Question

Prove that : $\frac{2 \cos^3} \theta - \cos \theta{\sin \theta - 2 \sin^3 \theta} = \cot \theta$

Hint:

Notice that both $(2 \cos^2 \theta - 1)$ and $(1 - 2 \sin^2 \theta)$ are actually formulas for $\cos 2\theta$.

Answer 1

Step 1: Understanding the Concept:
We start with the Left Hand Side (LHS) and simplify it using trigonometric identities, specifically the identity $\sin^2 \theta + \cos^2 \theta = 1$.
Step 2: Factoring the Expression:
LHS $= \frac{\cos \theta (2 \cos^2 \theta - 1)}{\sin \theta (1 - 2 \sin^2 \theta)}$
Step 3: Substitution and Simplification:
Substitute $\cos^2 \theta = 1 - \sin^2 \theta$ in the numerator: \[ = \frac{\cos \theta [2(1 - \sin^2 \theta) - 1]}{\sin \theta (1 - 2 \sin^2 \theta)} \] \[ = \frac{\cos \theta [2 - 2 \sin^2 \theta - 1]}{\sin \theta (1 - 2 \sin^2 \theta)} \] \[ = \frac{\cos \theta (1 - 2 \sin^2 \theta)}{\sin \theta (1 - 2 \sin^2 \theta)} \]
Step 4: Final Answer:
The common term $(1 - 2 \sin^2 \theta)$ cancels out: \[ = \frac{\cos \theta}{\sin \theta} = \cot \theta = \text{RHS} \] Hence proved.