Precalculus Quiz: Symmetry And Periodicity Of Trigonometric Functions
20 questions · exam conditions
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Symmetry And Periodicity Of Trigonometric FunctionsQuestion 1 of 20

If f(x)=sin⁡(x)+cos⁡(x)f(x) = \sin(x) + \cos(x) and g(x)=sin⁡(x)−cos⁡(x)g(x) = \sin(x) - \cos(x), which statement about the symmetry properties of these functions is correct?

Both f(x)f(x) and g(x)g(x) are even functions
f(x)f(x) is neither even nor odd, while g(x)g(x) is an odd function
f(x)f(x) is an even function, while g(x)g(x) is neither even nor odd
Both f(x)f(x) and g(x)g(x) are neither even nor odd functions
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Precalculus Quiz

Precalculus Quiz: Symmetry And Periodicity Of Trigonometric Functions

Practice Symmetry And Periodicity Of Trigonometric Functions in Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Symmetry And Periodicity Of Trigonometric Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for Precalculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

If f(x)=sin⁡(x)+cos⁡(x)f(x) = \sin(x) + \cos(x) and g(x)=sin⁡(x)−cos⁡(x)g(x) = \sin(x) - \cos(x), which statement about the symmetry properties of these functions is correct?

  1. Both f(x)f(x) and g(x)g(x) are even functions
  2. f(x)f(x) is neither even nor odd, while g(x)g(x) is an odd function
  3. f(x)f(x) is an even function, while g(x)g(x) is neither even nor odd
  4. Both f(x)f(x) and g(x)g(x) are neither even nor odd functions (correct answer)
Explanation: To determine symmetry, we test if f(−x)=f(x)f(-x) = f(x) (even), f(−x)=−f(x)f(-x) = -f(x) (odd), or neither. For f(x)=sin⁡(x)+cos⁡(x)f(x) = \sin(x) + \cos(x): f(−x)=sin⁡(−x)+cos⁡(−x)=−sin⁡(x)+cos⁡(x)f(-x) = \sin(-x) + \cos(-x) = -\sin(x) + \cos(x). This equals neither f(x)f(x) nor −f(x)-f(x). For g(x)=sin⁡(x)−cos⁡(x)g(x) = \sin(x) - \cos(x): g(−x)=sin⁡(−x)−cos⁡(−x)=−sin⁡(x)−cos⁡(x)g(-x) = \sin(-x) - \cos(-x) = -\sin(x) - \cos(x). This equals neither g(x)g(x) nor −g(x)-g(x). Therefore, both functions are neither even nor odd.

Question 2

A function g(x)g(x) satisfies g(x+4π)=g(x)g(x + 4\pi) = g(x) for all xx, and g(x)=sin⁡(kx)g(x) = \sin(kx) for some constant k>0k > 0. What is the smallest possible positive value of kk?

  1. 14\frac{1}{4}
  2. 12\frac{1}{2} (correct answer)
  3. 11
  4. 22
Explanation: For g(x)=sin⁡(kx)g(x) = \sin(kx) to have period 4π4\pi, we need 2πk=4π\frac{2\pi}{k} = 4\pi. Solving: 2π=4πk2\pi = 4\pi k, so k=12k = \frac{1}{2}. We can verify: g(x+4π)=sin⁡(12(x+4π))=sin⁡(x2+2π)=sin⁡(x2)=g(x)g(x + 4\pi) = \sin(\frac{1}{2}(x + 4\pi)) = \sin(\frac{x}{2} + 2\pi) = \sin(\frac{x}{2}) = g(x). The other values of kk would give different periods: k=14k = \frac{1}{4} gives period 8π8\pi, k=1k = 1 gives period 2π2\pi, and k=2k = 2 gives period π\pi.

Question 3

Based on the unit circle, the point at angle θ\theta is (cos⁡(θ),sin⁡(θ))(\cos(\theta),\sin(\theta)) and the point at angle −θ-\theta reflects across the xx-axis. If sin⁡(π6)=12\sin\left(\frac{\pi}{6}\right)=\frac{1}{2}, what is sin⁡(−π6)\sin\left(-\frac{\pi}{6}\right)?

