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The average value of sin2^@,sin4^@, sin6...

The average value of `sin2^@,sin4^@, sin6^@.........sin180^@` is `(i) 1/90 cos1^0 (ii) 1/90 sin1^0 (iii) 1/90cot1^0` (iv) none of these

Text Solution

Verified by Experts

The correct Answer is:
`(1)/(90)(cot1^(@))`
There are 90 terms and the average A is
`A=(1)/(90)sum_(r=1)^(90)sin2r`
`A=(1)/(90)xx(sin(90^(@)xx(2)/(2)))/(sin1^(@))xxsin[(2^(@)+180^(@))/(2)]`
`=(1)/(90)xx(1)/(sin1^(@))xxsin(90^(@)+1^(@))`
`=(1)/(90)xx(cos1^(@))/(sin1^(@))=(1)/(90)(cot 1^(@))`
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