Let α,β be such that π<α−β<3π. If Sin α+Sin⁡β=−2165 and Cos⁡α+Cos⁡β=−2765 then the value of cos⁡α−β2 is

sin⁡α+sin⁡β=−2165-----(1) cos⁡α+cos⁡β=−2765----(2) Squaring Eq. (i), we get ⇒sin2⁡α+sin2⁡β+2sin⁡αsin⁡β=-21652 Squaring Eq. (ii), we get ⇒cos2⁡α+cos2⁡β+2cos⁡αcos⁡β=-27652 Adding Eqs. (iii) and (iv), we get 2+2cos⁡(α−β)=1(65)2(27)2+(21)2=1(65)2(729+441)=1(65)2(1170)=18651+cos⁡α−β=9652cos2⁡α−β2=965cos⁡α−β2=−3130 ∵π<α−β<3π⇒π2<α−β2<3π2⇒cos⁡α−β2<0

Let α,β be such that π<α−β<3π. If Sin α+Sin⁡β=−2165 and Cos⁡α+Cos⁡β=−2765 then the value of cos⁡α−β2 is

sin⁡α+sin⁡β=−2165-----(1) cos⁡α+cos⁡β=−2765----(2) Squaring Eq. (i), we get ⇒sin2⁡α+sin2⁡β+2sin⁡αsin⁡β=-21652 Squaring Eq. (ii), we get ⇒cos2⁡α+cos2⁡β+2cos⁡αcos⁡β=-27652 Adding Eqs. (iii) and (iv), we get 2+2cos⁡(α−β)=1(65)2(27)2+(21)2=1(65)2(729+441)=1(65)2(1170)=18651+cos⁡α−β=9652cos2⁡α−β2=965cos⁡α−β2=−3130 ∵π<α−β<3π⇒π2<α−β2<3π2⇒cos⁡α−β2<0

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$Let α, β be such that π < α - β < 3 π . If Sin α + Sin β = \frac{- 21}{65} and Cos α + Cos β = \frac{- 27}{65} then the value of \cos \frac{α - β}{2} is$

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$\frac{- 3}{\sqrt{130}}$

$\frac{- 6}{65}$

$\frac{6}{65}$

$\frac{3}{\sqrt{130}}$

answer is A.

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Detailed Solution

$\sin α + \sin β = - \frac{21}{65} - - - - - (1) \cos α + \cos β = - \frac{27}{65} - - - - (2)$

$Squaring Eq. (i), we get ⇒ \sin^{2} α + \sin^{2} β + 2 \sin α \sin β = {(\frac{- 21}{65})}^{2} Squaring Eq. (ii), we get ⇒ \cos^{2} α + \cos^{2} β + 2 \cos α \cos β = {(\frac{- 27}{65})}^{2} Adding Eqs. (iii) and (iv), we get \begin{array}{l} 2 + 2 \cos (α - β) = \frac{1}{(65)^{2}} [(27)^{2} + (21)^{2}] \\ = \frac{1}{(65)^{2}} (729 + 441) \\ = \frac{1}{(65)^{2}} (1170) = \frac{18}{65} \\ 1 + \cos α - β) = \frac{9}{65} \\ 2 \cos^{2} \frac{α - β}{2} = \frac{9}{65} \\ \cos \frac{α - β}{2} = - \frac{3}{\sqrt{130}} \end{array} [∵ π < α - β < 3 π \Rightarrow \frac{π}{2} < \frac{α - β}{2} < \frac{3 π}{2} \Rightarrow \cos (\frac{α - β}{2}) < 0]$