1.5 - Introduction to Inverse Trigonometric Functions
Contents
Objectives:
- Identify in which quadrants an angle whose sine/cosine/tangent is given can lie.
- Use inverse trigonometric functions and a calculator to solve for an angle.
- Understand how (and why) to find multiple solutions to problems requiring the use of an inverse trigonometric function.
- Definitions
- arcsine, arccosine, arctangent
Lesson Guide
Finding Angles on the Unit Circle
Briefly introduce students to the idea of inverse trigonometric functions and why we might want them: up until now, our computations have involved a given angle $\theta$, from which we have calculated corresponding trigonometric function values. What if instead, we know the value of a trigonometric function and want to find a corresponding angle $\theta$? The geometry of this question is crucial to understanding how to solve it. For example, if you know $y$, how many angles $\theta$ satisfy $\sin(\theta) = y$ and in which quadrants do they lie?
Have students complete Problem 1.
Problem 1 did not involve actually solving for an angle. Introduce students to the idea of how to solve for a standard angle on the unit circle using an example or two like the following:
-Example 1: Find two distinct angles $\theta$ such that $\sin(\theta)=1/2$.
Note that $\sin(\theta)=1/2$ corresponds to standard angles lying on the unit circle. In this equation, we are looking for angles on the unit circle whose corresponding $y$-value is $1/2$. We use the unit circle to find: $\theta=\pi/6$ or $\theta=5\pi/6$.
Have students complete Problem 2.
Non-Standard Angles on the Unit Circle
Not all angles are standard angles on the unit circle. Use an example to introduce the process of finding $\theta$ when it is a non-standard angle on the unit circle. A sample example is given below:
- Example 2: Find two distinct angles $\theta$ such that $\sin(\theta)=2/3$.
Because $\sin(\theta)$ is positive, $\theta$ may lie in either quadrant I or II. However, no standard angle on the unit circle produces $\sin(\theta)=2/3$. We first need an additional tool - namely, inverse trigonometric functions.
$\arcsin(2/3)=\sin^{-1}(2/3)=\theta$ implies that $\sin(\theta)=2/3$ (note that there are two forms of stating the inverse trig function).
*Note that $\sin^{-1}(\theta)$ is not equal to $\frac{1}{\sin(\theta)}$.
Using a calculator, we compute that $\sin^{-1}(2/3)\approx 0.73=\theta$. (Make sure everyone is in radian mode on their calculators!)
However, as with the first example of $\sin(\theta)=1/2$, we need to find two solutions on the interval $[0,2\pi]$. Unfortunately, the calculator only gives us one solution. How can we find another? Sketch the unit circle and show where the other angle should lie, as shown in the figure below. This agrees with our statement that $\theta$ may lie in either quadrant I or II. We conclude that $\theta=0.73$ and $\theta=\pi-0.73\approx 2.41$ are our two distinct angles within $[0,2\pi]$ satisfying $\sin(\theta)=2/3$. Remind students that they can check their work by evaluating $\sin(0.73)$ and $\sin(2.41)$, and making sure the solutions are close to $2/3$ (remind them that rounding will make the solutions approximate).
Have students complete Problems 3-5. Discuss Problem 5 as a class.
Have students complete the worksheet.
Comments
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