JEE MAIN - Mathematics (2004 - No. 9)
Let $$\alpha ,\,\beta $$ be such that $$\pi < \alpha - \beta < 3\pi $$.
If $$sin{\mkern 1mu} \alpha + \sin \beta = - {{21} \over {65}}$$ and $$\cos \alpha + \cos \beta = - {{27} \over {65}}$$ then the value of $$\cos {{\alpha - \beta } \over 2}$$ :
If $$sin{\mkern 1mu} \alpha + \sin \beta = - {{21} \over {65}}$$ and $$\cos \alpha + \cos \beta = - {{27} \over {65}}$$ then the value of $$\cos {{\alpha - \beta } \over 2}$$ :
$${{ - 6} \over {65}}\,\,$$
$${3 \over {\sqrt {130} }}$$
$${6 \over {65}}$$
$$ - {3 \over {\sqrt {130} }}$$
Explanation
Given $$sin{\mkern 1mu} \alpha + \sin \beta = - {{21} \over {65}}$$ .........(1)
and $$\cos \alpha + \cos \beta = - {{27} \over {65}}$$ ........(2)
Square and add (1) and (2) you will get
$$2\left( {1 + \cos \alpha \cos \beta + \sin \alpha \sin \beta } \right)$$$$ = {{{{\left( {21} \right)}^2} + {{\left( {27} \right)}^2}} \over {{{\left( {65} \right)}^2}}}$$
$$ \Rightarrow $$ $$2\left( {1 + \cos \left( {\alpha - \beta } \right)} \right) = {{1170} \over {{{\left( {65} \right)}^2}}}$$
$$ \Rightarrow $$ $$4{\cos ^2}{{\alpha - \beta } \over 2}$$$$ = {{1170} \over {{{\left( {65} \right)}^2}}}$$
$$ \Rightarrow $$ $${\cos ^2}{{\alpha - \beta } \over 2}$$$$ = {9 \over {130}}$$
$$\therefore$$ $$\cos {{\alpha - \beta } \over 2} = \pm {3 \over {\sqrt {130} }}$$
[ But $$\cos {{\alpha - \beta } \over 2} \ne + {3 \over {\sqrt {130} }}$$
as $$\pi < \alpha - \beta < 3\pi $$
$$ \Rightarrow $$ $${\pi \over 2} < {{\alpha - \beta } \over 2} < {{3\pi } \over 2}$$
$$ \Rightarrow $$ $$\cos {{\alpha - \beta } \over 2} < 0$$ ]
So $$\cos {{\alpha - \beta } \over 2} = - {3 \over {\sqrt {130} }}$$
and $$\cos \alpha + \cos \beta = - {{27} \over {65}}$$ ........(2)
Square and add (1) and (2) you will get
$$2\left( {1 + \cos \alpha \cos \beta + \sin \alpha \sin \beta } \right)$$$$ = {{{{\left( {21} \right)}^2} + {{\left( {27} \right)}^2}} \over {{{\left( {65} \right)}^2}}}$$
$$ \Rightarrow $$ $$2\left( {1 + \cos \left( {\alpha - \beta } \right)} \right) = {{1170} \over {{{\left( {65} \right)}^2}}}$$
$$ \Rightarrow $$ $$4{\cos ^2}{{\alpha - \beta } \over 2}$$$$ = {{1170} \over {{{\left( {65} \right)}^2}}}$$
$$ \Rightarrow $$ $${\cos ^2}{{\alpha - \beta } \over 2}$$$$ = {9 \over {130}}$$
$$\therefore$$ $$\cos {{\alpha - \beta } \over 2} = \pm {3 \over {\sqrt {130} }}$$
[ But $$\cos {{\alpha - \beta } \over 2} \ne + {3 \over {\sqrt {130} }}$$
as $$\pi < \alpha - \beta < 3\pi $$
$$ \Rightarrow $$ $${\pi \over 2} < {{\alpha - \beta } \over 2} < {{3\pi } \over 2}$$
$$ \Rightarrow $$ $$\cos {{\alpha - \beta } \over 2} < 0$$ ]
So $$\cos {{\alpha - \beta } \over 2} = - {3 \over {\sqrt {130} }}$$
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