To understand this section, you should have mastered basic trigonometry.

These sections are also directly related to inverse trig. functions:

Let's take a look at this triangle and try to calculate the lengths of the unknown side and the measures of the angles,

Well, the length of the hypotenuse is easy to find using the Pythagorean theorem. It's just

$$h^2 = 21^2 + 18^2$$

Taking the square root of both sides, we find

$$h = 30 \; cm$$

But what now? There are no *angles*. We can easily calculate the sine, cosine and tangent of angle **a**, they're just

$$sin(\color{magenta}{a}) = \frac{18}{30}$$

$$cos(\color{magenta}{a}) = \frac{24}{30}$$

$$tan(\color{magenta}{a}) = \frac{18}{24}$$

What we need to find those angles are inverse functions, functions that undo the action of the trig functions on both sides of the equation. For example,

$$ \begin{align} \text{If } \; f(x) &= sin(x), \; \text{then} \\[5pt] f^{-1}(x) &= sin^{-1}(x) \end{align}$$

You should recall the properties of inverse functions from your study of functions in general, and from learning about the inverse relationship between log and exponential functions. So the inverse relationship between sin(x) and its inverse is:

$$ \begin{align} sin^{-1}(sin(x)) &= x \; \; \color{#E90F89}{\text{and}} \\[5pt] sin(sin^{-1}(x)) &= x \end{align}$$

So let's take that sine relationship from the little table above and use the inverse sine:

$$ \begin{align} \text{If } \; sin(a) &= \frac{18}{30} \\ \\ \text{then } \; sin^{-1}(sin(a)) &= sin^{-1} \left( \frac{18}{30} \right) \end{align}$$

The left side is just our angle **a**

$$\text{so } \; a = sin^{-1}\left( \frac{18}{30} \right)$$

And we find the right side by using the **sin ^{-1}** button on any scientific calculator:

$$a = 36.9˚$$

The inverse trig functions are generally the second [2^{nd}] functions of the trig buttons. Take a look at your own calculator to confirm this.

You should also get in the habit of using the π button instead of typing in 3.14159 ... for π. It's always a good idea to use as much precision in your numbers as possible while you do a calculation, then round the number down to a reasonable number of digits when you report it.

Most calculators don't have [sec], [csc] and [cot] functions because it's easy enough to type in 1/sin(x), 1/cos(x), or 1/tan(x).

**Pro tip**: Make sure you always know which angle units your calculator is set to use: radians or degrees. Know how to switch between them.

Inverse functions are good for ** undoing** the action of a function on an independent variable. One way to think about inverse functions is that

and **inverse functions** take us from range to domain. In trigonometry, we need inverse functions to solve for the values of angles when all we have are the sides of a triangle.

But because of the periodic nature of the trig functions, we need to be very careful about their definitions and restrictions, and some quirks of their use.

The inverse sine function, $\text{sin}^{-1}(x)$, sometimes called the arcsine, $\text{arcsin}(x)$, is shown in this figure

The black curve is the sine function. We know that the graph of its inverse is simply the reflection of the graph of $\text{sin}(x)$ across the line $y=x$, as shown (see functions).

But this presents a problem: The red curve is multiple-valued: It has more than one value for each element of the domain. In order to overcome this, we truncate the domain of the function between -1 and 1, which restricts its range between $-\pi/2$ and $\pi/2$

This necessary restriction can have some implications on some calculations, and we'll deal with those below. The next graph is a blow-up of that green region.

Because the range of the inverse sine function is $[-\pi/2, \; \pi/2]$, (on the unit circle, $-\pi/2$ is the same as $3\pi/2$) it can never yield an angle greater than $\pi/2$ and less than $3\pi/2$. That's a problem, but not as much as you might think. Study the figure below to get comfortable with the domain and range of $\text{sin}^{-1}(x)$.

Find the polar coordinates of the point (-3, 2).

We know this point is in the second quadrant of the graph. We can construct a right triangle in that quadrant, as shown:

We need to find the polar coordinates, $(r, \, \theta)$. First,

$$r = \sqrt{x^2 + y^2} = \sqrt{3^2 + 2^2} = \sqrt{13}$$

Then we can use the inverse sine function to find the angle $\phi$:

$$ \begin{align} \text{sin}(\phi) &= \frac{2}{\sqrt{13}} \\[5pt] \phi &= \text{sin}^{-1} \left( \frac{2}{\sqrt{13}} \right) \\[5pt] &= 33.69^{\circ} \end{align}$$

Now $\theta = 180^{\circ} - \phi = 146.3^{\circ}$, so our polar coordinates are:

$$(r, \, \theta) = (\sqrt{13}, \, 146.3^{\circ})$$

The case is just about the same for the $\text{cos}^{-1}(x)$, or $\text{arccos}(x)$ function. In this case we restrict the domain in the same way, but now the range is $[0, \, \pi]$.

