Simplifying an Expression involving sin^-1 x

Question

If 0<x<1, write cos(sin^-1 x) as an algebraic expression in x.

Answer 1

A good exercise:

 

  Sin² ( tan ^-1 b/a ) + C0S² ( COT ^-1 b/a)  =1 

 

  Let X = tan^-1 a/b

   

Draw a right triangle:        Mark side opposite to an acute angle(X) a, other log b 

                                         

                                           then hypotenuse ;√( a^2 + b^2)

 

     Sin X = b/ √( a^2 + b^2)

 

     Sin ² X + Cos ² X = ( a² /√( a^2 + b^2)²  + b² / √( a^2 + b^2) ² 

                               

                                  =  ( a ² + b² ) / (a²  + b² ) =1                      

Answer 2

If 0<x<1, write cos(sin^-1(x) ) as an algebraic expression in x.

 

θ = sin^-1(x) ==> sin(θ) = x

 

0 < x < 1 ==> 0 < sin(θ) < 1, so θ lies in the first quadrant.

 

If x is a leg of a right triangle and θ is the acute angle across from it, the hypotenuse must be 1 because sin(θ) = x/1.

 

Using the Pythagorean Theorem the other leg is √(1-x^2), so:

 

cos(θ) = √(1-x^2)/1, so

 

cos(sin^-1(x) ) = √(1-x^2), 0<x<1.

 

E.g., if x = √(3)/2, then

cos(sin^-1(√(3)/2) ) = √(1-(√(3)/2)^2) = 1/2