Prove that cos(sin ^-1 x ) = ( sqrt ( 1 - x ^ 2 )

[peblez]

Hi, i am stuck on a question, it has answers but i don't really get the solution, can someone explain it to me. This is the question

Prove that cos(sin ^-1 x ) = ( sqrt ( 1 - x ^ 2 )

Solution:

Let y = sin ^ -1 x. Then - pi / 2 <_ y <_ pi / 2

cos y >_ 0

so cos ( sin ^-1 x) = cos y = ( sqrt ( 1 - sin ^ 2 y ) = sqrt ( 1 - x ^ 2 )

I dont understand the last step , can someone explain to me how they got the square root in there. thanks

 

[galactus]

Start with \(\displaystyle \L\\cos^{2}y=1-sin^{2}y\)

Now sub \(\displaystyle \L\\y=sin^{-1}x\):

\(\displaystyle \L\\cos^{2}(sin^{-1}x)=1-sin^{2}(sin^{-1}x)=1-x^{2}\)

Take square roots:

\(\displaystyle \L\\|cos(sin^{-1}x)|=\sqrt{1-x^{2}}\)

Since, \(\displaystyle \L\\\frac{-\pi}{2}\leq{sin^{-1}x}\leq{\frac{\pi}{2}}\),

we can see that \(\displaystyle \L\\cos(sin^{-1}x)\) is always positive.

So, we have:

\(\displaystyle \L\\cos(sin^{-1}x)=\sqrt{1-x^{2}}\)