Q116 / 180 · COMEDK 2026 (9 May Shift 2) · Mathematics
If $2\sin\theta = \left(x + \dfrac{1}{x}\right)$, then $\sin 3\theta + \dfrac{1}{2}\left(x^3 + \dfrac{1}{x^3}\right) =$
- Correct answerA$0$
- B$3$
- C$-1$
- D$1$
Solution
Given $x + \dfrac{1}{x} = 2\sin\theta$
Cubing both sides, we get:
$\left(x + \dfrac{1}{x}\right)^3 = (2\sin\theta)^3$
$x^3 + \dfrac{1}{x^3} + 3\left(x + \dfrac{1}{x}\right) = 8\sin^3\theta$
Substituting $x + \dfrac{1}{x} = 2\sin\theta$:
$x^3 + \dfrac{1}{x^3} + 3(2\sin\theta) = 8\sin^3\theta$
$x^3 + \dfrac{1}{x^3} = 8\sin^3\theta - 6\sin\theta$
$x^3 + \dfrac{1}{x^3} = -2(3\sin\theta - 4\sin^3\theta)$
Using the identity $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$:
$x^3 + \dfrac{1}{x^3} = -2\sin 3\theta$
Dividing by $2$:
$\dfrac{1}{2}\left(x^3 + \dfrac{1}{x^3}\right) = -\sin 3\theta$
Therefore, the given expression becomes:
$\sin 3\theta + \dfrac{1}{2}\left(x^3 + \dfrac{1}{x^3}\right) = \sin 3\theta - \sin 3\theta = 0$
Answer: $0$
Cubing both sides, we get:
$\left(x + \dfrac{1}{x}\right)^3 = (2\sin\theta)^3$
$x^3 + \dfrac{1}{x^3} + 3\left(x + \dfrac{1}{x}\right) = 8\sin^3\theta$
Substituting $x + \dfrac{1}{x} = 2\sin\theta$:
$x^3 + \dfrac{1}{x^3} + 3(2\sin\theta) = 8\sin^3\theta$
$x^3 + \dfrac{1}{x^3} = 8\sin^3\theta - 6\sin\theta$
$x^3 + \dfrac{1}{x^3} = -2(3\sin\theta - 4\sin^3\theta)$
Using the identity $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$:
$x^3 + \dfrac{1}{x^3} = -2\sin 3\theta$
Dividing by $2$:
$\dfrac{1}{2}\left(x^3 + \dfrac{1}{x^3}\right) = -\sin 3\theta$
Therefore, the given expression becomes:
$\sin 3\theta + \dfrac{1}{2}\left(x^3 + \dfrac{1}{x^3}\right) = \sin 3\theta - \sin 3\theta = 0$
Answer: $0$
Hence, the correct answer is option: 'A'.