Find #cos^(-1)x-cos^(-1)y#, if #x=1/4# and #y=2/3#? Trigonometry Inverse Trigonometric Functions Inverse Trigonometric Properties 1 Answer Cesareo R. · Shwetank Mauria Feb 7, 2017 #cos^-1(1/12(2+5sqrt(3)))# Explanation: Let #z=cos^-1x-cos^-1 y# then #cosz=cos(cos^-1x-cos^-1 y)# but #cos(a-b)=cosa cosb+sinasinb# so #cosz = cos(cos^-1x)cos(cos^-1y)+sin(cos^-1x)sin(cos^-1y)# but #cos(cos^-1x)=x# and #sin(cos^-1x)=sqrt(1-x^2)# so #cosz=x y + sqrt(1-x^2)sqrt(1-y^2)# so #cosz=1/4*2/3+sqrt(1-(1/4)^2)*sqrt(1-(2/3)^2)# = #1/6+sqrt(15/16)*sqrt(5/9)=1/6+5/12sqrt3# = #1/12(2+5sqrt(3))# so #z = cos^-1(1/12(2+5sqrt(3)))# Answer link Related questions How do you use the properties of inverse trigonometric functions to evaluate #tan(arcsin (0.31))#? What is #\sin ( sin^{-1} frac{sqrt{2}}{2})#? How do you find the exact value of #\cos(tan^{-1}sqrt{3})#? How do you evaluate #\sec^{-1} \sqrt{2} #? How do you find #cos( cot^{-1} sqrt{3} )# without a calculator? How do you rewrite #sec^2 (tan^{-1} x)# in terms of x? How do you use the inverse trigonometric properties to rewrite expressions in terms of x? How do you calculate #sin^-1(0.1)#? How do you solve the inverse trig function #cos^-1 (-sqrt2/2)#? How do you solve the inverse trig function #sin(sin^-1 (1/3))#? See all questions in Inverse Trigonometric Properties Impact of this question 1635 views around the world You can reuse this answer Creative Commons License