" (i) "sin theta+sin2 theta+sin3 theta=1+cos theta+cos2 theta

cos theta+sin theta=cos2 theta+sin2 theta

If theta=30° , show that 3sin theta- 4sin^3 theta= sin theta cos2 theta + costheta sin 2theta .

Evaluate the following determinants: (b) |(cos theta, -sin theta),(sin theta, cos theta)| = cos theta (cos theta) - sin theta(-sin theta) = cos^(2) theta + sin^(2) theta = 1

Simplify: (sin8 theta cos theta-sin6 theta cos3 theta)/(cos2 theta cos theta-sin3 theta sin4 theta)

(sin5 theta-2sin3 theta + sin theta) / (cos5 theta-cos theta) = tan theta

2 cos (theta - pi //2) + 3 sin (theta + pi//2) - (3sin theta + 2 cos theta ) = ? A. cos theta - sin theta B. sin theta - cos theta C. sin theta + cos theta D. cot theta - tan theta .

Simplify, costheta[[cos theta, sin theta],[-sin theta, cos theta]] + sin theta [[sin theta, -cos theta],[cos theta, sin theta]]

Let f_(n)(theta)=(sin theta+sin2 theta+sin3 theta++sin n theta)/(cos theta+cos2 theta+cos3 theta++cos n theta)

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