How do you find the $\mathrm{arcsin}\left(\mathrm{sin}\left(\frac{7\pi}{6}\right)\right)$ ?

meteraiqn
2022-01-22
Answered

How do you find the $\mathrm{arcsin}\left(\mathrm{sin}\left(\frac{7\pi}{6}\right)\right)$ ?

You can still ask an expert for help

gekraamdbk

Answered 2022-01-23
Author has **13** answers

The answer is:

$\mathrm{arcsin}\left(\mathrm{sin}\left(7\frac{\pi}{6}\right)\right)=-\frac{\pi}{6}$

The range of a function$\mathrm{arcsin}\left(x\right)$ is, by definition.

$-\frac{\pi}{2}\le \mathrm{arcsin}\left(x\right)\le \frac{\pi}{2}$

It means that we have to find an angle$\alpha$ that lies between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ and whose $\mathrm{sin}\left(\alpha \right)$ equals to a $\mathrm{sin}\left(7\frac{\pi}{6}\right)$ .

From trigonometry we know that

$\mathrm{sin}(\varphi +\pi )=-\mathrm{sin}\left(\varphi \right)$

for any angle$\varphi$ .

This is easy to see if use the definition of a sine as an ordinate of the end of a radius in the unit circle that forms an angle$\varphi$ with the X-axis (counterclockwise from the X-axis to a radius).

We also know that since is an odd function, that is$\mathrm{sin}(-\varphi )=-\mathrm{sin}\left(\varphi \right)$

We will use both properties as follows:

$\mathrm{sin}\left(7\frac{\pi}{6}\right)=\mathrm{sin}(\frac{\pi}{6}+\pi )=-\mathrm{sin}\left(\frac{\pi}{6}\right)=\mathrm{sin}(-\frac{\pi}{6})$

As we see, the angle$\alpha =-\frac{\pi}{6}$ first our conditions. It is in the range from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ and its sine equals to $\mathrm{sin}\left(7\frac{\pi}{6}\right)$ . Therefore, it's a correct answer to a problem.

The range of a function

It means that we have to find an angle

From trigonometry we know that

for any angle

This is easy to see if use the definition of a sine as an ordinate of the end of a radius in the unit circle that forms an angle

We also know that since is an odd function, that is

We will use both properties as follows:

As we see, the angle

enguinhispi

Answered 2022-01-24
Author has **15** answers

So:

$\mathrm{arcsin}={\mathrm{sin}}^{-1}$

${\mathrm{sin}}^{-1}\left(\mathrm{sin}\left(\frac{7\pi}{6}\right)\right)$

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the third quadrant.

${\mathrm{sin}}^{-1}(-\mathrm{sin}\left(\frac{\pi}{6}\right))$

The exact value of$\mathrm{sin}\left(\frac{\pi}{6}\right)$ is $\frac{1}{2}$ .

${\mathrm{sin}}^{-1}(-\frac{1}{2})$

The exact value of${\mathrm{sin}}^{-1}(-\frac{1}{2})$ is $-\frac{\pi}{6}$

The result can be shown in multiple forms.

Exact Form:

$-\frac{\pi}{6}$

Decimal Form:

$-0.52359877\dots$

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the third quadrant.

The exact value of

The exact value of

The result can be shown in multiple forms.

Exact Form:

Decimal Form:

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