Question

If $\sin(\theta-\alpha)$, $\sin~\theta$ and $\sin(\theta+\alpha)$ are in H.P., then the value of $\cos~2\theta$ is

• A. $1+4~\cos^{2}\frac{\alpha}{2}$
• B. $1-4~\cos^{2}\frac{\alpha}{2}$
• C. $-1-4~\cos^{2}\frac{\alpha}{2}$
• D. $-1+4~\cos^{2}\frac{\alpha}{2}$

Hint:

Trigonometry Tip: The identity $\sin(A+B)\sin(A-B) = \sin^2A - \sin^2B$ is a powerful shortcut that frequently appears in complex trigonometric equations.

Answer 1

The Correct Option is B

Concept: Trigonometry - Sequences and Series (Harmonic Progression) and Trigonometric Identities.

Step 1: Convert the H.P. condition to an A.P. condition. If three terms $a, b, c$ are in Harmonic Progression (H.P.), their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ must be in Arithmetic Progression (A.P.). Therefore, $\frac{1}{\sin(\theta-\alpha)}, \frac{1}{\sin~\theta}, \frac{1}{\sin(\theta+\alpha)}$ are in A.P. The condition for A.P. gives: $\frac{2}{\sin~\theta} = \frac{1}{\sin(\theta-\alpha)} + \frac{1}{\sin(\theta+\alpha)}$.

Step 2: Simplify the right side using a common denominator. Combine the fractions on the right side: $\frac{2}{\sin~\theta} = \frac{\sin(\theta+\alpha) + \sin(\theta-\alpha)}{\sin(\theta-\alpha)\sin(\theta+\alpha)}$.

Step 3: Apply trigonometric sum-to-product and product-to-difference formulas. The numerator expands to $2\sin\theta\cos\alpha$. The denominator expands to the standard identity $\sin^{2}\theta - \sin^{2}\alpha$. Substituting these back gives: $\frac{2}{\sin~\theta} = \frac{2\sin~\theta\cos~\alpha}{\sin^{2}\theta - \sin^{2}\alpha}$.

Step 4: Cross-multiply and rearrange the terms. Divide both sides by 2 and cross-multiply: $\sin^{2}\theta - \sin^{2}\alpha = \sin^{2}\theta\cos~\alpha$. Gather all $\sin^{2}\theta$ terms on one side: $\sin^{2}\theta(1 - \cos~\alpha) = \sin^{2}\alpha$.

Step 5: Use half-angle identities to solve for $\cos 2\theta$. We apply the half-angle identities: $1 - \cos~\alpha = 2\sin^{2}(\frac{\alpha}{2})$ and $\sin~\alpha = 2\sin(\frac{\alpha}{2})\cos(\frac{\alpha}{2})$.
Squaring the sine identity gives $\sin^{2}\alpha = 4\sin^{2}(\frac{\alpha}{2})\cos^{2}(\frac{\alpha}{2})$. Substitute these into the rearranged equation: $\sin^{2}\theta(2\sin^{2}\frac{\alpha}{2}) = 4\sin^{2}\frac{\alpha}{2}\cos^{2}\frac{\alpha}{2}$. Assuming $\sin(\frac{\alpha}{2}) \neq 0$, we divide both sides by $2\sin^{2}(\frac{\alpha}{2})$ to isolate $\sin^{2}\theta$, resulting in $\sin^{2}\theta = 2\cos^{2}(\frac{\alpha}{2})$.
The question asks for $\cos~2\theta$. We know the double-angle identity $\cos~2\theta = 1 - 2\sin^{2}\theta$. Substituting our derived value for $\sin^{2}\theta$ into this identity yields $\cos~2\theta = 1 - 2(2\cos^{2}\frac{\alpha}{2}) = 1 - 4\cos^{2}\frac{\alpha}{2}$. $$ \therefore \text{The value of } \cos~2\theta \text{ is } 1-4~\cos^{2}\frac{\alpha}{2}. $$