If cosα+cosβ=32 and sinα+sinβ=12 and θis the arithmetic mean of α and β then sin2θ+cos2θ=_____

cosα+cosβ=32→1 sinα+sinβ=12→22cosα+β2cosα−β2=32 2sinα+β2cosα−β2=1212=2cosα+β2cosα−β22sinα+β2cosα−β2=3212cotα+β2=3tanα+β2=13θ=α+β2⇒2θ=α+βsin2θ+cos2θ=sinα+β+cosα+β=2tanα+β21+tan2α+β2+1−tan2α+β21+tan2α+β2=231+19+1−191+19=610+810=1410=75

If cosα+cosβ=32 and sinα+sinβ=12 and θis the arithmetic mean of α and β then sin2θ+cos2θ=_____

cosα+cosβ=32→1 sinα+sinβ=12→22cosα+β2cosα−β2=32 2sinα+β2cosα−β2=1212=2cosα+β2cosα−β22sinα+β2cosα−β2=3212cotα+β2=3tanα+β2=13θ=α+β2⇒2θ=α+βsin2θ+cos2θ=sinα+β+cosα+β=2tanα+β21+tan2α+β2+1−tan2α+β21+tan2α+β2=231+19+1−191+19=610+810=1410=75

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If $\cos α + \cos β = \frac{3}{2}$ and $\sin α + \sin β = \frac{1}{2}$ and $θ$ is the arithmetic mean of α and $β$ then $\sin 2 θ + \cos 2 θ =_____$

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$\frac{3}{5}$

$\frac{4}{5}$

$\frac{7}{5}$

$\frac{8}{5}$

answer is C.

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Detailed Solution

$\begin{array}{l} \cos α + \cos β = \frac{3}{2} \to (1) \sin α + \sin β = \frac{1}{2} \to (2) \\ 2 \cos (\frac{α + β}{2}) \cos (\frac{α - β}{2}) = \frac{3}{2} 2 \sin (\frac{α + β}{2}) \cos (\frac{α - β}{2}) = \frac{1}{2} \\ \frac{(1)}{(2)} = \frac{2 \cos (\frac{α + β}{2}) \cos (\frac{α - β}{2})}{2 \sin (\frac{α + β}{2}) \cos (\frac{α - β}{2})} = \frac{\frac{3}{2}}{\frac{1}{2}} \\ \cot (\frac{α + β}{2}) = 3 \\ \tan (\frac{α + β}{2}) = \frac{1}{3} \\ θ = \frac{α + β}{2} \Rightarrow 2 θ = α + β \\ \sin 2 θ + \cos 2 θ = \sin (α + β) + \cos (α + β) \\ = \frac{2 \tan (\frac{α + β}{2})}{1 + \tan^{2} (\frac{α + β}{2})} + \frac{1 - \tan^{2} (\frac{α + β}{2})}{1 + \tan^{2} (\frac{α + β}{2})} \\ = \frac{\frac{2}{3}}{1 + \frac{1}{9}} + \frac{1 - \frac{1}{9}}{1 + \frac{1}{9}} \\ = \frac{6}{10} + \frac{8}{10} = \frac{14}{10} = \frac{7}{5} \end{array}$