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JEE Advanced
Exam
JEE Advanced
Subject
Mathematics
Chapter
Trigonometry

JEE Advanced · Mathematics · Trigonometry

Let f(θ)= (1+(4 θ)/( 6 θ)) (1+(4 2 θ)/( 5 θ)), then:

Let f(θ)=(1+4sin⁡θsin⁡6θ)\lleft(1+4sin⁡2θsin⁡5θ),f(\theta)= \left(1+\frac{4 \sin \theta}{\sin 6 \theta}\right)\lleft(1+\frac{4 \sin 2 \theta}{\sin 5 \theta}),then:

Options

  1. Af(π7)=25f\left(\frac{\pi}{7}\right)=25
  2. Bf(π7)=−25f\left(\frac{\pi}{7}\right)=-25
  3. Cf(2π7)=9f\left(\frac{2 \pi}{7}\right)=9
  4. Df(2π7)=−9f\left(\frac{2 \pi}{7}\right)=-9

Correct answer

C. f(2π7)=9f\left(\frac{2 \pi}{7}\right)=9

Step-by-step solution

f(π7)=5×5=25f\left(\frac{\pi}{7}\right)=5 \times 5=25 and f(2π7)=−3(−3)=9f\left(\frac{2 \pi}{7}\right)=-3(-3)=9
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