For non-negative integer n, let f (n )= _k=0^n sin ( k+1n+2 ) sin ( k+2n+2 ) _k=0^n 0.05em sin^2 ( k+1n+2 ) Assuming cos^-1x takes values in [0 , ] which of the following options is/are correct? (JEE…

Question

For non-negative integer n, let f (n )= _k=0^n sin ( k+1n+2 ) sin ( k+2n+2 ) _k=0^n 0.05em sin^2 ( k+1n+2 ) Assuming cos^-1x takes values in [0 , ] which of the following options is/are correct? (JEE Advanced 2019)

Accepted Answer

Correct answer: A, B, C. (a, b, c) f (n )= 0.05em2 sin ( k+1n+2 ) sin ( k+2n+2 ) 0.05em2 sin^2 ( k+1n+2 ) = 0.05em cos n+2- 0.05em cos ( 2k+3n+2 ) 0.05em 0.05em(1- cos ( 2k+2n+2 )) = (n+1 ) cos ( n+2 )- cos ( n+3n+2 ) sin ( n+1n+2 ) sin ( n+2 ) (n+1 )- cosπ sin ( n+1n+2 ) sin ( n+2 ) = (n+1 ) cos ( n+2 )- cos( + n+2) sin( - n+2) sin ( n+2 ) (n+1 )- cos ⁡ πsin( - n+2) sin ( n+2 ) = (n+1 ) cos ( n+2 )+ cos ( n+2 )n+1+1 = (n+2 ) cos ( n+2 )n+2= cos ( n+2 ) (a) f (4 )= cos ( 4+2 )= cos ( 6 )= 32. (b) = tan ( cos^-1f (6 ) )= tan ( cos^-1 ( cos ( 8 ) ) )= tan 8 tan 4= 2 tan 81- tan^2 8 ⇒1= 2 1- ^2 ⇒ ^2+2 -1=0 (c) sin (7 cos^-1f (5 ) ) = sin (7 cos^-1 ( cos 7 ) )= sin(7 7)= sinπ=0 (d) limn f (n )= limn cos ( n+2 )= cos (0 )=1

$For non-negative integer n, let f (n) = \frac{\sum _{k = 0}^{n} sin ( \frac{k + 1}{n + 2} π ) sin ( \frac{k + 2}{n + 2} π )}{\sum _{k = 0}^{n} sin ^{2} ( \frac{k + 1}{n + 2} π )} Assuming cos^{- 1} x takes values in [0, π] which of the following options is/are correct? (JEE Advanced 2019)$

Solution

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