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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,

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

$=cosx=LHS$

$QED$