How do you evaluate #cos^2(pi/8) - sin^2(pi/8)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer Nghi N. Apr 7, 2016 #sqrt2/2# Explanation: Apply the trig identity: #cos 2a = cos^2 a - sin^2 a# #cos^2 (pi/8) - sin^2 (pi/8) = cos (pi/4)# Trig table --> #cos (pi/4) = sqrt2/2# 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 17003 views around the world You can reuse this answer Creative Commons License