $\sin\theta\cos\theta$:

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

Then $\sin 2\theta = 2\sin\theta\cos\theta$, giving $2$. Slick — uses the sin-difference formula.

Common Mistake

Using $\sin 3\theta = 3\sin\theta\cos 2\theta$ — that’s wrong. The correct expansion is $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$. Memorise it as “3-cube becomes 3-cube minus 4-cube” (the "$3$" stays, the cube comes from cubing $\sin$).