How do you evaluate #sin (arccos (2/8))#? Trigonometry Inverse Trigonometric Functions Basic Inverse Trigonometric Functions 1 Answer Arunraju Naspuri Jun 23, 2015 #sin(arccos(2/8))#=#sqrt(15)/4# Explanation: Let #sin(cos^(-1)(2/8))=sin(x)##" " "color(red)((1))# From #color(red)((1))#; #x=cos^(-1)(2/8)# #cosx=2/8# #sinx=sqrt(1-cos^2(x)# #sinx=sqrt(1-(2/8)^2# #sinx=sqrt(1-4/64# #sinx=sqrt(64-4)/sqrt64# #sinx=sqrt60/sqrt64# #sinx=sqrt(15xx4)/8# #sinx=(2sqrt(15))/8# #sinx=sqrt(15)/4# The above sinx value substitute in #" " "color(red)((1))# #sin(arccos(2/8))#=#sqrt(15)/4# Answer link Related questions What are the Basic Inverse Trigonometric Functions? How do you use inverse trig functions to find angles? How do you use inverse trigonometric functions to find the solutions of the equation that are in... How do you use inverse trig functions to solve equations? How do you evalute #sin^-1 (-sqrt(3)/2)#? How do you evalute #tan^-1 (-sqrt(3))#? How do you find the inverse of #f(x) = \frac{1}{x-5}# algebraically? How do you find the inverse of #f(x) = 5 sin^{-1}( frac{2}{x-3} )#? What is tan(arctan 10)? How do you find the #arcsin(sin((7pi)/6))#? See all questions in Basic Inverse Trigonometric Functions Impact of this question 1990 views around the world You can reuse this answer Creative Commons License