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Trigonometry – II — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometry – II2 Marks
MCQ
$\sin^2\left(3^{\circ}\right)+\sin^2\left(6^{\circ}\right)+\sin^2\left(9^{\circ}\right)+\ldots+\sin^2\left(84^{\circ}\right)+\sin^2\left(87^{\circ}\right)+\sin^2\left(90^{\circ}\right)=$
  • ✓
    $\frac{31}{2}$
  • B
    $\frac{39}{2}$
  • C
    $\frac{59}{2}$
  • D
    36
✓

Answer

Correct option: A.
$\frac{31}{2}$
(A)
$\sin^2\left(3^{\circ}\right)+\sin^2\left(6^{\circ}\right)+\sin^2\left(9^{\circ}\right)+\ldots+\sin^2\left(84^{\circ}\right)$$+\sin^2\left(87^{\circ}\right)+\sin^2\left(90^{\circ}\right)$
$=\sin^2\left(3^{\circ}\right)+\sin^2\left(6^{\circ}\right)+\sin^2\left(9^{\circ}\right)+\ldots$$+\cos^2\left(6^{\circ}\right)+\cos^2\left(3^{\circ}\right)+\sin^2\left(90^{\circ}\right)$
$\ldots\left[\because\sin\left(90^{\circ}-\theta\right)=\cos\theta\right]$
$=\left[\sin^2\left(3^{\circ}\right)+\cos^2\left(3^{\circ}\right)\right]+\ldots$$+\left[\sin^2\left(42^{\circ}\right)+\cos^2\left(42^{\circ}\right)\right]+\sin^2\left(30^{\circ}\right)$$+\sin^2\left(45^{\circ}\right)+\sin^2\left(60^{\circ}\right)+\sin^2\left(90^{\circ}\right)$
$=(1+1+\ldots+1)+\left(\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2+(1)^2$
$=13+\frac{1}{4}+\frac{1}{2}+\frac{3}{4}+1$
$=\frac{31}{2}$

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