Inverse Trigonometric Functions - AP PreCalculus Study Guide

The Core Idea: Inverse Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent are periodic. This means that for a single output value (a ratio), there are infinitely many possible input values (angles). For example, $sin(x)=1/2$ is true for $x=\pi /6$, $x=5\pi /6$, $x=13\pi /6$, and so on. Because a single output corresponds to multiple inputs, these functions are not one-to-one and do not have a true inverse function over their entire domain.

To overcome this, we create inverse trigonometric functions by restricting the domain of the original function to a specific interval where the function is one-to-one (i.e., it passes the horizontal line test). The inverse trigonometric functions—arcsine, arccosine, and arctangent—are designed to answer the question: "What angle, within a specific restricted interval, produces this given trigonometric ratio?" The output of an inverse trigonometric function is a unique angle, known as the principal value.

Key Definitions and Properties

The three primary inverse trigonometric functions are defined by restricting the domains of their corresponding trigonometric functions. The domain of the original function becomes the range of the inverse, and the range of the original function becomes the domain of theinverse.

Arcsine Function: $arcsin(x)$ or $si{n}^{-1}(x)$

• Inverse of: The sine function, $f(x)=sin(x)$, with its domain restricted to $[-\pi /2,\pi /2]$.

• Meaning:$y=arcsin(x)$ is the unique angle $y$ in the interval $[-\pi /2,\pi /2]$ such that $sin(y)=x$.

• Domain:$[-1,1]$

• Range:$[-\pi /2,\pi /2]$ (Quadrants I and IV)

Arccosine Function: $arccos(x)$ or $co{s}^{-1}(x)$

• Inverse of: The cosine function, $f(x)=cos(x)$, with its domain restricted to $[0,\pi ]$.

• Meaning:$y=arccos(x)$ is the unique angle $y$ in the interval $[0,\pi ]$ such that $cos(y)=x$.

• Domain:$[-1,1]$

• Range:$[0,\pi ]$ (Quadrants I and II)

Arctangent Function: $arctan(x)$ or $ta{n}^{-1}(x)$

• Inverse of: The tangent function, $f(x)=tan(x)$, with its domain restricted to $(-\pi /2,\pi /2)$.

• Meaning:$y=arctan(x)$ is the unique angle $y$ in the interval $(-\pi /2,\pi /2)$ such that $tan(y)=x$.

• Domain:$(-\infty ,\infty )$ or all real numbers.

• Range:$(-\pi /2,\pi /2)$ (Quadrants I and IV)

Understanding Range Restrictions

The specific range of each inverse trigonometric function is critical. These ranges are chosen to ensure that for any valid input, there is exactly one output angle. This makes the inverse a true function.

• For $arcsin(x)$, the range $[-\pi /2,\pi /2]$ covers all possible values of the sine ratio from $-1$ to $1$ exactly once. Positive inputs $x$ yield angles in Quadrant I $(0,\pi /2]$, while negative inputs $x$ yield angles in Quadrant IV $[-\pi /2,0)$.

• For $arccos(x)$, the range $[0,\pi ]$ covers all possible values of the cosine ratio from $-1$ to $1$ exactly once. Positive inputs $x$ yield angles in Quadrant I $[0,\pi /2)$, while negative inputs $x$ yield angles in Quadrant II $(\pi /2,\pi ]$.

• For $arctan(x)$, the range $(-\pi /2,\pi /2)$ covers all possible values of the tangent ratio from $-\infty$ to $\infty$ exactly once. Positive inputs $x$ yield angles in Quadrant I $(0,\pi /2)$, while negative inputs $x$ yield angles in Quadrant IV $(-\pi /2,0)$. Note the use of parentheses, as the range does not include $-\pi /2$ or $\pi /2$ because the tangent function has vertical asymptotes at these values.

Core Concepts & Rules

• To define an inverse for a periodic function, the domain of the original function must be restricted to an interval where it is one-to-one.

• The function $y=arcsin(x)$ finds the unique angle $y$ in $[-\pi /2,\pi /2]$ for which $sin(y)=x$. Its domain is $[-1,1]$.

• The function $y=arccos(x)$ finds the unique angle $y$ in $[0,\pi ]$ for which $cos(y)=x$. Its domain is $[-1,1]$.

• The function $y=arctan(x)$ finds the unique angle $y$ in $(-\pi /2,\pi /2)$ for which $tan(y)=x$. Its domain is $(-\infty ,\infty )$.

• The input to an inverse trigonometric function is a ratio, and the output is an angle (in radians).

• The notation $si{n}^{-1}(x)$ means $arcsin(x)$; it does not mean $1/sin(x)$.

Step-by-Step Example 1: Evaluating an Inverse Trigonometric Function

Problem: Evaluate $arccos(-\surd 3/2)$ without a calculator.

Step 1: Understand the Goal

Let $\theta =arccos(-\surd 3/2)$. We are looking for the angle $\theta$ such that $cos(\theta )=-\surd 3/2$.

Step 2: Recall the Range Restriction

The definition of the arccosine function requires its output, $\theta$, to be in the interval $[0,\pi ]$. This means the angle must be in Quadrant I or Quadrant II.

Step 3: Determine the Quadrant

Since the cosine value $(-\surd 3/2)$ is negative, the angle $\theta$ must lie in Quadrant II.

Step 4: Find the Reference Angle

First, consider the positive value. The angle in Quadrant I whose cosine is $+\surd 3/2$ is $\pi /6$. This is our reference angle.

