Solving Trigonometric Equations Involving Inverses Lesson

Earlier in our course, we learned about composing a function with its inverse.
Let f be a one-to-one function. $$f(f^{-1}(x))=x$$ for every x in the domain of f^-1 $$f^{-1}(f(x))=x$$ for every x in the domain of f
In our lesson on Inverse Trigonometric Functions, we learned about the cancellation properties from composing a trigonometric function and its inverse. Here, we will review this concept as we will use it to solve equations in this lesson.
Recall that the sine function, cosine function, and tangent function are not one-to-one. We need to impose a domain restriction in each case, to create a one-to-one function with the same range.

Restricted Sine Function

$$y=\text{sin}(x), -\frac{π}{2}≤ x ≤ \frac{π}{2}$$ Domain: $$\left[-\frac{π}{2}, \frac{π}{2}\right]$$ Range: $$\left[-1, 1\right]$$

Arcsine Function (Inverse Sine Function)

$$y=\text{sin}^{-1}(x)$$ $$y=\text{arcsin}(x)$$ Domain: $$\left[-1, 1\right]$$ Range: $$\left[-\frac{π}{2}, \frac{π}{2}\right]$$

Sine Function and Arcsine Function

$$\text{sin}(\text{sin}^{-1}(x))=x$$ $$\text{for}$$ $$-1 ≤ x ≤ 1$$ $$\text{sin}^{-1}(\text{sin}(x))=x$$ $$\text{for}$$ $$-\frac{π}{2}≤ x ≤ \frac{π}{2}$$

Restricted Cosine Function

$$y=\text{cos}(x), 0 \le x \le \pi$$ Domain: $$\left[0, \pi\right]$$ Range: $$\left[-1, 1\right]$$

Arccosine Function (Inverse Cosine Function)

$$y=\text{cos}^{-1}(x)$$ $$y=\text{arccos}(x)$$ Domain: $$\left[-1, 1\right]$$ Range: $$\left[0, \pi \right]$$

Cosine Function and Arccosine Function

$$\text{cos}(\text{cos}^{-1}(x))=x$$ $$\text{for}$$ $$-1 ≤ x ≤ 1$$ $$\text{cos}^{-1}(\text{cos}(x))=x$$ $$\text{for}$$ $$0 ≤ x ≤ π$$

Restricted Tangent Function

$$y=\text{tan}(x), -\frac{π}{2}< x < \frac{π}{2}$$ Domain: $$\left(-\frac{π}{2}, \frac{π}{2}\right)$$ Range: $$(-\infty, \infty)$$

Arctangent Function (Inverse Tangent Function)

$$y=\text{tan}^{-1}(x)$$ $$y=\text{arctan}(x)$$ Domain: $$(-\infty, \infty)$$ Range: $$\left(-\frac{π}{2}, \frac{π}{2}\right)$$

Tangent Function and ArcTangent Function

$$\text{tan}(\text{tan}^{-1}(x))=x$$ $$\text{for}$$ $$x ∈ \mathbb{R}$$ $$\text{tan}^{-1}(\text{tan}(x))=x$$ $$\text{for}$$ $$-\frac{π}{2}< x < \frac{π}{2}$$

Unit Circle

The unit circle will be given here for reference:

