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Trigonometric Ratio and Identites question

2004 · Shift 0 · Q95
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Trigonometric Ratio and Identites question

2004 · Shift 0 · Q95

JEE MainMathematicsTrigonometric Ratio and IdentitesMCQ+4 / −1
Let α, β\alpha ,\,\betaα,β be such that π<α−β<3π\pi \lt \alpha - \beta \lt 3\piπ<α−β<3π. If sin α+sin⁡β=−2165sin{\mkern 1mu} \alpha + \sin \beta = - {{21} \over {65}}sinα+sinβ=−6521​ and cos⁡α+cos⁡β=−2765\cos \alpha + \cos \beta = - {{27} \over {65}}cosα+cosβ=−6527​ then the value of cos⁡α−β2\cos {{\alpha - \beta } \over 2}cos2α−β​ :
  1. A
    −665  {{ - 6} \over {65}}\,\,65−6​
  2. B
    3130{3 \over {\sqrt {130} }}130​3​
  3. C
    665{6 \over {65}}656​
  4. D
    −3130- {3 \over {\sqrt {130} }}−130​3​
View written solutionFree

Correct answer: D

  1. Use sum-to-product identities

Given sin⁡α+sin⁡β=−2165,cos⁡α+cos⁡β=−2765.\sin\alpha+\sin\beta=-\frac{21}{65},\qquad \cos\alpha+\cos\beta=-\frac{27}{65}.sinα+sinβ=−6521​,cosα+cosβ=−6527​.

Recall: sin⁡α+sin⁡β=2sin⁡α+β2cos⁡α−β2,\sin\alpha+\sin\beta=2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2},sinα+sinβ=2sin2α+β​cos2α−β​, cos⁡α+cos⁡β=2cos⁡α+β2cos⁡α−β2.\cos\alpha+\cos\beta=2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}.cosα+cosβ=2cos2α+β​cos2α−β​.

Let S=α+β2,D=α−β2.S=\frac{\alpha+\beta}{2},\qquad D=\frac{\alpha-\beta}{2}.S=2α+β​,D=2α−β​. Then 2sin⁡Scos⁡D=−2165,2\sin S\cos D=-\frac{21}{65},2sinScosD=−6521​, 2cos⁡Scos⁡D=−2765.2\cos S\cos D=-\frac{27}{65}.2cosScosD=−6527​.

  1. Eliminate SSS by squaring and adding

Squaring both equations and adding: 4cos⁡2D(sin⁡2S+cos⁡2S)=(2165)2+(2765)2.4\cos^2 D(\sin^2 S+\cos^2 S)=\left(\frac{21}{65}\right)^2+\left(\frac{27}{65}\right)^2.4cos2D(sin2S+cos2S)=(6521​)2+(6527​)2. Since sin⁡2S+cos⁡2S=1\sin^2 S+\cos^2 S=1sin2S+cos2S=1, 4cos⁡2D=212+272652.4\cos^2 D=\frac{21^2+27^2}{65^2}.4cos2D=652212+272​. Compute: 212=441,272=729,441+729=1170.21^2=441,\qquad 27^2=729,\qquad 441+729=1170.212=441,272=729,441+729=1170. So 4\cos^2 D=\frac{1170}{4225}= rac{18}{65}. Hence \cos^2 D=\frac{18}{260}= rac{9}{130}. Therefore, cos⁡D=±3130.\cos D=\pm \frac{3}{\sqrt{130}}.cosD=±130​3​.

  1. Determine the sign using the condition on α−β\alpha-\betaα−β

We are given π<α−β<3π.\pi<\alpha-\beta<3\pi.π<α−β<3π. Dividing by 222: π2<D<3π2.\frac{\pi}{2}<D<\frac{3\pi}{2}.2π​<D<23π​. In this interval, cos⁡D<0\cos D<0cosD<0.

Therefore, cos⁡α−β2=cos⁡D=−3130.\cos\frac{\alpha-\beta}{2}=\cos D=-\frac{3}{\sqrt{130}}.cos2α−β​=cosD=−130​3​.

  1. Match with the options

This corresponds to Option D.

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