How do you calculate #sin(2sin^-1(10x))#? Trigonometry Inverse Trigonometric Functions Inverse Trigonometric Properties 1 Answer Antoine Jul 3, 2015 #sin(2sin^(-1)(10x))=20xsqrt(1-100x^2)# Explanation: #"Let " y=sin(2sin^(-1)(10x))# Now, let #" "theta=sin^(-1)(10x)" "=>sin(theta)=10x# #=>y=sin(2theta)=2sinthetacostheta# Recall that: #" "cos^2theta=1-sin^2theta=>costheta=sqrt(1-sin^2theta)# #=>y=2sinthetasqrt(1-sin^2theta)# #=>y=2*(10x)sqrt(1-(10x)^2)=color(blue)(20xsqrt(1-100x^2))# 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 2225 views around the world You can reuse this answer Creative Commons License