AP Precalculus: Inverse Trigonometric Functions (Drill 1)

Question 1. What is the value of \( \arcsin\left(\dfrac{\sqrt{2}}{2}\right) \)?

• A) \( \dfrac{\pi}{6} \)
• B) \( \dfrac{\pi}{3} \)
• C) \( \dfrac{\pi}{4} \) ✓
• D) \( \dfrac{3\pi}{4} \)

Explanation: Choice C is correct. \( \arcsin\left(\frac{\sqrt{2}}{2}\right) \) asks: what angle in \( \left[-\frac{\pi}{2},\, \frac{\pi}{2}\right] \) has a sine value of \( \frac{\sqrt{2}}{2} \)? Since \( \sin\!\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \) and \( \frac{\pi}{4} \) is in \( \left[-\frac{\pi}{2},\, \frac{\pi}{2}\right] \), the answer is \( \frac{\pi}{4} \). Choice A is incorrect because \( \sin\!\left(\frac{\pi}{6}\right) = \frac{1}{2} \), not \( \frac{\sqrt{2}}{2} \). Choice B is incorrect because \( \sin\!\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \), not \( \frac{\sqrt{2}}{2} \). Choice D is incorrect because although \( \sin\!\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2} \) is true, the value \( \frac{3\pi}{4} \) lies outside the restricted range \( \left[-\frac{\pi}{2},\, \frac{\pi}{2}\right] \) of the arcsine function; arcsin always returns values in that interval.

Question 2. What is the range of \( f(x) = \arctan(x) \)?

• A) \( [-1,\, 1] \)
• B) \( [0,\, \pi] \)
• C) \( \left[-\dfrac{\pi}{2},\, \dfrac{\pi}{2}\right] \)
• D) \( \left(-\dfrac{\pi}{2},\, \dfrac{\pi}{2}\right) \) ✓

Explanation: Choice D is correct. The range of \( \arctan(x) \) is the open interval \( \left(-\frac{\pi}{2},\, \frac{\pi}{2}\right) \). As \( x \to +\infty \), \( \arctan(x) \to \frac{\pi}{2} \) from below, and as \( x \to -\infty \), \( \arctan(x) \to -\frac{\pi}{2} \) from above, but neither endpoint is ever attained. Choice A is incorrect because \( [-1,\, 1] \) is the range of \( \sin(x) \) and \( \cos(x) \); it describes output values of those functions, not of arctan. Choice B is incorrect because \( [0,\, \pi] \) is the range of \( \arccos(x) \), not \( \arctan(x) \). Choice C is incorrect because the closed interval \( \left[-\frac{\pi}{2},\, \frac{\pi}{2}\right] \) is the range of \( \arcsin(x) \); for arctan, the endpoints \( \pm\frac{\pi}{2} \) are horizontal asymptotes that are approached but never reached, so open brackets are required.

Question 3. What is the value of \( \cos\!\left(\arcsin\!\left(\dfrac{3}{5}\right)\right) \)?

• A) \( \dfrac{3}{4} \)
• B) \( \dfrac{3}{5} \)
• C) \( \dfrac{4}{5} \) ✓
• D) \( \dfrac{5}{3} \)

Explanation: Choice C is correct. Let \( \theta = \arcsin\!\left(\frac{3}{5}\right) \), so \( \sin(\theta) = \frac{3}{5} \) and \( \theta \in \left[-\frac{\pi}{2},\, \frac{\pi}{2}\right] \). Applying the Pythagorean identity: \( \cos^2(\theta) = 1 - \sin^2(\theta) = 1 - \frac{9}{25} = \frac{16}{25} \), so \( \cos(\theta) = \frac{4}{5} \). The positive root is taken because \( \theta \in \left[-\frac{\pi}{2},\, \frac{\pi}{2}\right] \) implies \( \cos(\theta) \geq 0 \). Choice A is incorrect because \( \frac{3}{4} \) is the ratio of the sine value to the cosine value, that is, \( \tan(\theta) \), not \( \cos(\theta) \). Choice B is incorrect because \( \frac{3}{5} \) is \( \sin(\theta) \); cosine has a different value. Choice D is incorrect because \( \frac{5}{3} \) is the reciprocal of \( \sin(\theta) \), which equals \( \csc(\theta) \), not \( \cos(\theta) \).

Question 4. What is the domain of \( f(x) = \arccos(x) \)?

• A) \( (-\infty,\, \infty) \)
• B) \( \left[-\dfrac{\pi}{2},\, \dfrac{\pi}{2}\right] \)
• C) \( [0,\, \pi] \)
• D) \( [-1,\, 1] \) ✓

Explanation: Choice D is correct. The function \( \arccos(x) \) is the inverse of cosine restricted to \( [0,\, \pi] \). Its domain consists of all values that cosine can produce: since \( \cos(\theta) \in [-1,\, 1] \) for all \( \theta \), the domain of \( \arccos \) is \( [-1,\, 1] \). For example, \( \arccos(2) \) has no real value. Choice A is incorrect because \( \arccos(x) \) is undefined for \( |x| > 1 \); the domain is not all real numbers. Choice B is incorrect because \( \left[-\frac{\pi}{2},\, \frac{\pi}{2}\right] \) is the range of \( \arcsin(x) \) and the restricted domain of sine, not the domain of arccosine. Choice C is incorrect because \( [0,\, \pi] \) is the range of \( \arccos(x) \), not its domain, a classic domain/range swap.

Question 5. A drone flying horizontally is 500 meters from an observer and at a height of h meters. The angle of elevation from the observer to the drone is given by \( \theta = \arctan\!\left(\dfrac{h}{500}\right) \). If the angle of elevation is \( \dfrac{\pi}{6} \), what is the height of the drone?

• A) \( \dfrac{500\sqrt{3}}{3} \) meters ✓
• B) \( 500\sqrt{3} \) meters
• C) 250 meters
• D) \( \dfrac{500}{6} \) meters

Explanation: Choice A is correct. Since arctan is the inverse of tan, applying tan to both sides of \( \frac{\pi}{6} = \arctan\!\left(\frac{h}{500}\right) \) gives \( \tan\!\left(\frac{\pi}{6}\right) = \frac{h}{500} \). Since \( \tan\!\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \), we get \( h = \frac{500}{\sqrt{3}} = \frac{500\sqrt{3}}{3} \) meters. Choice B is incorrect because \( 500\sqrt{3} \) uses \( \tan\!\left(\frac{\pi}{3}\right) = \sqrt{3} \) rather than \( \tan\!\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \), a common confusion between the 30° and 60° values. Choice C is incorrect because 250 would result from treating the angle as a ratio of \( \frac{1}{2} \) without applying the tangent function. Choice D is incorrect because \( \frac{500}{6} \) substitutes the angle value \( \frac{\pi}{6} \) directly as a plain fraction rather than computing \( \tan\!\left(\frac{\pi}{6}\right) \).