Chapter 7 Part 4: Inverse Trigonometric Functions and Equations

Welcome back to Professor Baker's Math Class! Today, we delved into the fascinating world of inverse trigonometric functions and how to use them to solve trigonometric equations. Let's recap the key concepts covered.

Inverse Trigonometric Functions

Remember that inverse trigonometric functions allow us to find the angle that corresponds to a given trigonometric ratio. It's all about reversing the process! Here's a quick review:

  • Inverse Sine Function: $y = \sin^{-1}x$ means that $\sin y = x$ and $y$ is in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
  • Inverse Cosine Function: $y = \cos^{-1}x$ means that $\cos y = x$ and $y$ is in $[0, \pi]$.
  • Inverse Tangent Function: $y = \tan^{-1}x$ means that $\tan y = x$ and $y$ is in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

Understanding the range restrictions is crucial for finding the correct values of inverse trigonometric functions. These restrictions ensure that the inverse functions are well-defined.

Examples of Finding Exact Values

Let's look at an example of finding the exact value using the inverse tangent function. For example, let's find the exact value of $\tan^{-1}(\tan(\frac{5\pi}{6}))$. First, note that $\tan(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{3}$. Therefore, $\tan^{-1}(-\frac{\sqrt{3}}{3}) = -\frac{\pi}{6}$.

Another example shown in the notes is finding the exact value of $\sin^{-1}(\sin(\frac{2\pi}{3}))$. Using the unit circle, we know that $\sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2}$. Therefore, $\sin^{-1}(\frac{\sqrt{3}}{2}) = \frac{\pi}{3}$.

Solving Trigonometric Equations

We also practiced solving trigonometric equations. Remember to use your knowledge of trigonometric identities and the unit circle to find all possible solutions within a given interval.

For example, let's solve the equation $2\sin(\theta) - \sqrt{3} = 0$. First, isolate $\sin(\theta)$ to get $\sin(\theta) = \frac{\sqrt{3}}{2}$. Then, find the angles $\theta$ in the interval $[0, 2\pi)$ where the sine function equals $\frac{\sqrt{3}}{2}$. These angles are $\theta = \frac{\pi}{3}$ and $\theta = \frac{2\pi}{3}$.

Another example in the notes finds all solutions to the equation $\csc(\theta) + 2 = 0$. First, we isolate $\csc(\theta)$ to find that $\csc(\theta) = -2$. Since $\csc(\theta) = \frac{1}{\sin(\theta)}$, we know that $\sin(\theta) = -\frac{1}{2}$. Therefore, $\theta = \frac{7\pi}{6} + 2k\pi$ or $\theta = \frac{11\pi}{6} + 2k\pi$.

Looking Ahead

Keep practicing these concepts to solidify your understanding. Remember, math is like building a house - each concept builds upon the previous one. Keep building, and you'll be amazed at what you can achieve!