Inverse Functions — Hard Practice Quiz
A Trigonometry cheat sheet for Inverse Functions — every key formula with its symbols defined — plus a hard-level practice quiz to test recall.
Formulas & key concepts
Inverse Sine (Arcsine)
Inverse Cosine (Arccosine)
Inverse Tangent (Arctangent)
Practice quiz
What is the value of $ \cos(\arctan(\frac{3}{4})) + \sin(\arccos(\frac{5}{13})) $?
- $ \frac{112}{65} $
- $ \frac{11}{13} $
- $ \frac{17}{13} $
- $ \frac{12}{5} $
Answer: $ \frac{112}{65} $
Solve for $ x $: $ \arctan(x) + \arctan(2x) = \frac{\pi}{4} $.
- $ x = \frac{-3 + \sqrt{17}}{4} $
- $ x = \frac{-3 - \sqrt{17}}{4} $
- $ x = \frac{1}{2} $
- $ x = \frac{1}{3} $
Answer: $ x = \frac{-3 + \sqrt{17}}{4} $
What is the domain of the function $ f(x) = \arcsin(\frac{x-1}{x+1}) $?
- $ [0, \infty) $
- $ (-1, \infty) $
- $ (-\infty, -1) \cup [0, \infty) $
- $ [-1, 1] $
Answer: $ [0, \infty) $
Which of the following statements is true regarding the principal values of inverse trigonometric functions?
- The range of $ \arcsin x $ is $ [0, \pi] $.
- The range of $ \arccos x $ is $ [0, \pi] $.
- The range of $ \arctan x $ is $ [-\frac{\pi}{2}, \frac{\pi}{2}] $.
- $ \arcsin x + \arccos x = \frac{\pi}{2} $ for all real $ x $.
Answer: The range of $ \arccos x $ is $ [0, \pi] $
Simplify the expression $ \sin(\arctan x + \arccos y) $.
- $ \frac{xy + \sqrt{1-y^2}}{\sqrt{1+x^2}} $
- $ \frac{x\sqrt{1-y^2} + y}{\sqrt{1+x^2}} $
- $ \frac{xy + \sqrt{1+x^2}\sqrt{1-y^2}}{\sqrt{1+x^2}} $
- $ \frac{x + y\sqrt{1+x^2}}{\sqrt{1+x^2}} $
Answer: $ \frac{xy + \sqrt{1-y^2}}{\sqrt{1+x^2}} $
If $ \arcsin x = \arccos(2x) $, what is the value of $ x $?
- $ \frac{1}{\sqrt{5}} $
- $ -\frac{1}{\sqrt{5}} $
- $ \frac{1}{2} $
- $ \frac{\sqrt{3}}{2} $
Answer: $ \frac{1}{\sqrt{5}} $
Consider the function $ f(x) = \arctan(x^2 - 4x + 5) $. What is the range of $ f(x) $?
- $ [\frac{\pi}{4}, \frac{\pi}{2}) $
- $ (-\frac{\pi}{2}, \frac{\pi}{2}) $
- $ [0, \frac{\pi}{2}) $
- $ [\frac{\pi}{4}, \pi) $
Answer: $ [\frac{\pi}{4}, \frac{\pi}{2}) $
If $ f(x) = \arcsin(\frac{x}{2}) $, what is $ f^{-1}(x) $?
- $ f^{-1}(x) = 2 \sin x $, for $ x \in [-\frac{\pi}{2}, \frac{\pi}{2}] $
- $ f^{-1}(x) = \frac{1}{2} \sin x $, for $ x \in [-1, 1] $
- $ f^{-1}(x) = \sin(2x) $, for $ x \in [-\frac{\pi}{2}, \frac{\pi}{2}] $
- $ f^{-1}(x) = 2 \arcsin x $, for $ x \in [-1, 1] $
Answer: $ f^{-1}(x) = 2 \sin x $, for $ x \in [-\frac{\pi}{2}, \frac{\pi}{2}] $
Evaluate $ \sin(\frac{1}{2} \arccos(\frac{1}{8})) $.
- $ \frac{\sqrt{7}}{4} $
- $ \frac{\sqrt{3}}{2} $
- $ \frac{1}{4} $
- $ \frac{\sqrt{15}}{4} $
Answer: $ \frac{\sqrt{7}}{4} $
If $ \arcsin x + \arcsin y = \frac{\pi}{2} $, and $ x, y \geq 0 $, what is the relationship between $ x $ and $ y $?
- $ x^2 + y^2 = 1 $
- $ x + y = 1 $
- $ x = y $
- $ x^2 - y^2 = 1 $
Answer: $ x^2 + y^2 = 1 $
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