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Mathematics Question on Trigonometric Functions

sin⁡25∘+sin⁡210∘+sin⁡215∘+....+sin⁡290∘\sin^2 5^{\circ} + \sin^2 10^{\circ} + \sin^2 15^{\circ} +....+ \sin^2 90^{\circ} =

A

88

B

99

C

1010

D

9.59.5

Answer

9.59.5

Explanation

Solution

sin⁡25∘+sin⁡210∘+sin⁡215∘+...+sin⁡290∘\sin^{2} 5^{\circ} +\sin^{2} 10^{\circ } +\sin^{2} 15^{\circ } +... +\sin ^{2} 90^{\circ }
=sin⁡25∘+sin⁡210∘+....+sin⁡235∘+sin⁡240∘+sin⁡245∘+sin⁡2(90∘−40∘)+sin⁡2(90∘−35∘)+....+sin⁡2(90∘−15)+sin⁡2(90∘−10∘)+sin⁡2(90∘−5∘)+sin⁡290∘= \sin ^{2} 5^{\circ } +\sin^{2} 10^{\circ }+....+\sin^{2} 35^{\circ } +\sin ^{2} 40^{\circ } + \sin ^{2} 45^{\circ } +\sin ^{2} \left(90^{\circ} - 40^{\circ}\right) +\sin ^{2} \left(90^{\circ } - 35^{\circ }\right)+....+\sin ^{2} \left(90^{\circ } - 15\right)+\sin ^{2} \left(90^{\circ } - 10^{\circ }\right) +\sin ^{2} \left(90^{\circ } - 5^{\circ }\right)+\sin ^{2} 90^{\circ }
=sin⁡25∘+sin⁡210∘+....+sin⁡235∘+sin⁡240∘+sin⁡245∘+cos⁡240∘+cos⁡235∘+....+cos⁡210∘+cos⁡25∘+sin⁡290∘= \sin ^{2} 5^{\circ}+\sin ^{2}10^{\circ} + ....+\sin ^{2} 35^{\circ} +\sin ^{2}40^{\circ}+\sin ^{2} 45^{\circ}+\cos ^{2} 40^{\circ}+\cos ^{2} 35^{\circ}+....+ \cos ^{2} 10^{\circ} +\cos ^{2}5^{\circ} +\sin ^{2} 90^{\circ}
=[(sin⁡25∘+cos⁡25∘)+(sin⁡210∘+cos⁡210∘)+......+sin⁡240∘+cos⁡240∘]+sin⁡245∘+sin⁡290∘= \left[\left( \sin^{2} 5^{\circ} + \cos^{2} 5^{\circ} \right)+\left( \sin ^{2} 10^{\circ } + \cos ^{2} 10^{\circ } \right) + ......+ \sin ^{2} 40^{\circ } + \cos ^{2} 40^{\circ }\right]+\sin^{2} 45^{\circ} + \sin^2 90^{\circ}
=[1+1+.....+1]+(12)2+1= [1 + 1 +.....+1 ] + \left( \frac{1}{\sqrt{2}}\right)^2 + 1
=8+12+1=192=9.5= 8 + \frac{1}{2} + 1 = \frac{19}{2} = 9.5