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JEE Main 2004
Exam
JEE Main
Year
2004
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Main 2004 · Mathematics · Trigonometric Ratios & Identities

Let α, β be such that π < α-β < 3 π . If α+ β=-(21)/(65) and α+...

Let α,β\alpha, \beta be such that π<α−β<3π\pi < \alpha-\beta < 3 \pi. If sin⁡α+sin⁡β=−2165\sin \alpha+\sin \beta=-\frac{21}{65} and cos⁡α+cos⁡β=−2765\cos \alpha+\cos \beta=-\frac{27}{65}, then the value of cos⁡α−β2\cos \frac{\alpha-\beta}{2} is

Options

  1. A−3130-\frac{3}{\sqrt{130}}
  2. B3130\frac{3}{\sqrt{130}}
  3. C665\frac{6}{65}
  4. D−665-\frac{6}{65}

Correct answer

A. −3130-\frac{3}{\sqrt{130}}

Step-by-step solution

sin⁡α+sin⁡β=−2165\sin \alpha+\sin \beta=-\frac{21}{65} and cos⁡α+cos⁡β=−2765\cos \alpha+\cos \beta=-\frac{27}{65}. Squaring and adding, we get 2+2cos⁡(α−β)=1170(65)2⇒cos⁡2(α−β2)=9130⇒cos⁡(α−β2)=−3130(∵π2<α−β2<3π2).\begin{aligned}& 2+2 \cos (\alpha-\beta)=\frac{1170}{(65)^2}\\ & \Rightarrow \cos ^2\left(\frac{\alpha-\beta}{2}\right)=\frac{9}{130} \Rightarrow \cos \left(\frac{\alpha-\beta}{2}\right)=\frac{-3}{\sqrt{130}} \quad\left(\because \frac{\pi}{2}< \frac{\alpha-\beta}{2}< \frac{3 \pi}{2}\right) . \end{aligned}
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