Inverse Trig Functions
Inverse Trig Functions
Inverse Functions: A Review
Definition: Two functions f(x) and g(x) are inverses if f(g(x)) = x and g(f(x)) = x.
Example 1: If f(x) = x^3, then f^{-1}(x) = \sqrt[3]{x}.
- f(f^{-1}(x)) = f(\sqrt[3]{x}) = (\sqrt[3]{x})^3 = x
- f^{-1}(f(x)) = f^{-1}(x^3) = \sqrt[3]{x^3} = x
Example 2: If g(x) = \frac{1}{x - 3}, then g^{-1}(x) = \frac{1}{x} + 3.
- g(g^{-1}(x)) = g(\frac{1}{x} + 3) = \frac{1}{(\frac{1}{x} + 3) - 3} = \frac{1}{\frac{1}{x}} = x
- g^{-1}(g(x)) = g^{-1}(\frac{1}{x - 3}) = \frac{1}{\frac{1}{x - 3}} + 3 = (x - 3) + 3 = x
Geometric Interpretation of Inverse Functions
Horizontal Line Test: A function has an inverse if and only if every horizontal line intersects its graph at most once.
Sketching the Inverse: To sketch the inverse of a function:
- Draw the line y = x.
- Reflect the graph of the original function across the line y = x.
Inverse Trigonometric Functions
Sine Function
sin(x) does not pass the horizontal line test over its entire domain.
Restricting the Domain: To define an inverse, restrict the domain of sin(x) to [-\frac{\pi}{2}, \frac{\pi}{2}].
Inverse Sine Function: Denoted as sin^{-1}(x) or arcsin(x).
Analogy: Similar to how x^2 has an inverse \sqrt{x} only when the domain of x^2 is restricted to [0, \infty).
* The inverse is created by drawing the line y=x and rotating the curve around the line. The solid blue curve is the \sqrt{x}.
Graph of Sine Inverse
Passes through points (0, 0), (-1, -\frac{\pi}{2}) and (1, \frac{\pi}{2})
Obtained by restricting the original sine curve between [-\frac{\pi}{2}, \frac{\pi}{2}]
The y values for the original sine function in that range went from [-1, 1]
Domain and range get switched compared to the original function.
Properties of sin^{-1}(x)
Domain: [-1, 1]
Range: [-\frac{\pi}{2}, \frac{\pi}{2}]
Symmetry: Odd function.
- sin^{-1}(-x) = -sin^{-1}(x)
Cosine Function
cos(x) also fails the horizontal line test over its entire domain.
Restricting the Domain: Restrict the domain of cos(x) to [0, \pi].
On that region the function passes the horizontal line test, therefore an inverse function can be constructed by rotating it around the line y=x.
- Remember that it's tough to visualize it, but if you tilt your head 45 degrees, you would see that the red curve and the black curve are symmetric around the blue dotted line.
Properties of cos^{-1}(x)
Domain: [-1, 1]
Range: [0, \pi]
Symmetry: Neither even nor odd.
Tangent Function
tan(x) fails the horizontal line test over its entire domain.
Restricting the Domain: Restrict the domain of tan(x) to (-\frac{\pi}{2}, \frac{\pi}{2}).
When you rotate the curve you have to consider that asymptotes are also properties of the function.
- Every property of x gets switched with every property of y.
- A point on the x axis would go to a point on the y axis. A vertical asymptote would go to a horizontal asymptote, etcetera.
Graph of Tan Inverse
- Curve goes through zero and tapers off a horizontal asymptotes at \frac{\pi}{2} and -\frac{\pi}{2}
Properties of tan^{-1}(x)
Domain: (-\infty, \infty)
Range: (-\frac{\pi}{2}, \frac{\pi}{2})
Symmetry: Odd function.
- tan^{-1}(-x) = -tan^{-1}(x)
Other Inverse Trig Functions
- sec^{-1}(x), csc^{-1}(x), and cot^{-1}(x) exist, but are less commonly used.
Evaluating Inverse Trig Functions
- The trig functions in calc one, should be well known for calc two.
- The tan^{-1}(x) function seems to pop up a little bit more than the other ones, a little more than sin^{-1}(x) or cos^{-1}(x).
General Approach
Check the Domain: Ensure the input value is within the domain of the inverse trig function.
Set Equal to \theta: Let the inverse trig function equal an angle \theta.
Apply Trig Function: Apply the corresponding trigonometric function to both sides.
Solve for \theta: Determine the angle \theta that satisfies the equation, considering the range of the inverse trig function.
Example 1: Evaluate sin^{-1}(\frac{1}{2})
\frac{1}{2} is in the domain of sin^{-1}(x) which is [-1, 1].
Let sin^{-1}(\frac{1}{2}) = \theta.
sin(sin^{-1}(\frac{1}{2})) = sin(\theta)
\frac{1}{2} = sin(\theta). What angle, \theta, has a sine of \frac{1}{2}?
- \theta = \frac{\pi}{6}
Example 2: Evaluate cos^{-1}(\frac{\sqrt{2}}{2})
\frac{\sqrt{2}}{2} is in the domain of cos^{-1}(x) which is [-1, 1].
What angle, \theta, has a cosine of \frac{\sqrt{2}}{2}?
- \theta = \frac{\pi}{4}
Example 3: Evaluate tan^{-1}(\sqrt{3})
\sqrt{3} is in the domain of tan^{-1}(x) which is (-\infty, \infty).
What angle, \theta, has a tangent of \sqrt{3}?
- \theta = \frac{\pi}{3}
Example 4: Evaluate cos^{-1}(\sqrt{3})
\sqrt{3} is not in the domain of cos^{-1}(x) which is [-1, 1].
Therefore, cos^{-1}(\sqrt{3}) does not exist.
Example 5: Evaluate cos^{-1}(\frac{\sqrt{3}}{2})
\frac{\sqrt{3}}{2} is in the domain of cos^{-1}(x) which is [-1, 1].
What angle, \theta, has a cosine of \frac{\sqrt{3}}{2}?
- \theta = \frac{\pi}{6}
Example 6: Evaluate sin^{-1}(-1)
-1 is in the domain of sin^{-1}(x) which is [-1, 1].
What angle, \theta, has a sine of -1?
- Range of sin^{-1}(x) is [-\frac{\pi}{2}, \frac{\pi}{2}].
- \theta = -\frac{\pi}{2}
Example 7: Evaluate sin^{-1}(-\frac{\sqrt{3}}{2})
-\frac{\sqrt{3}}{2} is in the domain of sin^{-1}(x) which is [-1, 1].
What angle, \theta, has a sine of -\frac{\sqrt{3}}{2}?
- Range of sin^{-1}(x) is [-\frac{\pi}{2}, \frac{\pi}{2}].
- \theta = -\frac{\pi}{3}
Example 8: Evaluate cos^{-1}(-\frac{\sqrt{3}}{2})
-\frac{\sqrt{3}}{2} is in the domain of cos^{-1}(x) which is [-1, 1].
What angle, \theta, has a cosine of -\frac{\sqrt{3}}{2}?
- Range of cos^{-1}(x) is [0, \pi].
- \theta = \frac{5\pi}{6}
Example 9: Evaluate tan^{-1}(-1)
-1 is in the domain of tan^{-1}(x) which is (-\infty, \infty).
What angle, \theta, has a tangent of -1?
- Range of tan^{-1}(x) is (-\frac{\pi}{2}, \frac{\pi}{2}).
- \theta = -\frac{\pi}{4}