  1. 12\frac{1}{2}
  2. −12-\frac{1}{2} (correct answer)
  3. 32\frac{\sqrt{3}}{2}
  4. −32-\frac{\sqrt{3}}{2}
Explanation: This question tests understanding of the even/odd symmetry properties of trigonometric functions. Sine and tangent are odd functions, meaning sin⁡(−θ)=−sin⁡(θ)\sin(-\theta) = -\sin(\theta) and tan⁡(−θ)=−tan⁡(θ)\tan(-\theta) = -\tan(\theta), while cosine is an even function, meaning cos⁡(−θ)=cos⁡(θ)\cos(-\theta) = \cos(\theta). On the unit circle, the point at angle θ\theta has coordinates (cos⁡(θ),sin⁡(θ))(\cos(\theta), \sin(\theta)), and the point at angle −θ-\theta has coordinates (cos⁡(−θ),sin⁡(−θ))=(cos⁡(θ),−sin⁡(θ))(\cos(-\theta), \sin(-\theta)) = (\cos(\theta), -\sin(\theta)), reflecting across the x-axis. For angle π6\frac{\pi}{6}, we use the odd property of sine: sin⁡(−π6)=−sin⁡(π6)=−(12)=−12\sin\left(-\frac{\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) = -\left(\frac{1}{2}\right) = -\frac{1}{2}. Choice B is correct because it applies the odd property correctly with the given value. Choice A incorrectly treats sine as an even function, using sin⁡(−θ)=sin⁡(θ)\sin(-\theta) = \sin(\theta), when sine is actually odd with sin⁡(−θ)=−sin⁡(θ)\sin(-\theta) = -\sin(\theta). The unit circle provides geometric intuition: reflecting across the x-axis (going from θ\theta to −θ-\theta) keeps the x-coordinate (cosine) the same but flips the y-coordinate (sine), explaining why cosine is even and sine is odd.

Question 4

Using periodic behavior, the smallest positive number pp such that tan⁡(θ+p)=tan⁡(θ)\tan(\theta+p)=\tan(\theta) for all θ\theta is called the period of tangent. What is the period of tan⁡(θ)\tan(\theta)?

  1. 2π2\pi
  2. π\pi (correct answer)
  3. π2\frac{\pi}{2}
  4. 4π4\pi
Explanation: This question tests understanding of the periodicity of trigonometric functions. The sine and cosine functions are periodic with period 2π, meaning sin(θ + 2π) = sin(θ) and cos(θ + 2π) = cos(θ), while tangent has a shorter period of π, meaning tan(θ + π) = tan(θ). The period of tangent is π because this is the smallest positive value p such that tan(θ + p) = tan(θ) for all θ; tangent's vertical asymptotes repeat every π, completing one full cycle in that interval. Choice B is correct because it uses the period correctly, identifying the fundamental repeat distance for tangent. Choice A uses the wrong period, claiming the period of tangent is 2π when it's actually π. For tangent, remember its period is π (not 2π) because tan(θ) = sin(θ)/cos(θ), and both sine and cosine change sign when moving π radians, making their ratio the same. When working with properties, apply them systematically: first reduce using periodicity (add/subtract multiples of the period), then apply even/odd symmetry, then evaluate using special angles or the unit circle.

Question 5

On the unit circle, the point at angle θ\theta is (cos⁡(θ),sin⁡(θ))(\cos(\theta),\sin(\theta)) and the point at angle −θ-\theta is its reflection across the xx-axis. Based on this symmetry and the fact that sine is an odd function, what is sin⁡(−θ)\sin(-\theta) in terms of sin⁡(θ)\sin(\theta)?​

  1. sin⁡(−θ)=sin⁡(θ)\sin(-\theta)=\sin(\theta)
  2. sin⁡(−θ)=−sin⁡(θ)\sin(-\theta)=-\sin(\theta) (correct answer)
  3. sin⁡(−θ)=cos⁡(θ)\sin(-\theta)=\cos(\theta)
  4. sin⁡(−θ)=∣sin⁡(θ)∣\sin(-\theta)=\left|\sin(\theta)\right|
Explanation: This question tests understanding of the odd symmetry property of trigonometric functions. Sine is an odd function, meaning sin(-θ) = -sin(θ), which can be visualized on the unit circle where the point at angle -θ is the reflection of the point at angle θ across the x-axis. On the unit circle, a point at angle θ has coordinates (cos(θ), sin(θ)), and the point at angle -θ has coordinates (cos(-θ), sin(-θ)) = (cos(θ), -sin(θ)), reflecting across the x-axis. Choice B is correct because it applies the odd property of sine: sin(-θ) = -sin(θ). Choice A incorrectly treats sine as an even function, using sin(-θ) = sin(θ), when sine is actually odd with sin(-θ) = -sin(θ). Key to symmetry and periodicity: remember that cosine is EVEN (cos(-x) = cos(x)), while sine and tangent are ODD (sin(-x) = -sin(x), tan(-x) = -tan(x)), and that sine and cosine repeat every 2π while tangent repeats every π.

