Sin 250-sin 2100-sin 2150-sin 2900-a1-2- Find the value of a

We have, sin ^25^0+sin ^210^0+sin ^215^0+...+sin ^290^0=a1/2 = ( sin ^25^0+sin ^285^0 )+ ( sin ^210^0+sin ^280^0 )+...+ ( sin ^240^0 + #160;sin ^250 ...

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Byju's Answer

Standard XII

Mathematics

Integration by Parts

sin 250+sin 2...

Question

${sin}^{2} 5^{0} + {sin}^{2} 10^{0} + {sin}^{2} 15^{0} + . . . + {sin}^{2} 90^{0} = a \frac{1}{2} .$ Find the value of $a$

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Solution

We have, ${sin}^{2} 5^{0} + {sin}^{2} 10^{0} + {sin}^{2} 15^{0} + . . . + {sin}^{2} 90^{0} = a \frac{1}{2}$

$= ({sin}^{2} 5^{0} + {sin}^{2} 85^{0}) + ({sin}^{2} 10^{0} + {sin}^{2} 80^{0}) + . . . + ({sin}^{2} 40^{0} + {sin}^{2} 50^{0}) + {sin}^{2} 45^{0} + {sin}^{2} 90^{0}$

$= ({sin}^{2} 5^{0} + {cos}^{2} 5^{0}) + ({sin}^{2} 10^{0} + {cos}^{2} 10^{0}) + . . . + ({sin}^{2} 40^{0} + {cos}^{2} 40^{0}) + {sin}^{2} 45^{0} + {sin}^{2} 90^{0}$

$= (1 + 1 + 1 + 1 + 1 + 1 + 1 + 1) + {(\frac{1}{\sqrt{2}})}^{2} + 1 = 9 \frac{1}{2} =$ R.H.S.
$∴ a = 9$

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sin250+sin2100+sin2150+...+sin2900=a12. Find the value of a

We have, sin250+sin2100+sin2150+...+sin2900=a12=(sin250+sin2850)+(sin2100+sin2800)+...+(sin2400+sin2500)+sin2450+sin2900 =(sin250+cos250)+(sin2100+cos2100)+...+(sin2400+cos2400)+sin2450+sin2900 =(1+1+1+1+1+1+1+1)+(1√2)2+1=912= R.H.S.∴a=9

${sin}^{2} 5^{0} + {sin}^{2} 10^{0} + {sin}^{2} 15^{0} + . . . + {sin}^{2} 90^{0} = a \frac{1}{2} .$ Find the value of $a$