Question

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

Answer 1

`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)`