AP Precalculus -3.13 Trigonometry and Polar Coordinates- MCQ Exam Style Questions - Effective Fall 2023
AP Precalculus -3.13 Trigonometry and Polar Coordinates- MCQ Exam Style Questions – Effective Fall 2023
AP Precalculus -3.13 Trigonometry and Polar Coordinates- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.
▶️ Answer/Explanation
The polar form of a complex number is \( r(\cos \theta + i \sin \theta) \). For the point \((3,3)\):
\( r = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2} \)
\( \theta = \arctan\left(\frac{3}{3}\right) = \arctan(1) = \frac{\pi}{4} \)
Thus, the polar form is \( 3\sqrt{2}\left( \cos\frac{\pi}{4} + i \sin\frac{\pi}{4} \right) \), which matches option (A).
✅ Answer: (A)
▶️ Answer/Explanation
The four points are at \( (5,0), (0,5), (-5,0), (0,-5) \) in rectangular coordinates, corresponding to angles \( 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2} \).
In polar form \( r = 5 \), the angles are spaced \( \frac{\pi}{2} \) apart starting at \( 0 \). Option (C) starts at \( \frac{\pi}{2} \) and adds \( \frac{\pi}{2}k \) for \( k = 0,1,2,3 \), yielding \( \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi \) — but the given list starts from \( k=1 \) to \( k=4 \), which would give \( \pi, \frac{3\pi}{2}, 2\pi, \frac{5\pi}{2} \) (mod \( 2\pi \)), equivalent to \( \pi, \frac{3\pi}{2}, 0, \frac{\pi}{2} \) — the same four angles.
Thus (C) correctly generates the set of angles for the four points.
✅ Answer: (C)
▶️ Answer/Explanation
From the graph, \( P \) is in Quadrant II with rectangular coordinates \( (-2, 2) \).
– Distance from origin: \( r = \sqrt{(-2)^2 + 2^2} = \sqrt{8} = 2\sqrt{2} \)
– Reference angle: \( \arctan\left(\frac{2}{|-2|}\right) = \arctan(1) = \frac{\pi}{4} \)
In Quadrant II, \( \theta = \pi – \frac{\pi}{4} = \frac{3\pi}{4} \).
Thus polar coordinates: \( (2\sqrt{2}, \frac{3\pi}{4}) \).
✅ Answer: (D)
▶️ Answer/Explanation
For a point in polar coordinates \( (r, \theta) \), the same location can be represented as \( (-r, \theta + \pi) \) (or adding odd multiples of \( \pi \) to \( \theta \) while negating \( r \)).
Here, \( r = 5, \theta = \frac{\pi}{4} \).
Using \( \theta + \pi = \frac{\pi}{4} + \pi = \frac{5\pi}{4} \), we get \( (-5, \frac{5\pi}{4}) \).
✅ Answer: (B)
▶️ Answer/Explanation
From the diagram: \( P \) is at \( (4, \frac{7\pi}{6}) \).
Clockwise rotation by \( \frac{\pi}{2} \) subtracts from the angle: \( \theta_K = \frac{7\pi}{6} – \frac{\pi}{2} = \frac{7\pi}{6} – \frac{3\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3} \).
Polar coordinates of \( K \): \( (4, \frac{2\pi}{3}) \).
Convert to rectangular:
\( x = 4\cos\frac{2\pi}{3} = 4 \cdot \left(-\frac{1}{2}\right) = -2 \)
\( y = 4\sin\frac{2\pi}{3} = 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3} \)
Thus \( K = (-2, 2\sqrt{3}) \).
✅ Answer: (A)
▶️ Answer/Explanation
For \( z = 10 + 10i \), we have \( x = 10, y = 10 \).
\( r = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2} \)
\( \theta = \arctan\left(\frac{10}{10}\right) = \arctan(1) = \frac{\pi}{4} \) (Quadrant I)
Thus polar form: \( r(\cos\theta + i\sin\theta) = 10\sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right) \), which matches option (D).
✅ Answer: (D)
▶️ Answer/Explanation
Convert from polar \((r,\theta)\) to rectangular \((x,y)\) using:
\( x = r \cos \theta \), \( y = r \sin \theta \).
