Lesson Objectives
• Learn how to work with the composition of trigonometric functions
• Learn how to solve trigonometric equations using inverses
• Learn how to solve inverse trigonometric equations

## How to Solve Trigonometric Equations Involving Inverses

Earlier in our course, we learned about function composition and how to show that two functions are inverses. We know that two functions f(x) and g(x) are inverses if: $$f(g(x))=x$$ and $$g(f(x))=x$$ This tells us that: $$f^{-1}(f(x))=x$$ and $$f(f^{-1}(x))=x$$ As a simple example. $$f(x)=x^3$$ $$f^{-1}(x)=\sqrt[3]{x}$$ $$f^{-1}(f(x))=\sqrt[3]{x^3}=x$$ $$f(f^{-1}(x))=(\sqrt[3]{x})^3=x$$

### Unit Circle

The unit circle will be given here for reference:

### Trigonometric Function Composition

In our lesson on inverse trigonometric functions, we covered trigonometric function composition. In some cases, we need to compose a trigonometric function with its inverse. When this situation occurs, we have to consider the range of our given inverse trigonometric function.
Example #1: Find the exact value of each expression. $$\text{cos}^{-1}\left(\text{cos}\frac{π}{6}\right)$$ In some cases, the cosine function and its inverse will undo each other. We must be careful that the angle given on the inside is within the domain of the restricted cosine function.
Restricted Cosine Function $$y=\text{cos}(x), 0 ≤ x ≤ π$$ Here, our value of $\frac{π}{6}$ falls within our restricted domain. In this case, the answer will just be the angle: $$\require{cancel}\cancel{\text{cos}^{-1}}\left(\cancel{\text{cos}}\frac{π}{6}\right)=\frac{π}{6}$$ If we did this the long way, we would start with the inside operation: $$\text{cos}^{-1}\left(\text{cos}\frac{π}{6}\right)$$ $$=\text{cos}^{-1}\left(\frac{\sqrt{3}}{2}\right)$$ We can finish up the problem by applying the inverse cosine function. $$=\text{cos}^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{π}{6}$$ Here, either method works since the original angle was within the domain of the restricted cosine function or the range of the inverse cosine function. Let's now look at an example that is not as straight forward.
Example #2: Find the exact value of each expression. $$\text{cos}^{-1}\left(\text{cos}\frac{5π}{4}\right)$$ Here, we can see that our angle is not within the domain of the restricted cosine function. When this occurs, we can't simply cancel and take the angle as our answer. Let's work from the inside out. $$\text{cos}^{-1}\left(\text{cos}\frac{5π}{4}\right)$$ $$=\text{cos}^{-1}\left(-\frac{\sqrt{2}}{2}\right)$$ Now, we can use our inverse cosine function. Again, we must think about the domain of the restricted cosine function or the range of the inverse cosine function. This means we will be working with angles in quadrants I and II only. $$=\text{cos}^{-1}\left(-\frac{\sqrt{2}}{2}\right)=\frac{3π}{4}$$ Let's look at an example that can't be found from the unit circle.
Example #3: Find the exact value of each expression. $$\text{sin}^{-1}\left(\text{sin}\left(\frac{6π}{5}\right)\right)$$ Here, we can see that our angle is not within the domain of our restricted sine function.
Restricted Sine Function $$y=\text{sin}(x), -\frac{π}{2}≤ x ≤ \frac{π}{2}$$ Let's work from the inside out. First, let's consider our angle: $$\frac{6π}{5}$$ This angle lies in quadrant III and has a reference angle of: $$\frac{π}{5}$$ Since sine is negative in quadrant III, we want to choose an angle in quadrant IV in the domain of our restricted sine function. Recall that we want a negative angle measure here or a rotation that is clockwise. This gives us an angle of: $$-\frac{π}{5}$$ Since this value is now in the domain of the restricted sine function, we can show our answer as: $$\text{sin}^{-1}\left(\text{sin}\left(\frac{6π}{5}\right)\right)$$ $$=\cancel{\text{sin}^{-1}}\left(\cancel{\text{sin}}\left(-\frac{π}{5}\right)\right)=-\frac{π}{5}$$

