Q179 / 270 · NDA 2026 (Phase 1) · Mathematics
If $\cos\theta=\dfrac{1}{3}$, then what is the value of $\sin\left(\dfrac{\theta}{2}\right)\sin\left(\dfrac{3\theta}{2}\right)$ ?
- Correct answerA$\dfrac{5}{9}$
- B$\dfrac{7}{9}$
- C$\dfrac{10}{9}$
- D$\dfrac{11}{9}$
Solution
Using the trigonometric identity $2\sin A \sin B = \cos(A-B) - \cos(A+B)$, we have:
$\sin\left(\dfrac{\theta}{2}\right)\sin\left(\dfrac{3\theta}{2}\right) = \dfrac{1}{2}\left[\cos\left(\dfrac{3\theta}{2} - \dfrac{\theta}{2}\right) - \cos\left(\dfrac{3\theta}{2} + \dfrac{\theta}{2}\right)\right]$
$= \dfrac{1}{2}[\cos\theta - \cos(2\theta)]$
Using the double angle formula $\cos(2\theta) = 2\cos^2\theta - 1$, we get:
$\cos(2\theta) = 2\left(\dfrac{1}{3}\right)^2 - 1 = \dfrac{2}{9} - 1 = -\dfrac{7}{9}$
Substituting the values of $\cos\theta$ and $\cos(2\theta)$:
$\sin\left(\dfrac{\theta}{2}\right)\sin\left(\dfrac{3\theta}{2}\right) = \dfrac{1}{2}\left[\dfrac{1}{3} - \left(-\dfrac{7}{9}\right)\right]$
$= \dfrac{1}{2}\left[\dfrac{3}{9} + \dfrac{7}{9}\right]$
$= \dfrac{1}{2}\left[\dfrac{10}{9}\right] = \dfrac{5}{9}$
Answer: $\dfrac{5}{9}$
$\sin\left(\dfrac{\theta}{2}\right)\sin\left(\dfrac{3\theta}{2}\right) = \dfrac{1}{2}\left[\cos\left(\dfrac{3\theta}{2} - \dfrac{\theta}{2}\right) - \cos\left(\dfrac{3\theta}{2} + \dfrac{\theta}{2}\right)\right]$
$= \dfrac{1}{2}[\cos\theta - \cos(2\theta)]$
Using the double angle formula $\cos(2\theta) = 2\cos^2\theta - 1$, we get:
$\cos(2\theta) = 2\left(\dfrac{1}{3}\right)^2 - 1 = \dfrac{2}{9} - 1 = -\dfrac{7}{9}$
Substituting the values of $\cos\theta$ and $\cos(2\theta)$:
$\sin\left(\dfrac{\theta}{2}\right)\sin\left(\dfrac{3\theta}{2}\right) = \dfrac{1}{2}\left[\dfrac{1}{3} - \left(-\dfrac{7}{9}\right)\right]$
$= \dfrac{1}{2}\left[\dfrac{3}{9} + \dfrac{7}{9}\right]$
$= \dfrac{1}{2}\left[\dfrac{10}{9}\right] = \dfrac{5}{9}$
Answer: $\dfrac{5}{9}$
Hence, the correct answer is option: 'A'.