sin theta*sin(60-theta)*sin(60+theta)=

Prove that: |sintheta."sin"(60-theta).sin(60+theta)|lt=1/4 for all values of theta .

Prove that abs(sinthetasin(60^@-theta)sin(60^@+theta)) le1/4 for all values of theta .

If sin(60^(@)+theta)+sin(60^(@)-theta)+sin theta=0 , where 0 (A) 0 (B) 1 (C) 2 (D) 3

Prove that sinthetasin(60^(@)+theta)sin(60^(@)-theta)=1/4sin3theta Hence deduce that sin20^(@)sin40^(@)sin60^(@)sin80^(@)=3//16 (OCT-97,INST.93)

Prove that: sin(60^0-theta)cos(30^0+theta)+cos(60^0-theta)sin(30^0+theta)=1.

(sin theta+2sin3 theta+sin5 theta)/(sin3 theta+2sin5 theta+sin7 theta)=(sin3 theta)/(sin5 theta)

(sin theta+2sin3 theta+sin5 theta)/(sin3 theta+2sin5 theta+sin7 theta)=(sin3 theta)/(sin5 theta)

Prove that :(sin theta+2sin3 theta+sin5 theta)/(sin3 theta+2sin5 theta+sin7 theta)=(sin3 theta)/(sin5 theta)

sin 7 theta * sin theta+sin11theta * sin 3theta= A) sin 4 theta * sin 10theta B) cos 4 theta * cos 10 theta C) sin 8 theta * sin 4 theta D) cos 8 theta * cos 14 theta

(sin3 theta-sin theta sin^(2)(2 theta))/(sin theta+sin2 theta cos theta)=cos x then x=