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High School Math Grades 9-12, Ages 13-18

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#1 h Jun 19, 2016, 2:01 AM • 3 Y

Y by Adventure10, Mango247, gauss202

The graph of the function $f(x)=\sin^{-1}(2\sin{x})$ consists of the union of disjoint pieces. Compute the distance between the endpoints of any one piece

This post has been edited 1 time. Last edited by Math1331Math, Jun 19, 2016, 2:01 AM
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#2 h May 13, 2025, 12:18 PM • 1 Y

Y by gauss202

Upon inspection we require $-1 \le 2\sin(x) \le 1$ for the inverse sine function to be properly defined. This yields:

$-\frac{1}{2} \le \sin(x) \le \frac{1}{2} \Rightarrow -\frac{\pi}{6} \le x \le \frac{\pi}{6}$ ;

and $f:\left[-\frac{\pi}{6},\frac{\pi}{6}\right] \rightarrow \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , which gives us the endpoints $\left(-\frac{\pi}{6},-\frac{\pi}{2}\right); \left(\frac{\pi}{6},\frac{\pi}{2}\right)$ for the piece that is symmetric with respect to the origin. Thus, the straight linear distance between these endpoints computes to:

$D = \sqrt{\left(2 \cdot \frac{\pi}{6}\right)^2+\left(2\cdot \frac{\pi}{2}\right)^2} = 2\pi\sqrt{\frac{1}{36}+\frac{1}{4}} = \textcolor{red}{\frac{\pi\sqrt{10}}{3}}$ .

This post has been edited 2 times. Last edited by Mathzeus1024, May 13, 2025, 1:01 PM

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#3 h May 15, 2025, 2:48 PM

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This is basically $f(x) = arcsin (2sinx)$ , right?

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#4 h May 16, 2025, 1:45 AM

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BlackOctopus23 wrote:

This is basically $f(x) = arcsin (2sinx)$ , right?

Yes, $\sin^{-1}$ is $\arcsin$

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#5 h May 16, 2025, 2:37 AM

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Why didn't he just say $\arcsin$ rather than $\sin^{-1}$ ?

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#6 h May 16, 2025, 4:29 AM

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Why didn't you say $\sin^{-1}$ instead of $\arcsin$ ?

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MathPerson12321

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#7 h May 16, 2025, 4:34 AM

Y by

jasperE3 wrote:

Why didn't you say $\sin^{-1}$ instead of $\arcsin$ ?

it doesnt matter much you still understand the notation at first glance

Mathzeus1024 wrote:

Upon inspection we require $-1 \le 2\sin(x) \le 1$ for the inverse sine function to be properly defined. This yields:

$-\frac{1}{2} \le \sin(x) \le \frac{1}{2} \Rightarrow -\frac{\pi}{6} \le x \le \frac{\pi}{6}$ ;

and $f:\left[-\frac{\pi}{6},\frac{\pi}{6}\right] \rightarrow \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , which gives us the endpoints $\left(-\frac{\pi}{6},-\frac{\pi}{2}\right); \left(\frac{\pi}{6},\frac{\pi}{2}\right)$ for the piece that is symmetric with respect to the origin. Thus, the straight linear distance between these endpoints computes to:

$D = \sqrt{\left(2 \cdot \frac{\pi}{6}\right)^2+\left(2\cdot \frac{\pi}{2}\right)^2} = 2\pi\sqrt{\frac{1}{36}+\frac{1}{4}} = \textcolor{red}{\frac{\pi\sqrt{10}}{3}}$ .

thats what i did too

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#8 h May 16, 2025, 2:43 PM

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Okay. Thank you.

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