Here \( r = 10 \) and \( \theta = \frac{5\pi}{6} \).
\( \cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2} \), \( \sin \frac{5\pi}{6} = \frac{1}{2} \).
Thus \( x = 10 \cdot \left(-\frac{\sqrt{3}}{2}\right) = -5\sqrt{3} \),
\( y = 10 \cdot \frac{1}{2} = 5 \).
✅ Answer: (A)
▶️ Answer/Explanation
1. Convert to Rectangular:
\(x = r\cos\theta = 1 \cdot \cos(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2}\)
\(y = r\sin\theta = 1 \cdot \sin(\frac{5\pi}{6}) = \frac{1}{2}\)
2. Match Coordinate Pair:
\((-\frac{\sqrt{3}}{2}, \frac{1}{2})\)
(Note: Option A matches the calculation in the provided answer key).
✅ Answer: (A)
▶️ Answer/Explanation
1. Analyze Point D:
Point D has coordinates \((-2, \frac{7\pi}{4})\).
First, locate the angle \(\theta = \frac{7\pi}{4}\). This angle is in Quadrant IV (between \(\frac{3\pi}{2}\) and \(2\pi\)).
2. Apply the Negative Radius:
A negative radius (\(r = -2\)) means we reflect the point through the origin to the opposite quadrant.
The quadrant opposite to Quadrant IV is Quadrant II.
3. Evaluate Other Points:
A: \((2, -\frac{\pi}{4})\) is in QIV.
B: \((2, -\frac{3\pi}{4})\) is in QIII.
C: \((-2, \frac{5\pi}{4})\) \(\to\) Angle is QIII \(\to\) Reflect to QI.
✅ Answer: (D)
▶️ Answer/Explanation
Rectangular coordinates \((3,3)\) correspond to complex number \( z = 3 + 3i \).
Polar coordinates:
\[ r = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2} \]
\[ \theta = \arctan\left(\frac{3}{3}\right) = \arctan(1) = \frac{\pi}{4} \quad (\text{in first quadrant}) \]
Polar form: \( z = r(\cos \theta + i\sin \theta) = 3\sqrt{2} \cos\left(\frac{\pi}{4}\right) + i \cdot 3\sqrt{2} \sin\left(\frac{\pi}{4}\right) \).
This matches option (A).
✅ Answer: (A)
▶️ Answer/Explanation
First, calculate the radius $r$ using $r = \sqrt{x^2 + y^2} = \sqrt{(-\frac{3\sqrt{3}}{2})^2 + (\frac{3}{2})^2} = 3$.
The point is in Quadrant II since $x < 0$ and $y > 0$.
Find the standard angle: $\tan \theta = \frac{y}{x} = \frac{3/2}{-3\sqrt{3}/2} = -\frac{1}{\sqrt{3}}$, which gives $\theta = \frac{5\pi}{6}$.
The problem requires $r < 0$, so we use $r = -3$.
To keep the same location with a negative radius, add $\pi$ to the angle: $\frac{5\pi}{6} + \pi = \frac{11\pi}{6}$.
The resulting polar coordinate is $\left( -3, \frac{11\pi}{6} \right)$.
Therefore, the correct option is b.
▶️ Answer/Explanation
The correct option is (A).
To convert rectangular coordinates \( (x, y) = (3, 3) \) to polar form, we must find the modulus \( r \) and the argument \( \theta \).
1. Calculate the modulus \( r \) using the formula \( r = \sqrt{x^2 + y^2} \).
2. Substitute the given values: \( r = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} \).
3. Simplify the radical: \( \sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2} \).
4. Calculate the angle \( \theta \) using \( \tan(\theta) = \frac{y}{x} \).
5. Substitute values: \( \tan(\theta) = \frac{3}{3} = 1 \). Since the point is in the first quadrant, \( \theta = \frac{\pi}{4} \).
6. The polar form of a complex number is \( z = r \cos(\theta) + i (r \sin(\theta)) \).
7. Substituting \( r = 3\sqrt{2} \) and \( \theta = \frac{\pi}{4} \), we get: \( \left(3\sqrt{2} \cos \left(\frac{\pi}{4}\right)\right) + i \left(3\sqrt{2} \sin \left(\frac{\pi}{4}\right)\right) \).
