Express the left hand side as
sin3theta=sin(theta+2theta)
now expand the right side of this equation using color(blue)"Addition formula"
color(red)(|bar(ul(color(white)(a/a)color(black)(sin(A±B)=sinAcosB±cosAsinB)color(white)(a/a)|)))
rArrsin(theta+2theta)=sinthetacos2theta+costhetasin2theta.......(A)
color(red)(|bar(ul(color(white)(a/a)color(black)(cos2theta=cos^2theta-sin^2theta=2cos^2theta-1=1-2sin^2theta)color(white)(a/a)|)))
The right hand side is expressed only in terms of sintheta's
so we use cos2theta=1-2sin^2theta........(1)
color(red)(|bar(ul(color(white)(a/a)color(black)(sin2theta=2sinthetacostheta)color(white)(a/a)|)))........(2)
Replace cos2theta" and " sin2theta by the expansions (1) and (2)
into (A)
sin(theta+2theta)=sinthetacolor(red)((1-2sin^2theta))+costhetacolor(red)((2sinthetacostheta)
and expanding brackets gives.
sin(theta+2theta)=sintheta-2sin^3theta+2sinthetacos^2theta....(B)
color(red)(|bar(ul(color(white)(a/a)color(black)(cos^2theta+sin^2theta=1rArrcos^2theta=1-sin^2theta)color(white)(a/a)|)))
Replace cos^2theta=1-sin^2theta" into (B)"
rArrsin(theta+2theta)=sintheta-2sin^3theta+2sintheta(1-sin^2theta)
and expanding 2nd bracket gives.
sin(theta+2sintheta)=sintheta-2sin^3theta+2sintheta-2sin^3theta
Finally, collecting like terms.
sin3theta=3sintheta-4sin^3theta="R.H.S hence proven"
