Question : The value of the expression $\sin^{2}1^{\circ}+\sin^{2}11^{\circ}+\sin^{2}21^{\circ}+\sin^{2}31^{\circ}+\sin^{2}41^{\circ}+\sin^{2}45^{\circ}+\sin^{2}49^{\circ}+\sin^{2}59^{\circ}+\sin^{2}69^{\circ}+\sin^{2}79^{\circ}+\sin^{2}89^{\circ}$ is:
Option 1: $0$
Option 2: $5\frac{1}{2}$
Option 3: $5$
Option 4: $4\frac{1}{2}$
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Correct Answer: $5\frac{1}{2}$
Solution :
$\sin^{2}1^{\circ}+\sin^{2}11^{\circ}+\sin^{2}21^{\circ}+\sin^{2}31^{\circ}+\sin^{2}41^{\circ}+\sin^{2}45^{\circ}+\sin^{2}49^{\circ}+\sin^{2}59^{\circ}+\sin^{2}69^{\circ}+\sin^{2}79^{\circ}+\sin^{2}89^{\circ}$
Using the identity $\sin(90^{\circ} - \theta) = \cos\theta$, we get,
$\sin^{2}49^{\circ}=\sin^{2}(90^\circ-41^{\circ}) = \cos^2{41}^\circ$
$\sin^{2}59^{\circ}=\sin^{2}(90^\circ-31^{\circ}) = \cos^2{31}^\circ$
$\sin^{2}69^{\circ}=\sin^{2}(90^\circ-21^{\circ}) = \cos^2{21}^\circ$
$\sin^{2}79^{\circ}=\sin^{2}(90^\circ-11^{\circ}) = \cos^2{11}^\circ$
$\sin^{2}89^{\circ}=\sin^{2}(90^\circ-1^{\circ}) = \cos^2{1}^\circ$
$=\sin^{2}1^{\circ}+\sin^{2}11^{\circ}+\sin^{2}21^{\circ}+\sin^{2}31^{\circ}+\sin^{2}41^{\circ}+\sin^{2}45^{\circ}+\cos^{2}41^{\circ}+\cos^{2}31^{\circ}+\cos^{2}21^{\circ}+\cos^{2}11^{\circ}+\cos^{2}1^{\circ}$Each pair adds up to $1$ because $\sin^{2}\theta + \cos^{2}\theta = 1$
$=5+\sin^{2}45^{\circ}$
$=5+\frac{1}{2}$
$=\frac{11}{2}=5\frac{1}{2}$
Hence, the correct answer is $5\frac{1}{2}$.
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