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How to do you prove this identity?

 

\(\tan \dfrac{\theta}{2} = \dfrac{\sin \theta}{1 + \cos \theta}\)

 Aug 4, 2020
 #1
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If you are allowed to start with the identity:  tan( x/2 )  =  ( 1 - cos(x) ) / sin(x)  then: 

 

tan( x/2 )  =  ( 1 - cos(x) ) / sin(x)  =  

 

multiply both the numerator and denominator by  ( 1 + cos(x) )

 

=  ( 1 - cos(x) ) / sin(x)  ·  ( 1 + cos(x) ) / ( 1 + cos(x) )

 

=  [ ( 1 - cos(x) ) · ( 1 + cos(x) ) ]  / [ sin(x) · ( 1 + cos(x) ) ]

 

=  [ 1 - cos2(x) ] / [ sin(x) · ( 1 + cos(x) ) ]

 

=  sin2(x) / [ sin(x) · ( 1 + cos(x) ) ]

 

=  sin(x) / ( 1 + cos(x) )

 Aug 4, 2020

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