Q55 / 150 · MHT CET 2026 (17 April Shift 2) · Mathematics
$\dfrac{(\cos 2\theta + i\sin 2\theta)^7}{(\cos 4\theta + i\sin 4\theta)^3} =$
- Correct answerA$\cos 2\theta + i\sin 2\theta$
- B$(\cos 2\theta + i\sin 2\theta)^4$
- C$\cos 4\theta + i\sin 4\theta$
- D$(\cos 4\theta + i\sin 4\theta)^{-2}$
Solution
Using De Moivre's theorem, $(\cos \theta + i\sin \theta)^n = \cos n\theta + i\sin n\theta$.
The numerator is $(\cos 2\theta + i\sin 2\theta)^7 = \cos 14\theta + i\sin 14\theta$.
The denominator is $(\cos 4\theta + i\sin 4\theta)^3 = \cos 12\theta + i\sin 12\theta$.
The given expression becomes $\dfrac{\cos 14\theta + i\sin 14\theta}{\cos 12\theta + i\sin 12\theta}$.
Using the property $\dfrac{\cos \alpha + i\sin \alpha}{\cos \beta + i\sin \beta} = \cos(\alpha - \beta) + i\sin(\alpha - \beta)$, we get:
$\cos(14\theta - 12\theta) + i\sin(14\theta - 12\theta) = \cos 2\theta + i\sin 2\theta$.
Answer: $\cos 2\theta + i\sin 2\theta$
The numerator is $(\cos 2\theta + i\sin 2\theta)^7 = \cos 14\theta + i\sin 14\theta$.
The denominator is $(\cos 4\theta + i\sin 4\theta)^3 = \cos 12\theta + i\sin 12\theta$.
The given expression becomes $\dfrac{\cos 14\theta + i\sin 14\theta}{\cos 12\theta + i\sin 12\theta}$.
Using the property $\dfrac{\cos \alpha + i\sin \alpha}{\cos \beta + i\sin \beta} = \cos(\alpha - \beta) + i\sin(\alpha - \beta)$, we get:
$\cos(14\theta - 12\theta) + i\sin(14\theta - 12\theta) = \cos 2\theta + i\sin 2\theta$.
Answer: $\cos 2\theta + i\sin 2\theta$
Hence, the correct answer is option: 'A'.