Question #ba1cf

Question

2198 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-25T14:56:24

#LHS=cosx#

#=cos(x/2+x/2)#

#=cos(x/2)cos(x/2)-sin(x/2)sin(x/2)#

#=cos^2(x/2)-sin^2(x/2)#

#=(cos^2(x/2)-sin^2(x/2))/1#

#=(cos^2(x/2)-sin^2(x/2))/ (cos^2(x/2)+sin^2(x/2))#

#=(cos^2(x/2)/cos^2(x/2)-sin^2(x/2)/cos^2(x/2))/ (cos^2(x/2)/cos^2(x/2)+sin^2(x/2)/cos^2(x/2))#

#=(1-tan^2(x/2))/(1+tan^2(x/2))=RHS#

Answer 2 · 2017-02-25T15:02:35

See proof below

We use

#cos^2x+sin^2x=1#

#sin^2x=1-cos^2x#

#sin^2(x/2)=1-cos^2(x/2)#

Dividing by #cos^2(x/2)#

#sin^2(x/2)/cos^2(x/2)=(1-cos^2(x/2))/cos^2(x/2)#

#tan^2(x/2)=(1-cos^2(x/2))/cos^2(x/2)#

Therefore,

#=1-(1-cos^2(x/2))/cos^2(x/2)#

#=(cos^2(x/2)-1+cos^2(x/2))/(cos^2(x/2))#

#=(2cos^2(x/2)-1)/(cos^2(x/2))=cosx/cos^2(x/2)#

And,

#1+tan^2(x/2)=1+(1-cos^2(x/2))/cos^2(x/2)#

#=(cos^2(x/2)+1-cos^2(x/2))/(cos^2(x/2))#

#=1/cos^2(x/2)#

Therefore,

#RHS=(1-tan^2(x/2))/(1+tan^2(x/2))=(cosx/cos^2(x/2))/(1/cos^2(x/2))#

#=cosx=LHS#

#QED#