MCQ
The value of $\sin^25^\circ+\sin^210^\circ+\sin^215^\circ+\ ...\ +\sin^285^\circ+\sin^290^\circ$ is:
- A$7$
- B$8$
- ✓$9.5$
- D$10$
✓
Answer
Correct option: C.
$9.5$
We have:
$\sin^25^\circ+\sin^210^\circ+\sin^215^\circ+...+\sin^285^\circ+\sin^290^\circ$
$=\sin^25^\circ+\sin^210^\circ+\sin^215^\circ+...+\sin^2(90^\circ-10^\circ)+\sin^2(90^\circ-5^\circ)+\sin^290^\circ$
$=\sin^25^\circ+\sin^210^\circ+\sin^215^\circ+...+\cos^210^\circ+\cos^25^\circ+\sin^290^\circ$
$=(\sin^25^\circ+\cos^25^\circ)+(\sin^210^\circ+\cos^210^\circ)+(\sin^215^\circ+\cos^215^\circ)$
$+(\sin^220^\circ+\cos^220^\circ)+(\sin^225^\circ+\cos^225^\circ)+(\sin^230^\circ+\cos^230^\circ)$
$+(\sin^235^\circ+\cos^235^\circ)+(\sin^240^\circ+\cos^240^\circ)+\sin^245^\circ+\sin^290^\circ$
$=1+1+1+1+1+1+1+1+\Big(\frac{1}{\sqrt2}\Big)+(1)^2$ $[\because \sin^2\theta+\cos^2\theta=1]$
$=8+\frac{1}{2}+1$
$=9.5$
$\sin^25^\circ+\sin^210^\circ+\sin^215^\circ+...+\sin^285^\circ+\sin^290^\circ$
$=\sin^25^\circ+\sin^210^\circ+\sin^215^\circ+...+\sin^2(90^\circ-10^\circ)+\sin^2(90^\circ-5^\circ)+\sin^290^\circ$
$=\sin^25^\circ+\sin^210^\circ+\sin^215^\circ+...+\cos^210^\circ+\cos^25^\circ+\sin^290^\circ$
$=(\sin^25^\circ+\cos^25^\circ)+(\sin^210^\circ+\cos^210^\circ)+(\sin^215^\circ+\cos^215^\circ)$
$+(\sin^220^\circ+\cos^220^\circ)+(\sin^225^\circ+\cos^225^\circ)+(\sin^230^\circ+\cos^230^\circ)$
$+(\sin^235^\circ+\cos^235^\circ)+(\sin^240^\circ+\cos^240^\circ)+\sin^245^\circ+\sin^290^\circ$
$=1+1+1+1+1+1+1+1+\Big(\frac{1}{\sqrt2}\Big)+(1)^2$ $[\because \sin^2\theta+\cos^2\theta=1]$
$=8+\frac{1}{2}+1$
$=9.5$
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