How do you solve the inverse trig function sin(sin^-1 (1/3))sin(sin−1(13))? Trigonometry Inverse Trigonometric Functions Inverse Trigonometric Properties 1 Answer Alan P. May 25, 2015 sin^-1(1/3)sin−1(13) or arcsin(1/3)arcsin(13) is the angle thetaθ for which sin(theta) = 1/3sin(θ)=13 Therefore sin(sin^-1(1/3))sin(sin−1(13)) = sin(theta)=sin(θ) for the value of thetaθ for which sin(theta) = 1/3sin(θ)=13 That is sin(sin^-1(1/3)) = 1/3sin(sin−1(13))=13 Answer link Related questions How do you use the properties of inverse trigonometric functions to evaluate tan(arcsin (0.31))tan(arcsin(0.31))? What is \sin ( sin^{-1} frac{sqrt{2}}{2})sin(sin−1√22)? How do you find the exact value of \cos(tan^{-1}sqrt{3})cos(tan−1√3)? How do you evaluate \sec^{-1} \sqrt{2} sec−1√2? How do you find cos( cot^{-1} sqrt{3} )cos(cot−1√3) without a calculator? How do you rewrite sec^2 (tan^{-1} x)sec2(tan−1x) 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)sin−1(0.1)? How do you solve the inverse trig function cos^-1 (-sqrt2/2)cos−1(−√22)? How do you solve the inverse trig function arcsin (sin 5pi/6)arcsin(sin5π6)? See all questions in Inverse Trigonometric Properties Impact of this question 4850 views around the world You can reuse this answer Creative Commons License