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Question

The average value of sin 2°, sin 4°, sin 6°, . . . , sin 180° is

A

190cot 1°
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B

190tan 1°
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C

190tan 89°
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D

190cot 89°
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Solution

The correct options are
A
190cot 1°
C
190tan 89°
Here, we represent the sum of sines as:
S=sin 2°+sin 4°+ . .+ sin 180°.
Now, using the expression for sum of sines
S=sin(α)+sin(α+β)+ . . +sin(α+(n1)β)S=sin(nβ2)sin(β2)sin(α+(n1)β2).
Now, for our series, we have α=2° , β=2° and n=90.
So, we write sum as:
S=sin(90×2°2)sin(2°2)sin(2°+(901)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=190cot 1°=190tan 89°.

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