How do you simplify #cos(2 arcsin x)#? Trigonometry Inverse Trigonometric Functions Inverse Trigonometric Properties 1 Answer Cesareo R. Jun 1, 2016 #cos(2arcsin(x))=1-2x^2# Explanation: Using #cos(a+b)=Cos(a) Cos(b) - Sin(a) Sin(b)#. If #a = b# then #cos(2a)=1-2sin^2(a)# so #cos(2arcsin(x))=1-2x^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 35117 views around the world You can reuse this answer Creative Commons License