Q4(vii):

Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (vii) $\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$

Given: An trigonometric expression involving an acute angle $\theta$: $\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta}$.

To Prove: $\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$.

Step 1: Analyze the Left-Hand Side (LHS)

The given expression is: $LHS = \frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta}$.

Step 2: Factor out common terms from the numerator and denominator

In the numerator, $\sin \theta$ is common to both terms. In the denominator, $\cos \theta$ is common to both terms.

Factoring the numerator: $\sin \theta - 2 \sin^3 \theta = \sin \theta (1 - 2 \sin^2 \theta)$.

Factoring the denominator: $2 \cos^3 \theta - \cos \theta = \cos \theta (2 \cos^2 \theta - 1)$.

Substituting these back into the LHS expression:

$LHS = \frac{\sin \theta (1 - 2 \sin^2 \theta)}{\cos \theta (2 \cos^2 \theta - 1)}$

Step 3: Apply the Trigonometric Identity

Recall the fundamental Pythagorean identity: $\sin^2 \theta + \cos^2 \theta = 1$.

From this identity, we can express $1$ as $(\sin^2 \theta + \cos^2 \theta)$.

Substitute this into the numerator term $(1 - 2 \sin^2 \theta)$:

$1 - 2 \sin^2 \theta = (\sin^2 \theta + \cos^2 \theta) - 2 \sin^2 \theta$

$= \cos^2 \theta - \sin^2 \theta$

Now, substitute this into the denominator term $(2 \cos^2 \theta - 1)$:

$2 \cos^2 \theta - 1 = 2 \cos^2 \theta - (\sin^2 \theta + \cos^2 \theta)$

$= 2 \cos^2 \theta - \sin^2 \theta - \cos^2 \theta$

$= \cos^2 \theta - \sin^2 \theta$

Step 4: Simplify the expression

Substitute the simplified terms back into the LHS:

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

Since $\theta$ is an acute angle and the expression is defined, $(\cos^2 \theta - \sin^2 \theta) \neq 0$. Therefore, we can cancel the common factor $(\cos^2 \theta - \sin^2 \theta)$ from the numerator and denominator:

$LHS = \frac{\sin \theta}{\cos \theta}$

Step 5: Apply the Quotient Identity

Using the quotient identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$:

$LHS = \tan \theta$

Conclusion:

Since $LHS = RHS$, the identity is proven.

Final Answer: $\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$ is proved.