Question 6

Using the periodicity property of cosine, cos⁡(θ+2π)=cos⁡(θ)\cos(\theta+2\pi)=\cos(\theta). What is cos⁡(−7π4)\cos\left(-\frac{7\pi}{4}\right) expressed using symmetry and/or periodicity as an equivalent standard-angle cosine value?

  1. cos⁡(π4)\cos\left(\frac{\pi}{4}\right) (correct answer)
  2. −cos⁡(π4)-\cos\left(\frac{\pi}{4}\right)
  3. sin⁡(π4)\sin\left(\frac{\pi}{4}\right)
  4. cos⁡(3π4)\cos\left(\frac{3\pi}{4}\right)
Explanation: This question tests understanding of both symmetry and periodicity of trigonometric functions. The sine and cosine functions are periodic with period 2π, meaning sin(θ + 2π) = sin(θ) and cos(θ + 2π) = cos(θ), while tangent has a shorter period of π, meaning tan(θ + π) = tan(θ). To evaluate cos(-7π/4), we first use the even property: cos(-7π/4) = cos(7π/4), then periodicity: 7π/4 = -π/4 + 2π, so cos(7π/4) = cos(-π/4) = cos(π/4). Choice A is correct because it combines both properties in proper sequence with specific values. Choice B incorrectly treats cosine as an odd function, giving cos(-θ) = -cos(θ), when cosine is actually even with cos(-θ) = cos(θ). When working with properties, apply them systematically: first reduce using periodicity (add/subtract multiples of the period), then apply even/odd symmetry, then evaluate using special angles or the unit circle. The unit circle provides geometric intuition: reflecting across the x-axis (going from θ to -θ) keeps the x-coordinate (cosine) the same but flips the y-coordinate (sine), explaining why cosine is even and sine is odd.

Question 7

On the unit circle, the point at angle θ\theta is (cos⁡(θ),sin⁡(θ))(\cos(\theta),\sin(\theta)), and the point at angle −θ-\theta is its reflection across the xx-axis. Based on this symmetry, what is sin⁡(−θ)\sin(-\theta) in terms of sin⁡(θ)\sin(\theta)?​​​

  1. sin⁡(−θ)=sin⁡(θ)\sin(-\theta)=\sin(\theta)
  2. sin⁡(−θ)=1sin⁡(θ)\sin(-\theta)=\dfrac{1}{\sin(\theta)}
  3. sin⁡(−θ)=−sin⁡(θ)\sin(-\theta)=-\sin(\theta) (correct answer)
  4. sin⁡(−θ)=cos⁡(θ)\sin(-\theta)=\cos(\theta)
Explanation: This question tests understanding of the even/odd symmetry properties of trigonometric functions. On the unit circle, a point at angle θ has coordinates (cos(θ), sin(θ)), and the point at angle -θ has coordinates (cos(-θ), sin(-θ)) = (cos(θ), -sin(θ)), reflecting across the x-axis. On the unit circle, point (cos(θ), sin(θ)) at angle θ reflects across the x-axis to point (cos(θ), -sin(θ)) at angle -θ, showing why the x-coordinate (cosine) stays the same while the y-coordinate (sine) changes sign. Choice C is correct because it applies the odd property of sine correctly: sin(-θ) = -sin(θ). Choice A incorrectly treats sine as an even function, using sin(-θ) = sin(θ), when sine is actually odd with sin(-θ) = -sin(θ). The unit circle provides geometric intuition: reflecting across the x-axis (going from θ to -θ) keeps the x-coordinate (cosine) the same but flips the y-coordinate (sine), explaining why cosine is even and sine is odd. Key to symmetry and periodicity: remember that cosine is EVEN (cos(-x) = cos(x)), while sine and tangent are ODD (sin(-x) = -sin(x), tan(-x) = -tan(x)), and that sine and cosine repeat every 2π while tangent repeats every π.

Question 8

On the unit circle, the point at angle θ\theta is (cos⁡(θ),sin⁡(θ))(\cos(\theta),\sin(\theta)) and the point at angle −θ-\theta reflects across the xx-axis. Based on this symmetry, what is sin⁡(−θ)\sin(-\theta) in terms of sin⁡(θ)\sin(\theta)?

