How do you prove # [1 - cos(x)]/[sin(x)] = tan(x/2)#?

2 Answers
Shwetank Mauria
Oct 1, 2016

Please see below.

Explanation:

As #cosx=1-2sin^2(x/2)# and #sinx=2sin(x/2)cos(x/2)#

Hence #(1-cosx)/sinx#

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

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

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

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

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

= #tan(x/2)#

Tiago Hands
Oct 1, 2016

Please follow the instructions below:

Explanation:

#sin(x/2+x/2)#

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

#=sin(x/2){cos(x/2)+cos(x/2)}#

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

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

#=sin(x)#

This is because:

#sin(A+B)=sin(A)cos(B)+cos(A)sin(B)#

#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)#

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

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

#=cos(x)#

This is because:

#cos(A+B)=cos(A)cos(B)-sin(A)sin(B)#

Which means that:

#LHS#

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

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

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

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

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

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

#=tan(x/2)#

#=RHS#

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