Again, this will have very real implications for our later calculations. Namely, the inverse sin, cos and tan functions will give, in some cases, answers that need to be adjusted by adding 180˚ or $\pi$ radians.

Because the range of the inverse cosine function is $[0, \, \pi]$, it can never yield an angle greater than $\pi$ and less than $2\pi$. If a quick graph of the point in question shows that it is in one of the lower quadrants (white region of the circle below), you will have to add 180˚ or $\pi$ radians to the output of the $\text{cos}^{-1}(x)$ function. Study the figure below to get comfortable with the domain and range of $\text{cos}^{-1}(x)$.

The $\text{tan}^{-1}(x)$ function, or $\text{arctan}(x)$, is a little different but the basic ideas are the same.

The black curves represent the asymptotic tangent curves with which we're familiar. If we perform our reflection over the line $y = x$, we get a vertical stack of curves, which are of course multiple-valued; we have to pick one.

We choose, by convention, the central curve, which makes the domain of $\text{tan}^{-1}(x)$ $[\infty, \, \infty]$ and the range $\left[ -\frac{\pi}{2}, \, \frac{\pi}{2} \right]$.

Now let's look at the problems these domain and range restrictions can have on each of the three main inverse trig functions.

The domain of the inverse tangent function is $[-\infty, \, \infty]$ but its range is $[-\pi/2, \pi/2]$. The inverse tangent function has the same range restriction as the sine function, so if you're seeking angles to go with points in the 2^{nd} and 3^{rd} quadrants (white region in the graph below), you'll have to add 180˚ or $\pi$ rad to the result you get from the $\text{tan}^{-1}$ function.

Find the two missing angles in this right triangle:

**Solution**

It's overkill, but just to illustrate the point, here are both angles found using each of the functions $\text{sin}(x)$, $\text{cos}(x)$ and $\text{tan}(x)$.

For efficiency, you'd really want to find the third angle by subtracting the first from 90˚, of course.

For each, we make the appropriate SOH-CAH-TOA expression, such as $\text{sin}(a) = \frac{5}{9.43}$, then apply the inverse function to both sides ($\text{sin}^{-1}(x)$ in this case), to find the angle. The (rounded) solutions are $a = 32^{\circ}$ and $b = 58^{\circ}$, no matter how you get them.

$$\text{sin}(a) = \frac{5}{9.43}$$

$$\text{sin}^{-1}[\text{sin}(a)] = \text{sin}^{-1} \left( \frac{5}{9.43} \right)$$

$$a = 32^{\circ}$$

$$\text{sin}(b) = \frac{8}{9.43}$$

$$\text{sin}^{-1}[\text{sin}(b)] = \text{sin}^{-1} \left( \frac{8}{9.43} \right)$$

$$b = 58^{\circ}$$

$$\text{cos}(a) = \frac{8}{9.43}$$

$$\text{cos}^{-1}[\text{cos}(a)] = \text{cos}^{-1} \left( \frac{8}{9.43} \right)$$

$$a = 32^{\circ}$$

$$\text{cos}(b) = \frac{5}{9.43}$$

$$\text{cos}^{-1}[\text{cos}(b)] = \text{cos}^{-1} \left( \frac{5}{9.43} \right)$$

$$b = 58^{\circ}$$

$$\text{tan}(a) = \frac{5}{8}$$

$$\text{tan}^{-1}[\text{tan}(a)] = \text{tan}^{-1} \left( \frac{5}{8} \right)$$

$$a = 32^{\circ}$$

$$\text{tan}(b) = \frac{8}{5}$$

$$\text{tan}^{-1}[\text{tan}(b)] = \text{tan}^{-1} \left( \frac{8}{5} \right)$$

$$b = 58^{\circ}$$

Find the measure of angle θ in each of the right triangles.

1. |
## SolutionUse the definition of the sine: $$ \begin{align} sin(\theta) &= \frac{210}{320} \\[5pt] \theta &= sin^{-1}\left( \frac{210}{320} \right) \\[5pt] &= 41˚ \end{align}$$
$$ \begin{align} cos(\theta) &= \frac{241}{320} \\[5pt] \theta &= cos^{-1}\left( \frac{241}{320} \right) \\[5pt] &= 41˚ \end{align}$$ |

