# Basic Inverse Trigonometric Functions

Introduction to Inverse Trig Functions (1 of 2): Why a whole new topic?

Tip: This isn't the place to ask a question because the teacher can't reply.

1 of 4 videos by Eddie W.

## Key Questions

• An inverse trigonometric function takes, as its input, a ratio and produces the angle that the original trig function associates with it.

For instance, if $\setminus \sin \left(\setminus \frac{\setminus \pi}{3}\right) = \setminus \frac{\setminus \sqrt{3}}{2}$, then ${\sin}^{- 1} \left(\setminus \frac{\setminus \sqrt{3}}{2}\right) = \setminus \frac{\setminus \pi}{3}$.

However, it's important to keep in mind that, unlike the original trig function, all angles are not permissible. Since, in order to be a function, there must be only one member of the range associated with each member of the domain, we have to be careful. For example, you should know that

$\setminus \cos \left(\setminus \frac{\setminus \pi}{6}\right) = \setminus \cos \left(\setminus \frac{11 \setminus \pi}{6}\right) = \setminus \cos \left(\setminus \frac{13 \setminus \pi}{6}\right) = \setminus \cos \left(\setminus \frac{23 \setminus \pi}{6}\right) = \setminus \cdots = \setminus \frac{\setminus \sqrt{3}}{2}$

Yet, we don't want to evaluate $\setminus {\cos}^{- 1} \left(\setminus \frac{\setminus \sqrt{3}}{2}\right)$ and obtain $\setminus \frac{\setminus \pi}{6}$, $\setminus \frac{11 \setminus \pi}{6}$, $\setminus \frac{13 \setminus \pi}{6}$, $\setminus \frac{23 \setminus \pi}{6}$, and so on. This would mean that inverse cosine is not a function.

Therefore, we limit the range of each inverse function so that there can be only one output for each input. For the three primary inverse functions, there domains and ranges are as follows:

• ${\sin}^{- 1} \left(x\right)$: Domain: $\left[- 1 , 1\right]$; Range: $\left[- \setminus \frac{\setminus \pi}{2} , \setminus \frac{\setminus \pi}{2}\right]$
• ${\cos}^{- 1} \left(x\right)$: Domain: $\left[- 1 , 1\right]$; Range: $\left[0 , \setminus \pi\right]$
• ${\tan}^{- 1} \left(x\right)$: Domain: $\left[- \setminus \infty , \setminus \infty\right]$; Range: $\left[- \setminus \frac{\setminus \pi}{2} , \setminus \frac{\setminus \pi}{2}\right]$
• The basic inverse trigonometric functions are used to find the missing angles in right triangles. While the regular trigonometric functions are used to determine the missing sides of right angled triangles, using the following formulae:

$\sin \theta$ = opposite $\div i \mathrm{de}$hypotenuse
$\cos \theta$ = adjacent $\div i \mathrm{de}$ hypotenuse
$\tan \theta$ = opposite $\div i \mathrm{de}$ adjacent

the inverse trigonometric functions are used to find the missing angles, and can be used in the following way:

For example, to find angle A, the equation used is:

${\cos}^{-} 1$ = side b $\div i \mathrm{de}$ side c

• $\sin x = n$ if and only if $x = \arcsin n + 2 \pi k$ for some integer $k$

$\cos x = n$ if and only if $x = \arccos n + 2 \pi k$ for some integer $k$

$\tan x = n$ if and only if $x = \arctan n + \pi k$ for some integer $k$

and so on.

So,: Solve $7 \sin x - 5 = 0$

$7 \sin x - 5 = 0$
$7 \sin x = 5$
$\sin x = \frac{5}{7}$

$x = \arcsin \left(\frac{5}{7}\right) + 2 \pi k$ for integer $k$.

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