• Find angles using inverse sine, cosine, and tangent functions

Evaluating Angles Using Inverse Trigonometric Functions

Inverse trigonometric functions are essential tools in mathematics for finding angles when the ratios of the sides of right triangles are known. These functions “undo” the calculations made by their respective trigonometric functions, translating a ratio back into an angle measurement. Inverse trigonometric functions allow us to retrieve the angle from a given trigonometric ratio. They provide crucial insights in fields ranging from engineering to physics, where angles need to be determined from known lengths or ratios.

• Inverse Sine (arcsin): If [latex]\sin{(y)}=x[/latex], then [latex]y={\sin}^{-1}{(x)}[/latex].
  • [latex]{\sin}^{-1}{(x)}[/latex] is defined for [latex]x[/latex] in the interval [latex][-1,1][/latex] and returns an angle [latex]y[/latex] in the interval [latex][-\frac{\pi}{2},\frac{\pi}{2}][/latex]
• Inverse Cosine (arccos): If [latex]\cos{(y)}=x[/latex], then [latex]y={\cos}^{-1}{(x)}[/latex].
  • [latex]{\cos}^{-1}{(x)}[/latex] is defined for [latex]x[/latex] in the interval [latex][-1,1][/latex] and returns an angle [latex]y[/latex] in the interval [latex][0,\pi][/latex]
• Inverse Tangent (arctan): If [latex]\tan{(y)}=x[/latex], then [latex]y={\tan}^{-1}{(x)}[/latex].
  • [latex]{\tan}^{-1}{(x)}[/latex] does not have restrictions on [latex]x[/latex] but returns an angle [latex]y[/latex] in the interval [latex][-\frac{\pi}{2},\frac{\pi}{2}][/latex]

inverse sine, cosine, and tangent

For angles in the interval [latex]\left[−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right][/latex], if [latex]\sin{(y)}=x[/latex], then [latex]{\sin}^{-1}{(x)}=y[/latex].

For angles in the interval [latex][0,\pi][/latex], if [latex]\cos{(y)}=x[/latex], then [latex]{\cos}^{-1}{(x)}=y[/latex].

For angles in the interval [latex]\left(−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right)[/latex], if [latex]\tan{(y)}=x[/latex], then [latex]{\tan}^{-1}{(x)}=y[/latex].

Be aware that [latex]{\sin}^{-1}{(x)}[/latex] denotes the inverse sine function, which is not the same as the reciprocal of sine, [latex]\frac{1}{\sin{(x)}}[/latex].

To accurately evaluate inverse trigonometric functions, particularly for special input values, it’s essential to recognize the outputs for standard angles and adjust these for specific cases. This is analogous to the processes used with original trigonometric functions, enhancing the understanding of their inverse counterparts.

With the inverse trigonometric functions, special angles such as [latex]\frac{\pi}{ 6} (30^\circ)\text{, }\frac{\pi}{ 4} (45^\circ),\text{ and } \frac{\pi}{ 3} (60^\circ)[/latex], and their reflections are used to find exact values, mirroring the process used for trigonometric functions.

Standard trigonometric functions—sine, cosine, and tangent—are used to find the ratio of sides in a right triangle given an angle. For instance, [latex]\sin{(x)}[/latex] represents the ratio of the opposite side to the hypotenuse, [latex]\cos{(x)}[/latex] is the adjacent side to the hypotenuse, and [latex]\tan{(x)}[/latex] is the opposite side to the adjacent side.

Angle	[latex]\sin(\theta)[/latex]	[latex]\cos(\theta)[/latex]	[latex]\tan(\theta)[/latex]
[latex]\frac{\pi}{ 6}(30^\circ)[/latex]	[latex]\frac{1}{2}[/latex]	[latex]\frac{\sqrt{3}}{2}[/latex]	[latex]\frac{1}{\sqrt{3}}[/latex]
[latex]\frac{\pi}{ 4}(45^\circ)[/latex]	[latex]\frac{\sqrt{2}}{2}[/latex]	[latex]\frac{\sqrt{2}}{2}[/latex]	[latex]1[/latex]
[latex]\frac{\pi}{ 3}(60^\circ)[/latex]	[latex]\frac{\sqrt{3}}{2}[/latex]	[latex]\frac{1}{2}[/latex]	[latex]\sqrt{3}[/latex]

How To: Evaluate an Inverse Trigonometric Function for Special Input Values

1. Identify the Corresponding Angle: Determine which angle [latex]x[/latex] produces an output equal to the input value for the inverse trigonometric function based on known trigonometric values.
2. Check for Validity: Ensure that the identified [latex]x[/latex] falls within the function’s defined range, and that it appropriately corresponds to the given inverse function (sine, cosine, or tangent).
3. Calculate the Inverse: For valid inputs, use the inverse function to compute the corresponding angle that the trigonometric ratio represents.

Evaluate each of the following.

a. [latex]\sin−1\left(\frac{1}{2}\right)[/latex]

b. [latex]\sin−1\left(−\frac{2}{\sqrt{2}}\right)[/latex]

c. [latex]\cos−1\left(−\frac{3}{\sqrt{2}}\right)[/latex]

d. [latex]\tan^{− 1}(1)[/latex]

Show Solution

a. Evaluating [latex]\sin^{−1}(\frac{1}{2})[/latex] is the same as determining the angle that would have a sine value of [latex]\frac{1}{2}[/latex]. In other words, what angle x would satisfy [latex]\sin(x)=\frac{1}{2}[/latex]? There are multiple values that would satisfy this relationship, such as [latex]\frac{\pi}{6}[/latex] and [latex]\frac{5\pi}{6}[/latex], but we know we need the angle in the interval [latex]\left[−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right][/latex], so the answer will be [latex]\sin^{−1}(\frac{1}{2})=\frac{\pi}{6}[/latex]. Remember that the inverse is a function, so for each input, we will get exactly one output.

b. To evaluate [latex]\sin^{−1}\left(−\frac{\sqrt{2}}{2}\right)[/latex], we know that [latex]\frac{5\pi}{4}[/latex] and [latex]\frac{7\pi}{4}[/latex] both have a sine value of [latex]−\frac{\sqrt{2}}{2}[/latex], but neither is in the interval [latex]\left[−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right][/latex]. For that, we need the negative angle coterminal with [latex]\frac{7\pi}{4}:\sin^{−1}\left(−\frac{\sqrt{2}}{2}\right)=−\frac{\pi}{4}[/latex].

c. To evaluate [latex]\cos^{−1}\left(−\frac{\sqrt{3}}{2}\right)[/latex], we are looking for an angle in the interval [latex][0,\pi][/latex] with a cosine value of [latex]−\frac{\sqrt{3}}{2}[/latex]. The angle that satisfies this is [latex]\cos^{−1}\left(−\frac{\sqrt{3}}{2}\right)=\frac{5\pi}{6}[/latex].

d. Evaluating [latex]\tan^{−1}(1)[/latex], we are looking for an angle in the interval [latex](−\frac{\pi}{2}\text{, }\frac{\pi}{2})[/latex] with a tangent value of 1. The correct angle is [latex]\tan^{−1}(1)=\frac{\pi}{4}[/latex].