How do you derive the trigonometric formulas for double and half angels for sin, cos, and tan? I.e: How do I derive something like sin(2x)=2(sinx)(cosx)?

1 Answer
Jun 16, 2018

Below

Explanation:

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)=cosxcosxsinxsinx=cos2xsin2x

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=12sin2(θ2)

2sin^2(theta/2)=1-costheta2sin2(θ2)=1cosθ

sin^2(theta/2)=1/2(1-costheta)sin2(θ2)=12(1cosθ)

color (red) (sin(theta/2)=+-sqrt(1/2(1-costheta)sin(θ2)=±12(1cosθ)

tan(2x)=tan(x+x)=(tanx+tanx)/(1-tan^2x)=color (red) ((2tanx)/(1-tan^2x)tan(2x)=tan(x+x)=tanx+tanx1tan2x=2tanx1tan2x

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(1cosθ)±12(cosθ+1)

tan(theta/2)=sqrt(1-costheta)/sqrt(1+costheta)tan(θ2)=1cosθ1+cosθ

tan(theta/2)=sqrt((1-costheta)(1+costheta))/(1+costheta)^2tan(θ2)=(1cosθ)(1+cosθ)(1+cosθ)2

color (red) (tan(theta/2)=sintheta/(1+costheta)tan(θ2)=sinθ1+cosθ