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प्रश्न
Prove that sin2 5° + sin2 10° .......... + sin2 85° + sin2 90° = `9 1/2`.
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उत्तर
LHS
= sin2 5° + sin2 10° + sin2 15° + sin2 20° + sin2 25° + sin2 30° + sin2 35°+ sin2 40° + sin 2 45° + sin2 50° + sin2 55° + sin2 60° + sin2 65° + sin2 70° + sin2 75° + sin2 80° + sin2 85° + sin2 90°.
= (sin2 5° + sin2 85°) + (sin2 10° + sin2 80°) + (sin2 15° + sin2 75° ) + (sin2 20°+ sin2 70°) + (sin2 25° + sin2 65°) + (sin2 30° + sin2 60°) + (sin2 35° + sin2 55°) + (sin2 40° + sin2 50°) + sin 2 45° + sin2 90°.
= (sin2 5° + cos2 5°) + (sin2 10° + cos2 10°) + (sin2 15° + cos2 15° ) + (sin2 20°+ cos2 20°) + (sin2 25° + cos2 25°) + (sin2 30° + cos2 30°) + (sin2 35° + cos2 35°) + (sin2 40° + cos2 40°) +`(1/sqrt2)^2 + (1)^2 ....[ ∵ sin(90° - θ) = cos θ ∵ sin 90° = 1 and sin 45° = 1/sqrt2 ] [∵ sin^2 θ + cos^2 θ = 1]`
= 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + `1/2` + 1
= `9 1/2`
= RHS
Hence proved.
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