16*2sin^(2)theta+sin^(2)2 theta=2

Solve: 2sin^(2)theta+sin^(2)2theta=2

Solve: 2sin^(2)theta+sin^(2)2theta=2

Solve: 2sin^(2)theta+sin^(2)2theta=2

If cos theta+cos^(2)theta=1 then sin^(2)theta+2sin^(2)theta+sin^(2)theta=

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

2cos theta-cos3 theta-cos5 theta-16cos^(3)theta sin^(2)theta=

2cos theta-cos3 theta-cos5 theta-16cos^(3)theta sin^(2)theta=

Prove that sin^(2)theta+sin^(2)2 theta+sin^(2)3 theta+....+sin^(2)n theta=(n)/(2)-(sin n theta cos(n+1)theta)/(2sin theta)