Thanks!!
Let $\displaystyle \sin^{-1}x = \theta$
$\displaystyle \Rightarrow \sin \theta = x$
so now we can construct a right triangle with an acute angle $\displaystyle \theta$, where the opposite side to that angle is $\displaystyle x$ and the hypotenuse is 1. thus, by Pythagoras' theorem, we can find the adjacent side, which will be $\displaystyle \sqrt {1 - x^2}$
Now, finally, we can say, $\displaystyle \cos \left( \sin^{-1} x \right) = \cos \theta = $?? that final piece is for you
you can try a similar method for the second
To Show: $\displaystyle \cos (\sin^{-1} x ) = \sqrt{1-x^2}$.
Do the following steps in order.
1)Define the function $\displaystyle f(x) = \cos(\sin^{-1} x )$.
2)Show that $\displaystyle [-1,1]$ is in the domain of this function.
3)Show that $\displaystyle f(x)$ is continous on $\displaystyle [-1,1]$.
4)Show that $\displaystyle f(x)$ is differenciable on $\displaystyle (-1,1)$.
5)Show that $\displaystyle f'(x) = \frac{x}{\sqrt{1-x^2}}$ on $\displaystyle (-1,1)$.
6)Notice that $\displaystyle \left( \sqrt{1-x^2} \right)' = f'(x)$ on $\displaystyle (-1,1)$
7)Therefore argue that $\displaystyle \sqrt{1-x^2}+C = f'(x)$ on $\displaystyle (-1,1)$.
8)Prove that $\displaystyle C=0$.
9)Therefore, $\displaystyle \cos (\sin^{-1} x ) = f(x) = \sqrt{1-x^2}$ on $\displaystyle (-1,1)$.
10)But $\displaystyle \cos (\sin^{-1} 1) = \sqrt{1-1^2}$ and $\displaystyle \cos (\sin^{-1} -1) = \sqrt{1-(-1)^2}$.
11)Finally conclude that $\displaystyle \cos(\sin^{-1} x ) = \sqrt{1-x^2}$ on $\displaystyle [-1,1]$.
This equation is not correct!
What if $\displaystyle x = \pi $.
The correct equation is much more difficult.
$\displaystyle \cos x = \mbox{Kriz}(x) \sqrt{1-\sin^2x}$.
Where $\displaystyle \mbox{Kriz}(x)$ is the so-called "Krizalid function" it is defined as follows:
$\displaystyle \mbox{Kriz}(x) = \left\{ \begin{array}{c}1 \mbox{ for }x\in [-\pi/2,\pi/2] \\ -1 \mbox{ otherwise } \end{array} \right.$ and $\displaystyle \mbox{Kriz}(x + 2\pi) = \mbox{Kriz}(x)$, i.e. the function is periodic.
So for example, we get $\displaystyle 1 \mbox{ on }[-\pi/2,\pi/2]$ it we add $\displaystyle 2\pi$ we get back $\displaystyle [3\pi/2,5\pi/2]$ is also $\displaystyle 1$. If we subtract $\displaystyle \pi$ we get back $\displaystyle [-5\pi/2,-3\pi/2]$ it is $\displaystyle 1$ here also. But $\displaystyle \pi $ is not among any of these periodic intervals. So it means $\displaystyle \mbox{Kriz}(\pi) = -1$, thus, $\displaystyle \cos \pi = \mbox{Kriz}(\pi) \sqrt{1 - 0^2} = -1$.
By Pythagoras' theorem. if the sides of a right-triangle are $\displaystyle a$, $\displaystyle b$ and $\displaystyle c$, where $\displaystyle c$ is the hypotenuse, then we have:
$\displaystyle a^2 + b^2 = c^2$
this means, to find one of the other legs, say $\displaystyle a$, we solve for it.
$\displaystyle a = \sqrt {c^2 - b^2}$
that is exactly what i did here. c was 1 and b was x