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Trigonometry — Mathematics STD 10 — Question

ICSE BoardEnglish MediumSTD 10MathematicsTrigonometry4 Marks
Question
Prove that $\sin^25^\circ + \sin^210^\circ$ .......... $+ \sin^285^\circ + \sin^290^\circ = 9 \frac{1}{2}$.
✓

Answer

$LHS$
$= \sin^25^\circ + \sin^210^\circ + \sin^215^\circ + \sin^220^\circ + \sin^225^\circ + \sin^230^\circ + \sin^235^\circ +$
$ \sin^240^\circ + in^2 45^\circ + \sin^250^\circ + \sin^255^\circ + \sin^260^\circ + $
$\sin^265^\circ + \sin^270^\circ + \sin^275^\circ + \sin^280^\circ + \sin^285^\circ + \sin^290^\circ .$
$= (\sin^25^\circ + \sin^285^\circ ) + (\sin^210^\circ + \sin^280^\circ ) + (\sin^215^\circ + \sin^275^\circ ) +$
$ (\sin^220^\circ + \sin^270^\circ ) + (\sin^225^\circ + \sin^265^\circ ) + (\sin^230^\circ + \sin^260^\circ )$
$ + (\sin^235^\circ + \sin^255^\circ ) + (\sin^240^\circ + \sin^250^\circ ) + \sin^2 45^\circ + \sin^290^\circ .$
$= (\sin^25^\circ + \cos^25^\circ ) + (\sin^210^\circ + \cos^210^\circ ) + (\sin^215^\circ + \cos^215^\circ )$
$ + (\sin^220^\circ + \cos^220^\circ ) + (\sin^225^\circ + \cos^225^\circ ) $
$+ (\sin^230^\circ + \cos^230^\circ ) + (\sin^235^\circ + \cos^235^\circ ) + (\sin^240^\circ + \cos^240^\circ ) + $
$\left(\frac{1}{\sqrt{2}}\right)^2+(1)^2 \ldots\left[\because \sin \left(90^{\circ}-\theta\right)=\cos \theta \because \sin 90^{\circ}=1\right.$ and $\left.\sin 45^{\circ}=\frac{1}{\sqrt{2}}\right]\left[\because \sin ^2 \theta+\cos ^2 \theta=1\right]$
$=1+1+1+1+1+1+1+1+\frac{1}{2}+1$
$=9 \frac{1}{2}$
$= RHS$
Hence proved.

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