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Let \alpha ,\beta   be such that  \pi < \alpha -\beta < 3\pi .If  \sin \alpha +\sin \beta = -21/65,  and 

\cos \alpha +\cos \beta = -27/65 then the value of  \cos \frac{\alpha -\beta }{2} is :

  • Option 1)


  • Option 2)


  • Option 3)


  • Option 4)



Answers (1)


As we learnt in 

Trigonometric Ratios of Submultiples of an Angle -

Trigonometric ratios of submultiples of an angle 1

- wherein

This shows the formulae for half angles and their doubles.


 \sin \alpha +\sin \beta =\frac{-21}{65}\Rightarrow 2\sin \frac{(\alpha +\beta )}{2}\cos \frac{(\alpha -\beta )}{2}=\frac{-21}{65}    -----------------(i)

\cos \alpha +\cos \beta =\frac{-27}{65}\Rightarrow 2\cos \frac{(\alpha +\beta )}{2}\cos \frac{(\alpha -\beta )}{2}=\frac{-21}{65} -------------------(ii)

Squaring and adding,

4\cos ^{2}\frac{(\alpha -\beta) }{2}[\sin ^{2}\frac{\alpha +\beta}{2} +\cos ^{2}\frac{\alpha +\beta}{2}]=\frac{21^{2}+27^{^{2}}}{65^{2}}

\Rightarrow \cos ^{2}\frac{(\alpha -\beta) }{2}=\frac{1170}{4\times 4225}\Rightarrow \cos ^{2}\frac{(\alpha -\beta) }{2}=\frac{1170}{4\times 4225}=\frac{9}{130}

\Rightarrow \cos\frac{(\alpha -\beta) }{2}=\frac{-3}{\sqrt{130}}

[\because \pi< \alpha -\beta < 3\pi\Rightarrow \frac{\pi}{2} \leq \frac{\alpha-\beta}{2} \leq \frac{3\pi}{2} \Rightarrow cos\left(\frac{\alpha-\beta}{2} \right )<0]


Option 1)


This is incorrect option

Option 2)


This is incorrect option

Option 3)


This is correct option

Option 4)


This is incorrect option

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