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AP EAPCET 19-Aug-2021 Shift 2 Paper

Section: Mathematics

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Question : 35 of 160

Marks: +1, -0

Mean of the values

\sin^{2} 10^{\circ}, \sin^{2} 20^{\circ}, \sin^{2} 30^{\circ}, \ldots, \sin^{2} 90^{\circ}

is ______

$\frac{5}{9}$
$\frac{2}{3}$
$\frac{7}{9}$
$\frac{1}{9}$

Solution:

Mean

=\frac{\sin^{2} 10^{\circ}+\sin^{2} 20^{\circ}+\sin^{2} 30^{\circ}+ \dots +\sin^{2} 90^{\circ}}{9}

\Rightarrow \sin (90^{\circ}-\theta)=\cos \theta

and

\sin^{2} \theta+\cos^{2} \theta=1

\because \sin^{2} 80^{\circ}=\cos^{2} 10^{\circ}, \sin 70^{\circ}=\cos^{2} 20^{\circ}

,

\sin^{2} 60=\cos^{2} 30, \sin^{2} 50^{\circ}=\cos^{2} 40^{\circ}

(\sin^{2} 10^{\circ}+\sin^{2} 80^{\circ})+(\sin^{2} 20^{\circ}+\sin^{2} 70^{\circ})

\Rightarrow \text{Mean}=\frac{+\dots+(\sin^{2} 40^{\circ}+\sin^{2} 50^{\circ})+\sin^{2} 90^{\circ}}{9}

=\frac{1+1+1+1+1}{9}

[\because \sin^{2} 80^{\circ}=\cos^{2} 10^{\circ}, \sin^{2} 70^{\circ}=\cos^{2} 20, \sin^{2} 60^{\circ}=\cos^{2} 30^{\circ} \sin^{2} 50^{\circ}=\cos^{2} 40^{\circ}]

=\frac{5}{9}

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