Remember that 2x=x+x2x=x+x
sin(2x)=sin(x+x)=sinxcosx+cosxsinx=color (red) (2sinxcosx)sin(2x)=sin(x+x)=sinxcosx+cosxsinx=2sinxcosx
cos(2x)=cos(x+x)=cosxcosx-sinxsinx=color (red) (cos^2x-sin^2xcos(2x)=cos(x+x)=cosxcosx−sinxsinx=cos2x−sin2x
Now let 2x=theta2x=θ so x=theta/2x=θ2
costheta=cos^2(theta/2)-sin^2(theta/2)cosθ=cos2(θ2)−sin2(θ2)
costheta=2cos^2(theta/2)-1cosθ=2cos2(θ2)−1
2cos^2(theta/2)=costheta+12cos2(θ2)=cosθ+1
cos^2(theta/2)=1/2(costheta+1)cos2(θ2)=12(cosθ+1)
color (red) (cos(theta/2)=+-sqrt(1/2(costheta+1)cos(θ2)=±√12(cosθ+1)
Similarly,
costheta=1-2sin^2(theta/2)cosθ=1−2sin2(θ2)
2sin^2(theta/2)=1-costheta2sin2(θ2)=1−cosθ
sin^2(theta/2)=1/2(1-costheta)sin2(θ2)=12(1−cosθ)
color (red) (sin(theta/2)=+-sqrt(1/2(1-costheta)sin(θ2)=±√12(1−cosθ)
tan(2x)=tan(x+x)=(tanx+tanx)/(1-tan^2x)=color (red) ((2tanx)/(1-tan^2x)tan(2x)=tan(x+x)=tanx+tanx1−tan2x=2tanx1−tan2x
Finally tan(theta/2)=sin(theta/2)/cos(theta/2)tan(θ2)=sin(θ2)cos(θ2)
tan(theta/2)=(+-sqrt(1/2(1-costheta)))/(+-sqrt(1/2(costheta+1))tan(θ2)=±√12(1−cosθ)±√12(cosθ+1)
tan(theta/2)=sqrt(1-costheta)/sqrt(1+costheta)tan(θ2)=√1−cosθ√1+cosθ
tan(theta/2)=sqrt((1-costheta)(1+costheta))/(1+costheta)^2tan(θ2)=√(1−cosθ)(1+cosθ)(1+cosθ)2
color (red) (tan(theta/2)=sintheta/(1+costheta)tan(θ2)=sinθ1+cosθ