How do you evaluate #sin((2pi)/7)/(1+cos((2pi)/7))#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 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)# Answer link Related questions How do you find the trigonometric functions of any angle? What is the reference angle? How do you use the ordered pairs on a unit circle to evaluate a trigonometric function of any angle? What is the reference angle for #140^\circ#? How do you find the value of #cot 300^@#? What is the value of #sin -45^@#? How do you find the trigonometric functions of values that are greater than #360^@#? How do you use the reference angles to find #sin210cos330-tan 135#? How do you know if #sin 30 = sin 150#? How do you show that #(costheta)(sectheta) = 1# if #theta=pi/4#? See all questions in Trigonometric Functions of Any Angle Impact of this question 2668 views around the world You can reuse this answer Creative Commons License