### Solving Trigonometric Equations Using Inverse Trigonometric Functions

Let's now use this idea of function composition to solve a few different types of trigonometric equations.
Example #4: Solve each equation for $\left[0, 2π\right)$. $$40 \hspace{.15em}\text{sin}(β - 3) + 1=6$$ Our goal is to isolate the β.
Subtract 1 from each side: $$40 \hspace{.15em}\text{sin}(β - 3)=5$$ Divide each side by 40: $$\text{sin}(β - 3)=\frac{1}{8}$$ Take the inverse sine of each side: $$\text{sin}^{-1}(\text{sin}(β - 3))=\text{sin}^{-1}\left(\frac{1}{8}\right)$$ $$\cancel{\text{sin}^{-1}}(\cancel{\text{sin}}(β - 3))=\text{sin}^{-1}\left(\frac{1}{8}\right)$$ $$β - 3=\text{sin}^{-1}\left(\frac{1}{8}\right)$$ Add 3 to each side: $$β=\text{sin}^{-1}\left(\frac{1}{8}\right) + 3$$ Let's approximate this value: $$β ≈ 3.125$$ Let's now consider the other solution for β.
If we turn to our inverse sine function: $$\text{sin}^{-1}\left(\frac{1}{8}\right) ≈ 0.1253$$ We also have an angle in quadrant II with a sine value of 1/8. This can be found by subtracting $π$ - 0.1253. $$π - 0.1253 ≈ 3.0162$$ If we need to give an exact value, we can show this as: $$π - \text{sin}^{-1}\left(\frac{1}{8}\right)$$ Let's use this to obtain our second solution. $$β=π - \text{sin}^{-1}\left(\frac{1}{8}\right) + 3$$ Let's write our solutions in solution set notation. $$\left\{β=\text{sin}^{-1}\left(\frac{1}{8}\right) + 3, β=π - \text{sin}^{-1}\left(\frac{1}{8}\right) + 3\right\}$$ $$\left\{≈3.1253, ≈ 6.0163\right\}$$ Example #5: Solve each equation for exact solutions. $$\frac{4}{3}\text{cos}^{-1}\left(\frac{x}{4}\right)=π$$ Multiply each side by 3/4: $$\text{cos}^{-1}\left(\frac{x}{4}\right)=\frac{3}{4}π$$ Take the cosine of each side: $$\text{cos}\left(\text{cos}^{-1}\left(\frac{x}{4}\right)\right)=\text{cos}\left(\frac{3}{4}π\right)$$ Simplify: $$\frac{x}{4}=\text{cos}\left(\frac{3}{4}π\right)$$ Multiply both sides by 4: $$x=4\text{cos}\left(\frac{3}{4}π\right)$$ Substitute: $$x=4 \cdot -\frac{\sqrt{2}}{2}$$ Simplify: $$x=-2\sqrt{2}$$ $$\{-2\sqrt{2}\}$$ Example #6: Solve each equation for exact solutions. $$\text{cos}^{-1}(x)=\text{sin}^{-1}\left(\frac{3}{5}\right)$$ Since x is involved with our inverse cosine function, let's begin by taking the cosine of each side: $$\text{cos}(\text{cos}^{-1}(x))=\text{cos}\left(\text{sin}^{-1}\left(\frac{3}{5}\right)\right)$$ $$x=\text{cos}\left(\text{sin}^{-1}\left(\frac{3}{5}\right)\right)$$ Let's make a subsitution here. $$\text{sin}^{-1}\left(\frac{3}{5}\right)=β$$ $$\text{sin}\hspace{.1em}β=\frac{3}{5}$$ $$x=\text{cos}\hspace{.1em}β$$ Since our inverse sine function gives us angles in quadrant I and IV only and the sine value is positive, we know our angle β is in quadrant I. $$\text{sin}\hspace{.15em}β=\frac{y}{r}=\frac{3}{5}$$ $$y=3, r=5, x=?$$ Use the Pythagorean theorem to find the x-value here. Note, this is not the same x from our equation. $$x^2 + y^2=r^2$$ $$x^2 + 9=25$$ $$x^2=16$$ Use the principal square root since β is in quadrant I: $$x=4$$ Let's now return to our equation: $$x=\text{cos}\hspace{.1em}β$$ $$x=\text{cos}\hspace{.1em}β=\frac{x}{r}=\frac{4}{5}$$ $$x=\frac{4}{5}$$ $$\left\{\frac{4}{5}\right\}$$