Trigonometric Function Composition

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 straightforward.
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 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: $$\frac{6π}{5}$$ $$π < \frac{6π}{5}< \frac{3π}{2}$$ 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 trigonometric equations. Note: we solved these earlier in the course using a different approach.
Example #4: Solve each equation for $\left[0, 2π\right)$. $$-5 + \frac{1}{2}\text{tan}\left(-2x + \frac{5π}{4}\right)=-\frac{11}{2}$$ Add 5 to each side: $$\frac{1}{2}\text{tan}\left(-2x + \frac{5π}{4}\right)=-\frac{1}{2}$$ Multiply both sides by 2: $$\text{tan}\left(-2x + \frac{5π}{4}\right)=-1$$ Take the inverse tangent of each side: $$\text{tan}^{-1}\left(\text{tan}\left(-2x + \frac{5π}{4}\right)\right)=\text{tan}^{-1}(-1)$$ Use the cancellation property: $$\cancel{\text{tan}^{-1}}\left(\cancel{\text{tan}}\left(-2x + \frac{5π}{4}\right)\right)=\text{tan}^{-1}(-1)$$ $$-2x + \frac{5π}{4}=\text{tan}^{-1}(-1)$$ $$-2x + \frac{5π}{4}=-\frac{π}{4}$$ Subtract 5$π$/4 away from each side: $$-2x=-\frac{3π}{2}$$ Divide both sides by -2: $$x=\frac{3π}{4}$$ For now, we have only one solution in QII. How do we get our other solutions in the interval from $[0, 2π)$?
Inspecting the equation, we can see the period is $\frac{π}{2}$.
Our solution in QI: $$x=\frac{3π}{4}- \frac{π}{2}=\frac{π}{4}$$ Our solution in QIII: $$x=\frac{3π}{4}+ \frac{π}{2}=\frac{5π}{4}$$ Our solution in QIV: $$x=\frac{5π}{4}+ \frac{π}{2}=\frac{7π}{4}$$ $$\left\{\frac{π}{4}, \frac{3π}{4}, \frac{5π}{4}, \frac{7π}{4}\right\}$$ Example #5: Solve each equation for $\left[0, 2π\right)$. $$40 \hspace{.15em}\text{sin}(β - 3) + 1=6$$ 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: $$\text{sin}^{-1}(\text{sin}(β - 3))=\text{sin}^{-1}\left(\frac{1}{8}\right)$$ Use the cancellation property: $$\cancel{\text{sin}^{-1}}(\cancel{\text{sin}}(β - 3))=\text{sin}^{-1}\left(\frac{1}{8}\right)$$ $$β - 3=\text{sin}^{-1}\left(\frac{1}{8}\right)$$ 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. $$π - 0.1253 ≈ 3.0163$$ 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. $$β=π - \text{sin}^{-1}\left(\frac{1}{8}\right) + 3$$ $$\left\{\text{sin}^{-1}\left(\frac{1}{8}\right) + 3, π - \text{sin}^{-1}\left(\frac{1}{8}\right) + 3\right\}$$ $$\left\{≈3.1253, ≈ 6.0163\right\}$$

Solving an Equation Involving an Inverse Trigonometric Function

Example #6: Solve each equation for exact solutions. $$\frac{3}{4}\text{cos}^{-1}\left(\frac{x}{2}\right)=\frac{π}{4}$$ Multiply each side by 4/3: $$\frac{4}{3}\cdot \frac{3}{4}\text{cos}^{-1}\left(\frac{x}{2}\right)=\frac{4}{3}\cdot \frac{π}{4}$$ $$\text{cos}^{-1}\left(\frac{x}{2}\right)=\frac{π}{3}$$ Take the cosine of each side: $$\text{cos}\left(\text{cos}^{-1}\left(\frac{x}{2}\right)\right)=\text{cos}\left(\frac{π}{3}\right)$$ Use the cancellation property: $$\cancel{\text{cos}}\left(\cancel{\text{cos}^{-1}}\left(\frac{x}{2}\right)\right)=\text{cos}\left(\frac{π}{3}\right)$$ $$\frac{x}{2}=\text{cos}\frac{π}{3}$$ Multiply both sides by 2: $$2 \cdot \frac{x}{2}=2 \cdot \text{cos}\frac{π}{3}$$ $$x=2 \cdot \text{cos}\frac{π}{3}$$ $$x=2 \cdot \frac{1}{2}$$ $$x=1$$ Example #7: Solve each equation for exact solutions. $$\frac{1}{2}\text{cos}^{-1}\left(\frac{x}{4}\right)=π$$ Multiply both sides by 2: $$\text{cos}^{-1}\left(\frac{x}{4}\right)=2π$$ The range of the inverse cosine function is from [0, $π$]. Therefore, there is no solution for this equation. Example #8: Solve each equation for exact solutions. $$\text{cos}^{-1}\left(\frac{3}{5}\right) - \text{sin}^{-1}(x)=0$$ Let's add arcsin(x) to each side: $$\text{cos}^{-1}\left(\frac{3}{5}\right)=\text{sin}^{-1}(x)$$ Take the sine of each side and use the cancellation property: $$\text{sin}\left(\text{cos}^{-1}\left(\frac{3}{5}\right)\right)=\cancel{\text{sin}}\left(\cancel{\text{sin}^{-1}}(x)\right)$$ $$x=\text{sin}\left(\text{cos}^{-1}\left(\frac{3}{5}\right)\right)$$ Let's make a substitution: $$\text{let}\hspace{.2em}θ=\text{cos}^{-1}\left(\frac{3}{5}\right)$$ $$\text{then}\hspace{.2em}\text{cos}\hspace{.1em}θ=\frac{3}{5}$$Our angle θ is in quadrant I since the inverse cosine function returns an angle in quadrant I when the argument is positive. $$x=\text{sin}\hspace{.1em}θ$$ How can we find the sine of θ? We know that the cosine of θ is 3/5 and θ is in quadrant I. $$\text{cos}\hspace{.1em}θ=\frac{x}{r}=\frac{3}{5}$$ $$\text{sin}\hspace{.1em}θ=\frac{y}{r}=\frac{y}{5}$$ This tells us that the adjacent or x is 3 and the hypotenuse or r is 5. To find the opposite or y, let's plug into our Pythagorean Theorem: $$x^2 + y^2=r^2$$ $$3^2 + y^2=5^2$$ $$y^2=16$$ We are in quadrant I, so y-values are positive. $$y=4$$ $$\text{sin}\hspace{.1em}θ=\frac{y}{r}=\frac{4}{5}$$ Return to our problem: $$x=\text{sin}\hspace{.1em}β$$ $$x=\frac{4}{5}$$