  1. sin⁡(−θ)=sin⁡(θ)\sin(-\theta)=\sin(\theta)
  2. sin⁡(−θ)=−sin⁡(θ)\sin(-\theta)=-\sin(\theta) (correct answer)
  3. sin⁡(−θ)=cos⁡(θ)\sin(-\theta)=\cos(\theta)
  4. sin⁡(−θ)=∣sin⁡(θ)∣\sin(-\theta)=\left|\sin(\theta)\right|
Explanation: This question tests understanding of the odd symmetry property of the sine function. On the unit circle, a point at angle θ has coordinates (cos(θ), sin(θ)), and the point at angle -θ has coordinates (cos(-θ), sin(-θ)) = (cos(θ), -sin(θ)), reflecting across the x-axis. Since the y-coordinate represents sine, and reflection across the x-axis changes the sign of the y-coordinate, we have sin(-θ) = -sin(θ), showing that sine is an odd function. Choice B is correct because it correctly identifies that sin(-θ) = -sin(θ), capturing the sign change from the reflection. Choice A incorrectly treats sine as an even function, using sin(-θ) = sin(θ), when sine is actually odd with sin(-θ) = -sin(θ). Key to symmetry and periodicity: remember that cosine is EVEN (cos(-x) = cos(x)), while sine and tangent are ODD (sin(-x) = -sin(x), tan(-x) = -tan(x)), and that sine and cosine repeat every 2π while tangent repeats every π.

Question 9

Using the periodicity of tangent, tan⁡(θ+π)=tan⁡(θ)\tan(\theta+\pi)=\tan(\theta), what is the period of y=tan⁡(x)y=\tan(x)?

  1. π2\frac{\pi}{2}
  2. 2π2\pi
  3. π\pi (correct answer)
  4. 4π4\pi
Explanation: This question tests understanding of the periodicity of trigonometric functions. The sine and cosine functions are periodic with period 2π, meaning sin(θ + 2π) = sin(θ) and cos(θ + 2π) = cos(θ), while tangent has a shorter period of π, meaning tan(θ + π) = tan(θ). The period of tangent is π because this is the smallest positive value p such that tan(θ + p) = tan(θ) for all θ; tangent's vertical asymptotes repeat every π. Choice C is correct because it uses the period correctly. Choice B incorrectly uses the wrong period, claiming the period of tangent is 2π when it's actually π. For tangent, remember its period is π (not 2π) because tan(θ) = sin(θ)/cos(θ), and both sine and cosine change sign when moving π radians, making their ratio the same. To evaluate trig functions at negative angles or angles beyond 2π, first use periodicity to reduce to the range [0, 2π), then use even/odd properties if the angle is negative, and finally use reference angles if needed.

Question 10

On the unit circle, the point at angle θ\theta is (cos⁡(θ),sin⁡(θ))(\cos(\theta),\sin(\theta)). Using the even/odd symmetry, what is cos⁡(−π3)\cos\left(-\frac{\pi}{3}\right)?

  1. −12-\frac{1}{2}
  2. 32\frac{\sqrt{3}}{2}
  3. 12\frac{1}{2} (correct answer)
  4. −32-\frac{\sqrt{3}}{2}
Explanation: This question tests understanding of the even/odd symmetry properties of trigonometric functions. Sine and tangent are odd functions, meaning sin⁡(−θ)=−sin⁡(θ)\sin(-\theta) = -\sin(\theta) and tan⁡(−θ)=−tan⁡(θ)\tan(-\theta) = -\tan(\theta), while cosine is an even function, meaning cos⁡(−θ)=cos⁡(θ)\cos(-\theta) = \cos(\theta). On the unit circle, a point at angle θ\theta has coordinates (cos⁡(θ),sin⁡(θ))(\cos(\theta), \sin(\theta)), and the point at angle −θ-\theta has coordinates (cos⁡(−θ),sin⁡(−θ))=(cos⁡(θ),−sin⁡(θ))(\cos(-\theta), \sin(-\theta)) = (\cos(\theta), -\sin(\theta)), reflecting across the x-axis. For angle π/3\pi/3, we use the even property of cosine: cos⁡(−π/3)=cos⁡(π/3)=12\cos(-\pi/3) = \cos(\pi/3) = \frac{1}{2}. Choice C is correct because it applies the even property correctly. Choice A incorrectly treats cosine as an odd function, giving cos⁡(−θ)=−cos⁡(θ)\cos(-\theta) = -\cos(\theta), when cosine is actually even with cos⁡(−θ)=cos⁡(θ)\cos(-\theta) = \cos(\theta). Key to symmetry and periodicity: remember that cosine is EVEN (cos⁡(−x)=cos⁡(x)\cos(-x) = \cos(x)), while sine and tangent are ODD (sin⁡(−x)=−sin⁡(x)\sin(-x) = -\sin(x), tan⁡(−x)=−tan⁡(x)\tan(-x) = -\tan(x)), and that sine and cosine repeat every 2π2\pi while tangent repeats every π\pi.

