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- Sep 23rd 2007, 04:48 PMqbkr212 Math Question Involving Inverse Trig
Thanks!!:D

- Sep 23rd 2007, 05:04 PMJhevon
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 - Sep 23rd 2007, 06:26 PMThePerfectHacker
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 $\displaystyle [-1,1]$. - Sep 23rd 2007, 06:32 PMKrizalid
Or we can take

$\displaystyle \cos x=\sqrt{1-\sin^2x}$, and evaluate $\displaystyle \cos(\arcsin x)$

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For the second problem, find the sine in terms of the tangent, the apply the same method as above. - Sep 23rd 2007, 06:46 PMThePerfectHacker
This equation is not correct!

What if $\displaystyle x = \pi $. (Fubar)

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, $\displaystyle \cos \pi = \mbox{Kriz}(\pi) \sqrt{1 - 0^2} = -1$. - Sep 23rd 2007, 07:02 PMqbkr21Re:
- Sep 23rd 2007, 07:15 PMThePerfectHacker
The hypothenuse is $\displaystyle 1$ the side is $\displaystyle x$ call the other side $\displaystyle y$ then $\displaystyle x^2+y^2 = 1$ by Pythagorus, so $\displaystyle y = \sqrt{1-x^2}$.

- Sep 23rd 2007, 07:16 PMJhevon
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 - Sep 23rd 2007, 07:23 PMqbkr21Re:
One last thing. If we are looking for cos(Ө) that would be adjacent/hypotenuse. However this wouldn't the solution I am looking for. Adjacent/Hypotenuse would instead give me

- Sep 23rd 2007, 07:31 PMJhevon
- Sep 23rd 2007, 07:32 PMqbkr21Re:
Sorry...I was thinking of Problem #2. Let me get to work on it so I can see if I have any questions.

Thanks,

-qbkr21 - Sep 24th 2007, 08:04 AMKrizalid
- Sep 24th 2007, 10:08 AMThePerfectHacker