{( 2 θ+i 2 θ)^{7}}{( 4 θ+i 4 θ)³}=

Correct Answer is : cos⁡2θ+isin⁡2θ 2 θ+i 2 θcos2θ+isin2θ

Correct Answer is : cos⁡2θ+isin⁡2θ 2 θ+i 2 θcos2θ+isin2θ

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MHTCET 17 Apr 2026 Shift 2 Paper

Question : 102 of 150

Marks: +1, -0

\frac{(\cos 2 \theta+i \sin 2 \theta)^{7}}{(\cos 4 \theta+i \sin 4 \theta)^{3}}=

Solution:

Concept:

De Moivre's theorem simplifies powers of complex numbers in polar form, and division of such numbers involves subtracting their arguments.

Explanation:

Apply De Moivre's theorem:

(\cos \theta + i\sin \theta)^n = \cos(n\theta) + i\sin(n\theta)

Numerator:

(\cos 2\theta + i\sin 2\theta)^7 = \cos(14\theta) + i\sin(14\theta)

Denominator:

(\cos 4\theta + i\sin 4\theta)^3 = \cos(12\theta) + i\sin(12\theta)

Dividing two complex numbers in polar form gives magnitude ratio

1/1 = 1

and angle difference

(14\theta - 12\theta)

Therefore,

\frac{\cos(14\theta) + i\sin(14\theta)}{\cos(12\theta) + i\sin(12\theta)} = \cos(2\theta) + i\sin(2\theta)

Answer:

Option A:

\cos 2\theta + i \sin 2\theta

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