`(sin3theta-sintheta*sin^2(2theta))/(sintheta+sin2theta*costheta)=cosx->x=`

(sintheta+sin2theta)/(1+costheta+cos2theta)

(sintheta+sin2theta)/(1+costheta+cos2theta)

Show that: (frac(sin3theta+2sin5theta+sin7theta)(sintheta+2sin3theta+sin5theta))=cos2theta+sin2thetacot3theta

(sintheta+sin 2theta)/(1+costheta+cos2theta)=

(sin3theta-cos3theta)/(sintheta+costheta)+1 =

(sintheta+sin2theta)/(1+costheta+cos2theta) =

(sintheta+sin2theta)/(1+costheta+cos2theta)=?

(sin5theta+sin2theta-sintheta)/(cos5theta+2cos3theta+2cos^2theta+costheta) is equal to

Simplify: cos theta[{:(costheta,sintheta),(-sintheta,costheta):}]+sintheta[{:(sin theta ,-costheta),(costheta, sintheta):}]