How do you prove #sin ((5pi)/4)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle 1 Answer ali ergin Jun 25, 2016 #sin((5pi)/4)=-sqrt2/2# Explanation: #sin((5pi)/4)=sin(pi+pi/4)# #sin(a+b)=sin a*cos b+cos a*sin b# #sin(pi+pi/4)=sin pi*cos (pi/4)+cos pi*sin(pi/4)# #sin pi=0" ; "cos pi=-1# #sin(pi/4)=sqrt 2/2" ; "cos (pi/4)=sqrt2/2# #sin((5pi)/4)=0*sqrt2/2-1*sqrt2/2# #sin((5pi)/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 1749 views around the world You can reuse this answer Creative Commons License