show that cos²5°+cos²10°+...........+cos²90°=17/2
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Given => cos²5° + cos²10° + ...... + cos²90° = 17/2
Solving LHS
=> cos²5° + cos²10° + ...... + cos²90°
(cos(90-∅) = sin∅)
=> cos²5° + cos²10° + ..... + cos²(90-10) + cos²(90-5)+ cos²90°
(cos90° = 0)
=> (cos²5° + sin²5°) + (cos²10°+sin²10°) + ..... + ( cos²40° + sin²40°) + cos²45°
(sin²∅ + cos²∅ = 1)
=> 1 + 1 +.... +1 + ( 1/√2)²
=> 8 + 1/2
=> (16 + 1)/2
=> 17/2
=> RHS
Hence, LHS = RHS
Solving LHS
=> cos²5° + cos²10° + ...... + cos²90°
(cos(90-∅) = sin∅)
=> cos²5° + cos²10° + ..... + cos²(90-10) + cos²(90-5)+ cos²90°
(cos90° = 0)
=> (cos²5° + sin²5°) + (cos²10°+sin²10°) + ..... + ( cos²40° + sin²40°) + cos²45°
(sin²∅ + cos²∅ = 1)
=> 1 + 1 +.... +1 + ( 1/√2)²
=> 8 + 1/2
=> (16 + 1)/2
=> 17/2
=> RHS
Hence, LHS = RHS
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