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tan3α – formula’s demonstration

Posted in Math Theory by Lia on 04/06/2013

For a list with trigonometric identities click here

 

For this demonstration we are going to use the identity:

 

[1]  tan({alpha} +{beta} +{gamma}) = {{tan{alpha} + tan{beta} + tan{gamma} - {tan{alpha}tan{beta}tan{gamma}}}/{1 - {tan{alpha}}{tan{beta}} - {tan{beta}}{tan{gamma}} - {tan{gamma}}{tan{alpha}}}} (see demonstration for this formula here)

 

Considering that  {alpha} = {beta} ={gamma} and replacing  {}{beta}{} and {}{gamma}{} with {}{alpha}{} in [1] we get:

 

tan{3}{alpha} = {{tan{alpha} + tan{alpha} + tan{alpha} - {tan{alpha}tan{alpha}tan{alpha}}}/{1 - {tan{alpha}}{tan{alpha}} - {tan{alpha}}{tan{alpha}} - {tan{alpha}}{tan{alpha}}}}

↓↓

tan{3{alpha}} = {{3tan{alpha} -tan^3{alpha}}/{1 - 3tan^2{alpha}}}

 

 

 

 

 

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