▶️ Answer/Explanation
To convert from polar coordinates \((r, \theta)\) to rectangular coordinates \((x, y)\), we use the formulas \(x = r \cos \theta\) and \(y = r \sin \theta\).
Given the point \(\left(1, \frac{5\pi}{6}\right)\), substitute \(r = 1\) and \(\theta = \frac{5\pi}{6}\) into the equations.
First, find the \(x\)-coordinate: \(x = 1 \cdot \cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}\).
Next, find the \(y\)-coordinate: \(y = 1 \cdot \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}\).
Combining these values gives the rectangular coordinates \(\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\).
Therefore, the correct option corresponds to (A).
▶️ Answer/Explanation
The correct option is (B).
The rectangular coordinates are given as \((x, y) = (1, -\sqrt{3})\).
First, calculate the modulus \(r\) using the formula \(r = \sqrt{x^2 + y^2}\).
Substituting the values: \(r = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2\).
Next, determine the argument \(\theta\) using \(\tan(\theta) = \frac{y}{x} = \frac{-\sqrt{3}}{1} = -\sqrt{3}\).
Since \(x\) is positive and \(y\) is negative, the point lies in the fourth quadrant.
The reference angle is \(\frac{\pi}{3}\), so the angle in the fourth quadrant corresponds to \(\theta = -\frac{\pi}{3}\).
Thus, the polar form is \(r(\cos \theta + i \sin \theta) = 2\left(\cos\left(-\frac{\pi}{3}\right) + i \sin\left(-\frac{\pi}{3}\right)\right)\).
▶️ Answer/Explanation
To find the radius, we must convert the given general equation to the standard form \((x-h)^2 + (y-k)^2 = r^2\) by completing the square.
Group the \(x\) and \(y\) terms: \((x^2 – 6x) + (y^2 + 8y) = 56\).
Add \((\frac{-6}{2})^2 = 9\) and \((\frac{8}{2})^2 = 16\) to both sides to complete the squares: \((x^2 – 6x + 9) + (y^2 + 8y + 16) = 56 + 9 + 16\).
Factor the perfect square trinomials: \((x – 3)^2 + (y + 4)^2 = 81\).
Identify \(r^2\) from the right side of the equation: \(r^2 = 81\).
Take the square root to find the radius: \(r = \sqrt{81} = 9\).
Therefore, the radius of the circle is \(9\), which corresponds to option (B).
▶️ Answer/Explanation
The relationship between arc length $s$, radius $r$, and angle $\theta$ (in radians) is given by $s = r\theta$.
In this problem, the arc length $s$ is given as $40$.
The angle $\theta$ is given as $\frac{\pi}{4}$ radians.
Substitute the values into the formula: $40 = r \cdot \frac{\pi}{4}$.
Solve for $r$ by multiplying both sides by $4$: $160 = r\pi$.
Divide by $\pi$ to isolate the radius: $r = \frac{160}{\pi}$.
The correct option is (D).
▶️ Answer/Explanation
The radius of the circle is $r = 5$, as seen from point $P(5,0)$.
The arc length $s$ is given by the formula $s = r\theta$.
Substituting the given values, $6 = 5\theta$, which means $\theta = \frac{6}{5}$ radians.
The coordinates of point $Q$ on a circle are $(r \cos\theta, r \sin\theta)$.
The distance of a point from the $y$-axis is the absolute value of its $x$-coordinate.
Therefore, the distance is $|5 \cos(\frac{6}{5})|$.
Since $\frac{6}{5}$ radians is in the first quadrant, the distance is $5 \cos\left(\frac{6}{5}\right)$.
Correct Option: (C)
▶️ Answer/Explanation
The formula for the arc length $s$ is given by $s = r\theta$, where $\theta$ is in radians.
The radius of the circle is given as $r = 3$.
The central angle subtending the arc is given as $\theta = 0.687\pi$.
Substitute the values into the formula: $s = 3 \times (0.687\pi)$.
Calculate the product: $s = 2.061\pi$.
Using the approximation $\pi \approx 3.14159$, the length is $s \approx 6.4748…$
Rounding to three decimal places, the arc length is $6.475$.
The correct option is (C).