2. |
## SolutionUse the definition of the sine: $$ \begin{align} sin(\theta) &= \frac{4.4}{10} \\[5pt] \theta &= sin^{-1}\left( \frac{4.4}{10} \right) \\[5pt] &= 26˚ \end{align}$$
$$ \begin{align} cos(\theta) &= \frac{9}{10} \\[5pt] \theta &= cos^{-1}\left( \frac{9}{10} \right) \\[5pt] &= 26˚ \end{align}$$ |

3. |
## SolutionUse the definition of the sine: $$ \begin{align} sin(\theta) &= \frac{0.438}{2.425} \\[5pt] \theta &= sin^{-1}\left( \frac{0.438}{2.425} \right) \\[5pt] &= 10.4˚ \end{align}$$
$$ \begin{align} cos(\theta) &= \frac{2.385}{2.425} \\[5pt] \theta &= cos^{-1}\left( \frac{2.385}{2.425} \right) \\[5pt] &= 10.4˚ \end{align}$$ |

4. |
## SolutionUse the definition of the sine: $$ \begin{align} sin(\theta) &= \frac{17.35}{25} \\[5pt] \theta &= sin^{-1}\left( \frac{17.35}{25} \right) \\[5pt] &= 43.8˚ \end{align}$$
$$ \begin{align} cos(\theta) &= \frac{18}{25} \\[5pt] \theta &= cos^{-1}\left( \frac{18}{25} \right) \\[5pt] &= 43.8˚ \end{align}$$ |

If you aren't yet familiar with the **law of cosines**, don't worry about this next section. You can find the LOC here.

Find the two missing angles in this non-right triangle:

For a problem like this, we'll use the law of cosines, which contains the term **-ab·cos(X)**, where **X** is one of the unknown angles.

We solve for **cos(X)** and find the angle using the inverse cosine function. Here are the law-of-cosines solutions for each of the angles **a**, **b** & **c**.

These solutions are overkill again. It's really only necessary to solve for one angle using the law of cosines. Then the law of sines and the sum of angles of a triangle (180˚) can be used to find the others. Still, it's reassuring when the angles found with only the LOC add to 180˚!

Find the measure of angle θ in each of the non-right triangles.

5. |
## SolutionUse the law of cosines: $$ \begin{align} c^2 &= a^2 + b^2 - 2ab cos(\theta) \\[5pt] 3.78^2 &= 4.14^2 + 2.80^2 \\[5pt] &- 2(4.14)(2.80)cos(\theta) \\[5pt] cos(\theta) &= \frac{3.78^2 - 2.8^2 - 4.14^2}{-2(4.14)(2.80)} \end{align}$$ $$ \begin{align} \theta &= cos^{-1} \left( \frac{3.78^2 - 2.8^2 - 4.14^2}{-2(4.14)(2.80)} \right) \\[5pt] \theta &= 62.5˚ \end{align}$$ |

6. |
## SolutionUse the law of cosines: $$ \begin{align} c^2 &= a^2 + b^2 - 2ab cos(\theta) \\[5pt] 241.5^2 &= 214.5^2 + 137^2 \\[5pt] &- 2(214.5)(137)cos(\theta) \\[5pt] cos(\theta) &= \frac{241.5^2 - 214.5^2 - 137^2}{-2(214.5)(137)} \end{align}$$ $$ \begin{align} \theta &= cos^{-1} \left( \frac{241.5^2 - 214.5^2 - 137^2}{-2(214.5)(137)} \right) \\[5pt] \theta &= 83.7˚ \end{align}$$ |

7. |
## SolutionUse the law of cosines: $$ \begin{align} c^2 &= a^2 + b^2 - 2ab cos(\theta) \\[5pt] 137.17^2 &= 87.58^2 + 73.08^2 \\[5pt] &- 2(87.58)(73.08)cos(\theta) \\[5pt] cos(\theta) &= \frac{137.17^2 - 87.58^2 - 73.08^2}{-2(87.58)(73.08)} \end{align}$$ $$ \begin{align} \theta &= cos^{-1} \left( \frac{137.17^2 - 87.58^2 - 73.08^2}{-2(87.58)(73.08)} \right) \\[5pt] \theta &= 115.6˚ \end{align}$$ |

8. |
## SolutionUse the law of cosines: $$ \begin{align} c^2 &= a^2 + b^2 - 2ab cos(\theta) \\[5pt] 3.31^2 &= 2.12^2 + 3.56^2 \\[5pt] &- 2(2.12)(3.56)cos(\theta) \\[5pt] cos(\theta) &= \frac{3.31^2 - 2.12^2 - 3.56^2}{-2(2.12)(3.56)} \end{align}$$ $$ \begin{align} \theta &= cos^{-1} \left( \frac{3.31^2 - 2.12^2 - 3.56^2}{-2(2.12)(3.56)} \right) \\[5pt] \theta &= 65.7˚ \end{align}$$ |

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