2 Math Question Involving Inverse Trig

**qbkr21** · September 23rd 2007, 04:48 PM

Thanks!!

**Jhevon** · September 23rd 2007, 05:04 PM

Originally Posted by qbkr21

Thanks!!

Let $\sin^{-1}x = \theta$

$\Rightarrow \sin \theta = x$

so now we can construct a right triangle with an acute angle $\theta$ , where the opposite side to that angle is $x$ and the hypotenuse is 1. thus, by Pythagoras' theorem, we can find the adjacent side, which will be $\sqrt {1 - x^2}$

Now, finally, we can say, $\cos \left( \sin^{-1} x \right) = \cos \theta =$ ?? that final piece is for you

you can try a similar method for the second

**ThePerfectHacker** · September 23rd 2007, 06:26 PM

To Show: $\cos (\sin^{-1} x ) = \sqrt{1-x^2}$ .

Do the following steps in order.

1)Define the function $f(x) = \cos(\sin^{-1} x )$ .

2)Show that $[-1,1]$ is in the domain of this function.

3)Show that $f(x)$ is continous on $[-1,1]$ .

4)Show that $f(x)$ is differenciable on $(-1,1)$ .

5)Show that $f'(x) = \frac{x}{\sqrt{1-x^2}}$ on $(-1,1)$ .

6)Notice that $\left( \sqrt{1-x^2} \right)' = f'(x)$ on $(-1,1)$

7)Therefore argue that $\sqrt{1-x^2}+C = f'(x)$ on $(-1,1)$ .

8)Prove that $C=0$ .

9)Therefore, $\cos (\sin^{-1} x ) = f(x) = \sqrt{1-x^2}$ on $(-1,1)$ .

10)But $\cos (\sin^{-1} 1) = \sqrt{1-1^2}$ and $\cos (\sin^{-1} -1) = \sqrt{1-(-1)^2}$ .

11)Finally conclude that $\cos(\sin^{-1} x ) = \sqrt{1-x^2}$ on $[-1,1]$ .

**Krizalid** · September 23rd 2007, 06:32 PM

Or we can take

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

--

For the second problem, find the sine in terms of the tangent, the apply the same method as above.

**ThePerfectHacker** · September 23rd 2007, 06:46 PM

Originally Posted by Krizalid

Or we can take

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

--

For the second problem, find the sine in terms of the tangent, the apply the same method as above.

This equation is not correct!

What if $x = \pi$ .

The correct equation is much more difficult.
$\cos x = \mbox{Kriz}(x) \sqrt{1-\sin^2x}$ .

Where $\mbox{Kriz}(x)$ is the so-called "Krizalid function" it is defined as follows:
$\mbox{Kriz}(x) = \left\{ \begin{array}{c}1 \mbox{ for }x\in [-\pi/2,\pi/2] \\ -1 \mbox{ otherwise } \end{array} \right.$ and $\mbox{Kriz}(x + 2\pi) = \mbox{Kriz}(x)$ , i.e. the function is periodic.

So for example, we get $1 \mbox{ on }[-\pi/2,\pi/2]$ it we add $2\pi$ we get back $[3\pi/2,5\pi/2]$ is also $1$ . If we subtract $\pi$ we get back $[-5\pi/2,-3\pi/2]$ it is $1$ here also. But $\pi$ is not among any of these periodic intervals. So it means $\mbox{Kriz}(\pi) = -1$ , thus, $\cos \pi = \mbox{Kriz}(\pi) \sqrt{1 - 0^2} = -1$ .

**qbkr21** · September 23rd 2007, 07:02 PM

Originally Posted by Jhevon

thus, by Pythagoras' theorem, we can find the adjacent side, which will be $\sqrt {1 - x^2}$
d

How are you finding $\sqrt {1 - x^2}$ ??

I set it up, but am just not seeing it

Thanks!

-qbkr21

**ThePerfectHacker** · September 23rd 2007, 07:15 PM

The hypothenuse is $1$ the side is $x$ call the other side $y$ then $x^2+y^2 = 1$ by Pythagorus, so $y = \sqrt{1-x^2}$ .

**Jhevon** · September 23rd 2007, 07:16 PM

Originally Posted by qbkr21

How are you finding $\sqrt {1 - x^2}$ ??

I set it up, but am just not seeing it

Thanks!

-qbkr21

By Pythagoras' theorem. if the sides of a right-triangle are $a$ , $b$ and $c$ , where $c$ is the hypotenuse, then we have:

$a^2 + b^2 = c^2$

this means, to find one of the other legs, say $a$ , we solve for it.

$a = \sqrt {c^2 - b^2}$

that is exactly what i did here. c was 1 and b was x

**qbkr21** · September 23rd 2007, 07:23 PM

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

**Jhevon** · September 23rd 2007, 07:31 PM

Originally Posted by qbkr21

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

umm, that is the answer. that's what you asked us to prove

**qbkr21** · September 23rd 2007, 07:32 PM

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

**Krizalid** · September 24th 2007, 08:04 AM

Originally Posted by ThePerfectHacker

This equation is not correct!

What if $x = \pi$ .

I just forgot to set the constraint.

--

I don't like the "Krizalid Function"

**ThePerfectHacker** · September 24th 2007, 10:08 AM

Originally Posted by Krizalid

I don't like the "Krizalid Function"

If you express it as a Fourier series. I looks much nice like that.

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