Solving an Inverse Trigonometric Equation Using an Identity

Let's look at one final problem that involves the use of an identity.
Example #9: 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: $$\text{cos}^{-1}(x)=β$$ This tells us that β is an angle from [0, $π$] that has a cosine value of x. $$\text{cos}\hspace{.1em}β=x$$ Let's update our equation: $$x=\text{sin}\left(β + \frac{π}{6}\right)$$ 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: $$\text{cos}\hspace{.1em}\frac{π}{6}=\frac{\sqrt{3}}{{2}}$$ $$\text{cos}\hspace{.1em}β=x$$ $$\text{sin}\hspace{.1em}\frac{π}{6}=\frac{1}{2}$$ $$x=\frac{\sqrt{3}}{2}\hspace{.1em}\text{sin}\hspace{.1em}β + \frac{x}{2}$$ Multiply both sides by 2: $$2x=\sqrt{3}\hspace{.1em}\text{sin}\hspace{.1em}β + x$$ Subtract x away from each side: $$x=\sqrt{3}\hspace{.1em}\text{sin}\hspace{.1em}β$$ How do we find the sine of β? Let's return to our substitution from above: $$\text{cos}^{-1}(x)=β$$ The inverse cosine function gives an angle from [0, $π$]. $$0 ≤ β ≤ π$$ Now, let's think about the equation: $$\text{sin}^{-1}(x)=β + \frac{π}{6}$$ The inverse sine function gives an angle from $\left[-\frac{π}{2}, \frac{π}{2}\right]$. $$-\frac{π}{2}≤ β + \frac{π}{6}≤ \frac{π}{2}$$ 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 for the hypotenuse, x for the adjacent, and y for the opposite. Let's use the Pythagorean Formula: $$x^2 + y^2=r^2$$ $$x^2 + y^2=1$$ $$y^2=1 - x^2$$ $$y=\sqrt{1 - x^2}$$ Let's use this to find the sine of β: $$\text{sin}\hspace{.1em}β=\frac{y}{r}=\frac{\sqrt{1 - x^2}}{1}$$ Now, let's plug into our equation: $$x=\sqrt{3}\hspace{.1em}\text{sin}\hspace{.1em}β$$ $$x=\sqrt{3}\hspace{.1em}\sqrt{1 - x^2}$$ Use the rule: $$\sqrt{a}\cdot \sqrt{b}=\sqrt{ab}$$ $$x=\sqrt{3(1 - x^2)}$$ $$x=\sqrt{3 - 3x^2}$$ Square both sides: $$x^2=3 - 3x^2$$ Add 3x² to both sides: $$4x^2=3$$ Divide both sides by 4: $$x^2=\frac{3}{4}$$ Use the square root property: $$x=\pm \sqrt{\frac{3}{4}}$$ Simplify: $$x=\pm \frac{\sqrt{3}}{2}$$ Here, we can consider what we worked with earlier. $$\text{cos}\hspace{.1em}β=x$$ $$0 ≤ β ≤ \frac{π}{3}$$ This tells us that x values are positive (x > 0) and we can throw out the negative solution. It is an extraneous solution that comes from squaring both sides of our equation. $$x=\frac{\sqrt{3}}{2}$$ Let's check: $$\text{sin}^{-1}(x) - \text{cos}^{-1}(x)=\frac{π}{6}$$ $$\text{sin}^{-1}\left(\frac{\sqrt{3}}{2}\right) - \text{cos}^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{π}{6}$$ $$\frac{π}{3}- \frac{π}{6}=\frac{π}{6}$$ $$\frac{π}{6}=\frac{π}{6}✓$$

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