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)?

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Answer 1 · 2018-06-16T07:17:14

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Remember that #2x=x+x#

#sin(2x)=sin(x+x)=sinxcosx+cosxsinx=color (red) (2sinxcosx)#

#cos(2x)=cos(x+x)=cosxcosx-sinxsinx=color (red) (cos^2x-sin^2x#

Now let #2x=theta# so #x=theta/2#

#costheta=cos^2(theta/2)-sin^2(theta/2)#

#costheta=2cos^2(theta/2)-1#

#2cos^2(theta/2)=costheta+1#

#cos^2(theta/2)=1/2(costheta+1)#

#color (red) (cos(theta/2)=+-sqrt(1/2(costheta+1)#

Similarly,

#costheta=1-2sin^2(theta/2)#

#2sin^2(theta/2)=1-costheta#

#sin^2(theta/2)=1/2(1-costheta)#

#color (red) (sin(theta/2)=+-sqrt(1/2(1-costheta)#

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

Finally #tan(theta/2)=sin(theta/2)/cos(theta/2)#

#tan(theta/2)=(+-sqrt(1/2(1-costheta)))/(+-sqrt(1/2(costheta+1))#

#tan(theta/2)=sqrt(1-costheta)/sqrt(1+costheta)#

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

#color (red) (tan(theta/2)=sintheta/(1+costheta)#