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How do you evaluate #sin((2pi)/7)/(1+cos((2pi)/7))#?

1 Answer

Shwetank Mauria

Mar 27, 2016

#sin((2pi)/7)/(1+cos((2pi)/7))=tan(pi/7)#

Explanation:

We use #sin2A=2sinAcosA# and #cos2A=2cos^2A-1#

hence #sin((2pi)/7)=2sin(pi/7)cos(pi/7)# and #cos((2pi)/7)=2cos^2(pi/7)-1#

Hence, #sin((2pi)/7)/(1+cos((2pi)/7))=(2sin(pi/7)cos(pi/7))/(1+2cos^2(pi/7)-1)# or

= #(2sin(pi/7)cos(pi/7))/(2cos^2(pi/7))# or

= #sin(pi/7)/(cos(pi/7))=tan(pi/7)#

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