### Solving an Inverse Trigonometric Equation Using an Identity

Let's look at one final problem that involves the use of an identity.
Example #7: Solve each equation for exact solutions. $$\text{sin}^{-1}(x) - \text{cos}^{-1}(x)=\frac{π}{6}$$ Let's isolate one of our inverse trigonometric functions: $$\text{sin}^{-1}(x)=\text{cos}^{-1}(x) + \frac{π}{6}$$ Take the sine of each side: $$x=\text{sin}\left(\text{cos}^{-1}(x) + \frac{π}{6}\right)$$ Let's make a substitution here: $$\text{cos}^{-1}(x)=β$$ $$x=\text{sin}\left(β + \frac{π}{6}\right)$$ Let's now use our sum identity for sine: $$x=\text{sin}\hspace{.1em}β \hspace{.1em}\text{cos}\frac{π}{6}+ \text{cos}\hspace{.1em}β \hspace{.1em}\text{sin}\hspace{.1em}\frac{π}{6}$$ Simplify: $$x=\text{sin}\hspace{.1em}β \hspace{.1em}\frac{\sqrt{3}}{2}+ \text{cos}\hspace{.1em}β \hspace{.1em}\frac{1}{2}$$ How do we find the sine of β and the cosine of β? Let's return to our substitution from above: $$\text{cos}^{-1}(x)=β$$ The inverse cosine function gives values from quadrant I or quadrant II. $$0 ≤ β ≤ π$$ Now, let's think about the equation: $$\text{sin}^{-1}(x)=\text{cos}^{-1}(x) + \frac{π}{6}$$ Again, let's use our β as a substitution: $$\text{sin}^{-1}(x)=β + \frac{π}{6}$$ The inverse sine function gives values from quadrant I or quadrant IV. $$-\frac{π}{2}≤ β + \frac{π}{6}≤ \frac{π}{2}$$ Let's subtract $\frac{π}{6}$ from each part: $$-\frac{π}{2}- \frac{π}{6}≤ β + \frac{π}{6}- \frac{π}{6}≤ \frac{π}{2}- \frac{π}{6}$$ $$-\frac{2π}{3}≤ β ≤ \frac{π}{3}$$ If we put these two restrictions together: $$0 ≤ β ≤ π$$ and $$-\frac{2π}{3}≤ β ≤ \frac{π}{3}$$ We can see that the intersection gives us: $$0 ≤ β ≤ \frac{π}{3}$$ Now, we know that β is in quadrant I.
Let's create a simple sketch using 1 as the r-value or the hypotenuse. Then we can just use x for the length of x and y for the length of y. Let's put y in terms of x: $$x^2 + y^2=1$$ $$y^2=1 - x^2$$ $$y=\sqrt{1 - x^2}$$ Let's use this to find the sine of β and the cosine of β $$\text{sin}\hspace{.1em}β=\frac{y}{r}=\frac{\sqrt{1 - x^2}}{1}$$ $$\text{cos}\hspace{.1em}β=\frac{x}{r}=\frac{x}{1}$$ Now, let's plug into our equation: $$x=\text{sin}\hspace{.1em}β \hspace{.1em}\frac{\sqrt{3}}{2}+ \text{cos}\hspace{.1em}β \hspace{.1em}\frac{1}{2}$$ $$x=\frac{\sqrt{3}}{2}\cdot \sqrt{1 - x^2}+ \frac{x}{2}$$ Multiply both sides by 2: $$2x=\sqrt{3}\cdot \sqrt{1 - x^2}+ x$$ Subtract x away from each side: $$x=\sqrt{3(1 - x^2)}$$ Square both sides: $$x^2=3(1 - x^2)$$ Simplify: $$x^2=3 - 3x^2$$ Add 3x2 to each side: $$4x^2=3$$ Divide each side by 4: $$x^2=\frac{3}{4}$$ Take the square root of each side. $$x=\pm \sqrt{\frac{3}{4}}$$ Simplify: $$x=\pm \frac{\sqrt{3}}{2}$$ Since we squared both sides, we need to check our solution. Note that taking the negative square root above yields a value that doesn't work in the original equation: $$\text{sin}^{-1}(x) - \text{cos}^{-1}(x)=\frac{π}{6}$$ $$\text{sin}^{-1}\left(\frac{\sqrt{3}}{2}\right) - \text{cos}^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{π}{6}$$ $$\frac{π}{3}- \frac{π}{6}=\frac{π}{6}$$ $$\frac{π}{6}=\frac{π}{6}✓$$ $$\left\{\frac{\sqrt{3}}{2}\right\}$$

#### Skills Check:

Example #1

Find the exact value. $$\text{tan}^{-1}\left(\text{tan}\left(-\frac{5π}{4}\right)\right)$$

A
$$\frac{5π}{4}$$
B
$$-\frac{π}{6}$$
C
$$-\frac{π}{4}$$
D
$$\frac{π}{2}$$
E
$$-\frac{3π}{4}$$

Example #2

Find the exact solution for x. $$\frac{3}{2}\hspace{.1em}\text{cos}^{-1}\left(\frac{x}{4}\right)=π$$

A
$$x=-2$$
B
$$x=\frac{2\sqrt{2}}{3}$$
C
$$x=-2\sqrt{2}$$
D
$$x=-\sqrt{3}$$
E
$$x=-\sqrt{13}$$

Example #3

Find the exact solution for x. $$6\text{cos}^{-1}(x)=5π$$

A
$$x=\frac{\sqrt{3}}{2}$$
B
$$x=\sqrt{5}$$
C
$$x=-\frac{\sqrt{5}}{2}$$
D
$$x=-\frac{π}{2}$$
E
$$x=-\frac{\sqrt{3}}{2}$$

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