The sum `S= sin theta + sin 2 theta + …+ sin n theta, ` equals

If sin 6theta = sin 4theta - sin 2theta , then theta =

Prove that , sin theta + sin 2theta + ……+ sin n theta = ( (sin( n theta/(2)) sin ((n+1)/(2))theta))/(sin theta/2) , for all n in N .

If theta in the first quadrant and 5 tan theta =4 " then " ( 5 sin theta - 3 sin theta)/( sin theta + 2 cos theta) is equal to

sin^(3)theta + sin theta - sin theta cos^(2)theta =

Sum to terms of the series sin theta sin2 theta+sin2 theta sin3 theta+sin3 theta sin4 theta+... is equal to

If ((1 + cos theta + i sin theta ) /(i+ sin theta + i cos theta ))= cos ntheta + i sin n theta then n is equal to :

Prove by Principle of Mathematical Induction, that : sin theta + sin 2 theta + sin 3 theta + ........ + sin n theta= (sin (n+1)theta/2 sin frac (n theta)(2))/(sin frac (theta)(2)) for all n in N.