# Trigonometry
$$\begin{aligned}g(x)&=tan(x -\frac{3\pi}{2})+6 \\\ x &= tan(g^{-1}(x) -\frac{3\pi}{2})+6 \\\x - 6 &= tan(g^{-1}(x) -\frac{3\pi}{2}) \\\tan^{-1}(x-6) &= g^{-1}(x) -\frac{3\pi}{2} \\\ g^{-1}(x) &= tan^{-1}(x-6)+\frac{3\pi}{2}\end{aligned}$$
$$tan$$
| domain | range |
| ------------------------------ | --------- |
| all reals | all reals |
| except $$\frac{\pi}{2} + \pi$$ | |
$$tan^{-1}$$
| domain | range |
| --------- | ---------------------------------- |
| all reals | $$(-\frac{\pi}{2},\frac{\pi}{2})$$ |
### What are the inverse trigonometric functions?
$$\arcsin(x), \text{or} \sin^{-1}(x)$$, is the inverse of $$\sin(x)$$.
$$\arccos(x), \text{or} \cos^{-1}(x)$$, is the inverse of $$\cos(x)$$.
$$\arctan(x), \text{or} \tan^{-1}(x)$$, is the inverse of $$\tan(x)$$.
### Range of the inverse trig functions
| Radians | Degrees |
| --------------------------------------------------------- | --------------------------------------------- |
| $$-\dfrac{\pi}{2}\leq\arcsin(\theta)\leq\dfrac{\pi}{2}$$ | $$-90^\circ\leq\arcsin(\theta)\leq 90^\circ$$ |
| $$0\leq\arccos(\theta)\leq\pi$$ | $$0^\circ\leq\arccos(\theta)\leq 180^\circ$$ |
| $$-\dfrac{\pi}{2}<\arctan(\theta)<\dfrac{\pi}{2}$$ | $$-90^\circ<\arctan(\theta)<90^\circ$$ |
The trigonometric functions aren't really invertible, because they have multiple inputs that have the same output. For example, $$\sin(0)=\sin(\pi)=0sin(0)$$. So what should be $$\sin^{-1}(0)$$?
In order to define the inverse functions, we have to restrict the domain of the original functions to an interval where they are invertible. These domains determine the range of the inverse functions.
The value from the appropriate range that an inverse function returns is called **the principal value** of the function.
#### **The sine value of all options is** $$0.98$$**. Which is the *****principal value***** of** $$\arcsin\left(0.98\right)$$**?**
The only measure out of the options that is within the interval $$\left\[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$$ is $$1.371$$.
Therefore, $$\arcsin\left(0.98\right)=1.37$$.
**The cosine value of all options is** $$0.32$$**. Which is the *****principal value***** of** $$\cos^{-1}\left(0.32\right)$$**?**
The only measure out of the options that is within the interval $$\left\[0,\pi\right]$$ is $$1.25$$.
Therefore, $$\cos^{-1}\left(0.32\right)=1.25$$.
**The tangent value of all options is** $$-21$$**. Which is the *****principal value***** of** $$\tan^{-1}\left(-21\right)$$**?**
The only measure out of the options that is within the interval $$\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$$ is $$-1.52$$.
Therefore, $$\tan^{-1}\left(-21\right)=-1.52$$.
### Inverse Trig Function Ranges
| Function
Name
| Function
Abbreviations
| Range of
Principal Values
|
| :---------------------: | :------------------------------: | :--------------------------------------------------------------------------------------------------------------------------------------------: |
| Arcsine | Arcsin $$x$$ or $$sin^{-1} x$$ | -1 <= x <= 1
-π/2 <= y <= π/2
|
| Arccosine | Arccos $$x$$ or $$cos^{ -1} x$$ | 1 <= x <= 1
0 <= y <= π
|
| Arctangent | Arccos $$x$$or $$tan^{-1}x$$ | x, all real numbers
-π/2 < y < π/2
|
| Arccotangent | Arccot $$x$$ or $$cot^{-1} x$$ | x, all real numbers
except 0 = π/2
-π/2 < y < π/2
|
| Arcsecant | Acrsec $$x$$ or $$sec^{-1} x$$ | x <= -1 and x >= 1
0 <= y < π/2 or π/2 < y <= π
|
| Arccosecant | Arccsc $$x$$ or $$csc^{-1} x$$ | x <= -1 and x >= 1
-π/2 <= y < 0 or 0 < y <= π/2
|
## [Inverse Trigonometric Functions Calculator](https://www.calculatorsoup.com/calculators/trigonometry/inversetrigonometricfunctions.php)
**Note**: If we have a $$y = cos\theta$$ and move to the right we are going counter clockwise around the unit circle and visa versa.
