Solution: 1 Mark each
When θ=30∘, we have 3 θ=90∘
(i)sin 3 θ=sin 90∘=1
3 sin θ−4 sin3 θ
=3 sin 30∘−4 sin3 30∘
=[3×12−4×(12)3]=(32−4×18)
=(32−12)=1
∴sin 3 θ=3 sin θ−4 sin3 θ
(ii)cos 3 θ=cos 90∘=0
4 cos3 θ−3 cos θ=4 cos3 30∘−3 cos 30∘
=[4×(√32)3−3×√32]
=(4×3√38−3√32)
=(3√32−3√32)=0
∴cos 3 θ=4 cos3 θ−3 cos θ