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NDA 2026
Exam
NDA
Year
2026
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2026 · Mathematics · Trigonometric Ratios & Identities

If θ=(1)/(3) , then what is the value of ((θ)/(2)) ((3θ)/(2)) ?

If cos⁡θ=13\cos\theta=\dfrac{1}{3}, then what is the value of sin⁡\sin\left(\dfrac{\theta}{2}\right)\sin(3θ2)\left(\dfrac{3\theta}{2}\right) ?

Options

  1. A59\dfrac{5}{9}
  2. B79\dfrac{7}{9}
  3. C109\dfrac{10}{9}
  4. D119\dfrac{11}{9}

Correct answer

A. 59\dfrac{5}{9}

Step-by-step solution

Using the trigonometric identity 2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2\sin A \sin B = \cos(A-B) - \cos(A+B), we have:

sin⁡(θ2)sin⁡(3θ2)=12[cos⁡(3θ2−θ2)−cos⁡(3θ2+θ2)]\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]

=12[cos⁡θ−cos⁡(2θ)]= \dfrac{1}{2}[\cos\theta - \cos(2\theta)]

Using the double angle formula cos⁡(2θ)=2cos⁡2θ−1\cos(2\theta) = 2\cos^2\theta - 1, we get:

cos⁡(2θ)=2(13)2−1=29−1=−79\cos(2\theta) = 2\left(\dfrac{1}{3}\right)^2 - 1 = \dfrac{2}{9} - 1 = -\dfrac{7}{9}

Substituting the values of cos⁡θ\cos\theta and cos⁡(2θ)\cos(2\theta):

sin⁡(θ2)sin⁡(3θ2)=12[13−(−79)]\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]

=12[39+79]= \dfrac{1}{2}\left[\dfrac{3}{9} + \dfrac{7}{9}\right]

=12[109]=59= \dfrac{1}{2}\left[\dfrac{10}{9}\right] = \dfrac{5}{9}

Answer: 59\dfrac{5}{9}
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