So,
$$cos\theta = 1 \ \Rightarrow 2\pi n \Longrightarrow \text{we can also write it as }0, 2\pi, 4\pi, 6\pi,... \newline cos\theta = -1 \Rightarrow 2\pi n + \pi \Longrightarrow -\pi, \pi, 3\pi, 5\pi,...$$
## Quiz's
### **Select one or more expressions that together represent all solutions to the equation. Your answer should be in *****radians*****.**
$$cos(x)=−0.1$$
#### Finding all solutions within $$-\pi\ $$\cos^{-1}(-0.1)=1.67$$
Now we can use the **identity** $$\cos(\theta)=\cos(-\theta)$$ to find that the second solution within the interval is $$-1.67$$
#### Extending to all solutions
We use the identity $$\cos(\theta)=\cos(\theta+2\pi)$$to extend the two solutions we found to all solutions.
> $$x=-1.67+n\cdot2\pi$$
> $$x=1.67+n\cdot2\pi$$
### **Solve the equation in the interval from** $$180^\circ$$ **to** $$720^\circ$$**. Your answer should be in *****degrees*****.**
$$cos(x)=−0.4$$
#### Finding all solutions within $$-180^\circ\ $$\cos^{-1}(-0.4)=113.58^\circ$$
Now we can use the identity $$\cos(\theta)=\cos(-\theta)$$ to find that the second solution within the interval is $$-113.58^\circ$$.
#### Finding the set of possible solutions
We use the identity $$\cos(\theta)=\cos(\theta+360^\circ)$$ to extend the two solutions we found to all solutions.
> $$x=113.58^\circ+n\cdot360^\circ$$
> $$x=-113.58^\circ+n\cdot360^\circ$$
Here, $$n$$ is any integer.
Finding the solutions between $$180^\circ$$ and $$720^\circ$$
Now that we have the sets of possible solutions, we can adjust the value of $$n$$ to find all the values that fall within our desired interval.Let us start with the first expression, $$113.58^\circ+n\cdot360^\circ$$.
| $$n$$ | $$113.58^\circ+n\cdot360^\circ$$ | Appropriate? |
| ----- | :------------------------------: | :------------------------: |
| $$0$$ | $$113.58^\circ$$ | $$\redD{\text{Too low}}$$ |
| $$1$$ | $$473.58^\circ$$ | $$\greenD{\text{Yes}}$$ |
| $$2$$ | $$833.58^\circ$$ | $$\redD{\text{Too high}}$$ |
Now the second expression, $$-113.58^\circ+n\cdot360^\circ$$.
| $$n$$ | $$-113.58^\circ+n\cdot360^\circ$$ | Appropriate? |
| ----- | :-------------------------------: | :------------------------: |
| $$0$$ | $$-113.58^\circ$$ | $$\redD{\text{Too low}}$$ |
| $$1$$ | $$246.42^\circ$$ | $$\greenD{\text{Yes}}$$ |
| $$2$$ | $$606.42^\circ$$ | $$\greenD{\text{Yes}}$$ |
| $$3$$ | $$966.42^\circ$$ | $$\redD{\text{Too high}}$$ |
#### Summary
* $$x=246.42^\circ$$
* $$x=473.58^\circ$$
* $$x=606.42^\circ$$
## Solutions
### Degree
#### **Select one or more expressions that together represent all solutions to the equation. Your answer should be in *****DEGREEs*****.** $$4\cos(10x)+2=2$$
Choose all answers that apply:
* $$-172^\circ+n\cdot36^\circ$$
* $$-90^\circ+n\cdot360^\circ$$
* $$-9^\circ+n\cdot36^\circ$$
* $$9^\circ+n\cdot36^\circ$$
* $$90^\circ+n\cdot360^\circ$$
* $$172^\circ+n\cdot90^\circ$$
#### Isolate the sinusoidal function
$$\begin{aligned}4\cos(10x)+2&=2\\\ 4\cos(10x)&=0\\\ \cos(10x)&=0\end{aligned}$$
#### Find all values of $$10x$$ between $$-180^\circ$$ and $$180^\circ$$
Our argument is $$10x$$. Let's use the calculator and round to the nearest hundredth.
$$\cos^{-1}(0)=90^\circ$$
Now we can use the **identity** $$\cos(\theta)=\cos(-\theta)$$ to find that the second solution within the interval is $$-90^\circ$$.
