The average value of sin2°,sin4°,sin6°,...,sin180°is
A
190cot1°
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B
190tan1°
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C
190tan89°
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D
190cot89°
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Solution
The correct options are A 190cot1° C 190tan89° Here, we represent the sum of sines as: S=sin2°+sin4°+..+sin180°. Now, using the expression for sum of sines S=sin(α)+sin(α+β)+..+sin(α+(n−1)β)⇒S=sin(nβ2)sin(β2)sin(α+(n−1)β2). Now, for our series, we have α=2°,β=2°andn=90. So, we write sum as: S=sin(90×2°2)sin(2°2)sin(2°+(90−1)2°2)⇒S=sin(90°)sin(1°)sin(180°+2°2)⇒S=sin(90°)sin(1°)sin(91°)⇒S=sin(90°)sin(1°)sin(91°)⇒S=1sin(1°)sin(90°+1°)⇒S=cos(1°)sin(1°)=cot(1°). Now, we have 90 terms so average of sum is given as: A=S90=190cot(1°) Also, cot(1°)=cot(90°−89°)=tan(89°)⇒A=190cot1°=190tan89°.