How do you evaluate #cos((13pi)/8)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 2 Answers P dilip_k May 14, 2016 #=1/2sqrt((2-sqrt2)# Explanation: using formula #costheta=sqrt(1/2(1+cos(2theta)))# #cos((13pi)/8)# #=sqrt(1/2(1+cos(2*(13pi)/8))) # #=sqrt(1/2(1+cos((13pi)/4))) # #=sqrt(1/2(1+cos(3pi+pi/4)) # #=sqrt(1/2(1-cos(pi/4)))# #=sqrt(1/2(1-1/sqrt2))# #=sqrt((sqrt2-1)/(2sqrt2)# #=sqrt((2-sqrt2)/(2xx2)# #=1/2sqrt((2-sqrt2)# Answer link Nghi N. · Nghi N May 14, 2016 #sqrt(2 - sqrt2)/2# Explanation: #cos ((13pi)/8) = cos ((3pi)/8 + cos ((16pi)/8)) = cos ((3pi)/8 + 2pi) = # #= cos ((3pi)/8).# Evaluate #cos ((3pi)/8)# by applying the trig identity: #cos 2a = 2cos^2 a - 1#. In this case, we get --> #2cos^2 ((3pi)/8) - 1 = cos ((6pi)/8) = cos ((3pi)/4) = -sqrt2/2# #2cos^2 ((3pi)/8) = 1 - sqrt2/2 = (2 - sqrt2)/2# #cos^2 ((3pi)/8) = (2 - sqrt2)/4# #cos ((3pi)/8) = +- sqrt(2 - sqrt2)/2# #cos ((13pi)/8) = cos ((3pi)/8) = sqrt(2 - sqrt2)/2# , because #cos ((3pi)/8)# is positive. 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 11543 views around the world You can reuse this answer Creative Commons License