#### Finding all solutions
Using the **identity** $$\cos(\theta)=\cos(\theta+360^\circ)$$, we can find all the solutions to our equation from the two angles we found above. The first solution gives us the following.
$$\begin{aligned}10x&=90^\circ+n\cdot360^\circ \\\x&=\dfrac{90^\circ+n\cdot360^\circ}{10}=9^\circ+n\cdot36^\circ\end{aligned}$$
Similarly, the second solution gives us the following.
$$\begin{aligned}10x&=-90^\circ+n\cdot360^\circ \\\x&=\dfrac{-90^\circ+n\cdot360^\circ}{10}=-9^\circ+n\cdot36^\circ\end{aligned}$$
#### Summary
Our solutions are as follows.
* $$x=-9^\circ+n\cdot36^\circ$$
* $$x=9^\circ+n\cdot36^\circ$$
### **Radian**
#### **Select one or more expressions that together represent all solutions to the equation. Your answer should be in *****RADIANs*****.** $$6\sin(20x)+1=1$$
#### Isolate the sinusoidal function
$$\begin{aligned}6\sin(20x)+1&=1\\\ 6\sin(20x)&=0\\\ \sin(20x)&=0\end{aligned}$$
#### Find all values of $$20x$$ between $$-\pi$$ and $$\pi$$
Our argument is $$20x$$.
$$\sin^{-1}(0)=0$$
Now we can use the identity $$\sin(\theta)=\sin(\pi-\theta)$$ to find that the other solutions within the interval are $$-\pi$$ and $$\pi$$. Since they are separated by $$2\pi$$, we can choose either value. Let's use the latter.
#### Finding all solutions
Using the identity $$\sin(\theta)=\sin(\theta+2\pi)$$, we can find all the solutions to our equation from the two angles we found above. The first solution gives us the following.
$$\begin{aligned}20x&=0+n\cdot2\pi \\\x&=\dfrac{n\cdot2\pi}{20}=n\cdot\dfrac{\pi}{10}\end{aligned}$$
Similarly, the second solution gives us the following.
$$\begin{aligned}20x&=\pi+n\cdot2\pi \x&=\dfrac{\pi+n\cdot2\pi}{20}=\dfrac{\pi}{20}+n\cdot\dfrac{\pi}{10}\end{aligned}$$
#### Summary
Our solutions are as follows.
* $$x=n\cdot\dfrac{\pi}{10}$$
* $$x=\dfrac{\pi}{20}+n\cdot\dfrac{\pi}{10}$$
$$\fbox{\large 16cos(8x)−2=6}$$
#### Isolate the sinusoidal function
$$\begin{aligned}16\cos(8x)-2&=6\\\ 16\cos(8x)&=8\\\ \cos(8x)&=0.5\end{aligned}$$
#### Find all values of $$8x$$ between $$-\pi$$ and $$\pi$$
Our argument is $$8x$$.
$$\cos^{-1}(0.5)=\dfrac{\pi}{3}$$
Now we can use the identity $$\cos(\theta)=\cos(-\theta)$$ to find that the second solution within the interval is $$-\dfrac{\pi}{3}$$
#### Finding all solutions
Using the identity $$\cos(\theta)=\cos(\theta+2\pi)$$, we can find all the solutions to our equation from the two angles we found above. The first solution gives us the following.
$$\begin{aligned}8x&=\dfrac{\pi}{3}+n\cdot2\pi \\\x&=\dfrac{\dfrac{\pi}{3}+n\cdot2\pi}{8}=\dfrac{\pi}{24}+n\cdot\dfrac{\pi}{4}\end{aligned}$$
Similarly, the second solution gives us the following.
$$\begin{aligned}8x&=-\dfrac{\pi}{3}+n\cdot2\pi \\\x&=\dfrac{-\dfrac{\pi}{3}+n\cdot2\pi}{8}=-\dfrac{\pi}{24}+n\cdot\dfrac{\pi}{4}\end{aligned}$$
#### Summary
Our solutions are as follows.
* $$x=-\dfrac{\pi}{24}+n\cdot\dfrac{\pi}{4}$$
* $$x=\dfrac{\pi}{24}+n\cdot\dfrac{\pi}{4}$$
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