If \cos \alpha+\cos \beta=a, and \sin \alpha+\sin \beta=b and \theta is the arithmetic mean between \alpha and \beta, then \sin 2 \theta+\cos 2 \theta is equal to

Correct Answer is : None of these

NIMCET 2008 Question Paper with solutions

Question : 26 of 120

Marks: +1, -0

\cos α+\cos β=a,

and

\sin α+\sin β=b

and

θ

is the arithmetic mean between

α

and

β

, then

\sin 2 θ+\cos 2 θ

is equal to

Solution:

Given

\cos α+\cos β=a

⇒ 2 \cos ( {α+β}/{2}) \cos ( {α-β}/{2})=a

⇒ 2 \cos θ \cos (\;{α-β}/{2})=a

.....(i)

[ ∵ θ=\;{α+β}/{2}( {AM}

α, β)]

Also,

\sin α+\sin β=b

⇒ 2 \sin ( {α+β}/{2}) \cos ( {α-β}/{2})=b

⇒ 2 \sin θ \cos ({α-β}/{2})=b

...(ii)

⇒ {a}/{\cos θ}= {b}/{\sin θ}

= {√{a^{2}+b^{2}}}/{√{\sin ^{2} θ+\cos ^{2} θ}}=√{a^{2}+b^{2}}

(from Eqs. (i) and (ii)]

⇒ \cos θ=\;{a}/{√{a^{2}+b^{2}}}

and

\sin θ= {b}/{√{a^{2}+b^{2}}}

⇒ \sin 2 θ= {2 a b}/{a^{2}+b^{2}}

\cos 2 θ= {a^{2}-b^{2}}/{a^{2}+b^{2}}

⇒ \sin 2 θ+\cos 2 θ= {a^{2}-b^{2}+2 a b}/{a^{2}+b^{2}}

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