Let `f_n(theta)=(costheta/2+cos2theta+cos(7theta)/2++cos(3n-2)^(theta/2))/(sintheta/2+sin2theta+sin(7theta)/2++sin(3n-2)^(theta/2))` then `f_3((3pi)/(16))=sqrt(2)-1` (b) `f_5(pi/(28))=sqrt(2)+1` `f_7(pi/(60))=(2+sqrt(3))` (d) none of these

Options

A. `f_(3)((3pi)/(16))=sqrt(2)-1`
B. `f_(5)((pi)/(28))=sqrt(2)+1`
C. `f_(7)((pi)/(60))=(2+sqrt(3))`
D. none of these.

Answer

A, B, C

Explanation

We have `f_(n)(theta)=cot((3n-1)(theta)/(4))`
`therefore f_(3)((3pi)/(16))=cot(8xx(1)/(4)xx(3pi)/(16))=cot(3pi)/(6)=sqrt(2)-1`
`f_(5)((pi)/(28))=cot(14xx(1)/(4)xx(pi)/(28))=cot((pi)/(8))=sqrt(2)+1`
`f_(7)((pi)/(60))=cot(20(1)/(4)xx(pi)/(60))=cot((pi)/(12))=(2+sqrt(3))`.