Question

Evaluate the function y=cos(sin^-1(x)) geometrically to find the answer in radical form. Set your answer equal to y.

Answer 1

What is really important here is the equation: \[
\sin^2(x)+\cos^2(x)=1
\]

Answer 2

thats true, i havent studied the trigonometric identities much yet.

Answer 3

We get \[
\sin^2(x) = 1-\cos^2(x)
\]

Answer 4

\[\sin(x)\sin(x)=1-\cos(x)\cos(x)\]

Answer 5

We also get \[
\cos^2(x) = 1-\sin^2(x)
\]Taking square root of both sides: \[
\cos(x) = \sqrt{1-\sin^2(x)}
\]
\[
y=\cos(\sin^{-1}(x)) = \sqrt{1-\sin^2(\sin^{-1}(x))}
\]Hmmmm.

Answer 6

See what I did there?

Answer 7

so
\[y=\cos(\sin ^{-1}(x))\]

Answer 8

eeeeeh, basically what you said. i dun wanna retype it .w. but yes, i think so.

Answer 9

\[
\cos(\color{red}x) = \sqrt{1-\sin^2(\color{red}x)} \\
\cos(\color{red}{\sin^{-1}(x)}) = \sqrt{1-\sin^2(\color{red}{\sin^{-1}(x)})}
\]

Answer 10

I hope you are not color blind.

Answer 11

and the whole thins is under the square root because in the original equation the inverse sine of x was the "x"...im not color blind :3