Let `f_n(theta)=(cos theta/2+cos 2theta + cos (7theta)/2+...+cos (3n-2) (theta/2))/(sin theta/2+sin 2theta + sin (7theta)/2+....+ sin (3n-2)(theta/2))` then `f_3((3pi)/16)`

Let f_n(theta)=(cos(theta/2)+cos2theta+cos((7theta)/2)+....+cos(3n-2)(theta/2))/(sin(theta/2)+sin2theta+sin((7theta)/2)+....+sin(3n-2)(theta/2)) then (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

Let f_n(theta)=(cos(theta/2)+cos2theta+cos((7theta)/2)+....+cos(3n-2)(theta/2))/(sin(theta/2)+sin2theta+sin((7theta)/2)+....+sin(3n-2)(theta/2)) then (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

Let f_n(theta)=(cos(theta/2)+cos2theta+cos((7theta)/2)+....+cos(3n-2)(theta/2))/(sin(theta/2)+sin2theta+sin((7theta)/2)+....+sin(3n-2)(theta/2)) then (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

Let f_(n)(theta)=(cos((theta)/(2))+cos2 theta+cos((7 theta)/(2))+...+cos(3n-2)((theta)/(2)))/(sin((theta)/(2))+sin2 theta+sin((7 theta)/(2))+...+sin(3n-2)((theta)/(2))) then (a)f_(3)((3 pi)/(16))=sqrt(2)-1(b)f_(5)((pi)/(28))=sqrt(2)+1(c)f_(7)((pi)/(60))=(2+sqrt(3))(d) none of these

2 cos theta - cos 3 theta - cos 5 theta - 16 cos^(3) theta sin^(2) theta =

Prove that ( cos theta + cos 2 theta + cos 3 theta + cos 4 theta)/(sin theta + sin 2 theta + sin 3 theta + sin 4 theta)= cot ""(5theta)/(2).

Prove that cos theta cos 2theta cos2^2 theta … cos 2^(n-1) theta = (sin2^n theta)/(2^n.sin theta) , for all n in N

cos (2 theta) cos ((theta) / (2)) - cos (3 theta) cos ((9 theta) / (2)) = sin (5 theta) sin ((5 theta) / (2))

Using induction, prove that cos theta · cos 2theta · cos 2^2 theta. ... cos 2^(n-1) theta =(sin2^n theta)/(2^n sin theta)