Step 5: Calculate the Angle in the Correct Quadrant

To find the angle in Quadrant II that has a reference angle of $\pi /6$, we calculate $\pi -(referenceangle)$.

$$\theta =\pi -\frac{\pi }{6}=\frac{6\pi }{6}-\frac{\pi }{6}=\frac{5\pi }{6}$$

Step 6: Final Answer

The angle $5\pi /6$ is in the required range $[0,\pi ]$ and has a cosine of $-\surd 3/2$.

Therefore, $arccos(-\surd 3/2)=5\pi /6$.

Step-by-Step Example 2: Finding an Input for a Function

Problem: A sinusoidal function is modeled by $f(x)=10sin(x)$ on the restricted domain $[-\pi /2,\pi /2]$. Determine the input value $x$ for which $f(x)=-7$.

Step 1: Set up the Equation

We are given $f(x)=-7$, so we can write the equation:

$$10\sin (x)=-7$$

Step 2: Isolate the Trigonometric Expression

Divide both sides by 10 to isolate $sin(x)$:

$$\sin (x)=-\frac{7}{10}=-0.7$$

Step 3: Apply the Inverse Function

To find the value of $x$, we need to "undo" the sine function. We apply the arcsine function to both sides.

$$x=\arcsin (-0.7)$$

Step 4: Verify the Solution is in the Domain

The problem specifies that the domain for $f(x)$ is $[-\pi /2,\pi /2]$. The range of the $arcsin$ function is also $[-\pi /2,\pi /2]$. Since the input to $arcsin$ ($-0.7$) is negative, the output angle will be in the interval $[-\pi /2,0)$, which is fully contained within the function's specified domain. Therefore, the principal value given by the arcsin function is the correct and only solution.

Step 5: Calculate the Final Answer

Since $-0.7$ is not a standard unit circle ratio, we use a calculator to find the approximate value.

$$x\approx -0.7754\text{ radians}$$

Using Your Calculator

A graphing calculator is essential for finding the values of inverse trigonometric functions when the input ratio is not from the unit circle.

Key Steps (TI-84 Style):

1. Set Mode to Radians: This is the most critical step. Press the [MODE] key, navigate down to the RADIAN DEGREE line, and ensure RADIAN is highlighted. Press `[ENTER]$andthen$[2nd][MODE](QUIT) to return to the home screen. 2. **Access Inverse Functions:** The inverse trigonometric functions are typically located as secondary functions above the $sin, $cos$, and $tan$ keys.

   • For $arcsin(x)$ or $si{n}^{-1}(x)$, press [2nd]$then$[sin]. - For $arccos(x) or $co{s}^{-1}(x)$, press [2nd]$then$[cos]. - For $arctan(x) or $ta{n}^{-1}(x)$, press [2nd]$then$[tan]. 3. **Perform the Calculation:** - To solve $x = arcsin(-0.7) from Example 2, you would type: [2nd]$[sin]$$(-)$$0.7$$)$[ENTER]`.

   • The calculator will display approximately $-\mathrm{0.77539...}$.

The calculator will always return the principal value, which is the single angle within the defined range of the inverse function.

AP Exam Quick Hit

Common Question Types

• Direct Evaluation (No Calculator): "What is the exact value of $arctan(-1)$?"

  • This tests your knowledge of both unit circle values and the restricted range of $arctan$. The answer must be $-\pi /4$, which is in the range $(-\pi /2,\pi /2)$. An answer of $3\pi /4$ or $7\pi /4$ would be incorrect.

• Solving for an Input (Calculator): "For the function $h(x)=20cos(x)+5$ on the interval $[0,\pi ]$, find the value of $x$ for which $h(x)=15$."

  • This requires algebraic manipulation ($20cos(x)=10$, so $cos(x)=1/2$) and then applying the inverse function ($x=arccos(1/2)$). While this example has a unit circle value, a similar problem could have $cos(x)=0.6$, requiring a calculator.

• Domain and Range Identification: "State the domain and range of the function $g(x)=arccos(x)$."

  • This is a direct recall of the function's definition. The domain is $[-1,1]$ and the range is $[0,\pi ]$.

Common Mistakes

• Choosing an Angle Outside the Range: For a problem like $arcsin(-1/2)$, a common error is to give the answer $11\pi /6$ or $7\pi /6$. While $sin(11\pi /6)=-1/2$, this angle is not within the required range of $[-\pi /2,\pi /2]$. The only correct answer is $-\pi /6$.

• Confusing Inverse and Reciprocal: Mistaking the notation $si{n}^{-1}(x)$ for the reciprocal, $(sin(x){)}^{-1}$. Remember that $si{n}^{-1}(x)$ is the inverse function $arcsin(x)$, while $(sin(x){)}^{-1}$ is the reciprocal function $csc(x)$. They are completely different.

• Calculator in Degree Mode: Forgetting to set the calculator to Radian mode. If you calculate $arccos(0.5)$ in Degree mode, you will get $60$, but the AP Precalculus exam almost always requires answers in radians. The correct radian answer is $\pi /3\approx 1.047$.

• Domain Violation: Attempting to evaluate an expression like $arcsin(1.5)$. The domain of $arcsin(x)$ and $arccos(x)$ is $[-1,1]$. Any input outside this interval is undefined. Your calculator will return a "DOMAIN ERROR".