Question 11

Using the periodicity property of sine, sin⁡(θ+2π)=sin⁡(θ)\sin(\theta+2\pi)=\sin(\theta) for all θ\theta. Based on this, what is sin⁡(π6+2π)\sin\left(\frac{\pi}{6}+2\pi\right)?

  1. sin⁡(π6)\sin\left(\frac{\pi}{6}\right) (correct answer)
  2. sin⁡(π6+π)\sin\left(\frac{\pi}{6}+\pi\right)
  3. −sin⁡(π6)-\sin\left(\frac{\pi}{6}\right)
  4. cos⁡(π6)\cos\left(\frac{\pi}{6}\right)
Explanation: This question tests understanding of the periodicity of trigonometric functions. The sine and cosine functions are periodic with period 2π, meaning sin(θ + 2π) = sin(θ) and cos(θ + 2π) = cos(θ), while tangent has a shorter period of π, meaning tan(θ + π) = tan(θ). Since sine has period 2π, we can write (π/6 + 2π) = π/6 + 2π, so sin(π/6 + 2π) = sin(π/6). Choice A is correct because it uses the period correctly. Choice B uses the wrong period, claiming the period of sine is π when it's actually 2π. Key to symmetry and periodicity: remember that cosine is EVEN (cos(-x) = cos(x)), while sine and tangent are ODD (sin(-x) = -sin(x), tan(-x) = -tan(x)), and that sine and cosine repeat every 2π while tangent repeats every π. To evaluate trig functions at negative angles or angles beyond 2π, first use periodicity to reduce to the range [0, 2π), then use even/odd properties if the angle is negative, and finally use reference angles if needed.

Question 12

Consider the angles π4\frac{\pi}{4} and 9π4\frac{9\pi}{4} on the unit circle. Since 9π4=π4+2π\frac{9\pi}{4}=\frac{\pi}{4}+2\pi, they are coterminal. Using periodicity, what is cos⁡(9π4)\cos\left(\frac{9\pi}{4}\right)?

  1. 00
  2. cos⁡(π4)\cos\left(\frac{\pi}{4}\right) (correct answer)
  3. −cos⁡(π4)-\cos\left(\frac{\pi}{4}\right)
  4. sin⁡(π4)\sin\left(\frac{\pi}{4}\right)
Explanation: This question tests understanding of the periodicity of trigonometric functions. The sine and cosine functions are periodic with period 2π, meaning sin(θ + 2π) = sin(θ) and cos(θ + 2π) = cos(θ), while tangent has a shorter period of π, meaning tan(θ + π) = tan(θ). Since cosine has period 2π, we can write (9π/4) = π/4 + 2π, so cos(9π/4) = cos(π/4). Choice B is correct because it uses the period correctly. Choice C forgets to apply the negative sign from the odd function property, giving the magnitude correct but the wrong sign. Key to symmetry and periodicity: remember that cosine is EVEN (cos(-x) = cos(x)), while sine and tangent are ODD (sin(-x) = -sin(x), tan(-x) = -tan(x)), and that sine and cosine repeat every 2π while tangent repeats every π. To evaluate trig functions at negative angles or angles beyond 2π, first use periodicity to reduce to the range [0, 2π), then use even/odd properties if the angle is negative, and finally use reference angles if needed.

Question 13

On the unit circle, the point at angle θ\theta is (cos⁡(θ),sin⁡(θ))(\cos(\theta),\sin(\theta)) and the point at angle −θ-\theta is its reflection across the xx-axis. Based on this symmetry, what is sin⁡(−θ)\sin(-\theta) in terms of sin⁡(θ)\sin(\theta)?

  1. sin⁡(−θ)=sin⁡(θ)\sin(-\theta)=\sin(\theta)
  2. sin⁡(−θ)=−sin⁡(θ)\sin(-\theta)=-\sin(\theta) (correct answer)
  3. sin⁡(−θ)=cos⁡(θ)\sin(-\theta)=\cos(\theta)
  4. sin⁡(−θ)=1sin⁡(θ)\sin(-\theta)=\dfrac{1}{\sin(\theta)}
Explanation: This question tests understanding of the even/odd symmetry properties of trigonometric functions. On the unit circle, a point at angle θ has coordinates (cos(θ), sin(θ)), and the point at angle -θ has coordinates (cos(-θ), sin(-θ)) = (cos(θ), -sin(θ)), reflecting across the x-axis. On the unit circle, point (cos(θ), sin(θ)) at angle θ reflects across the x-axis to point (cos(θ), -sin(θ)) at angle -θ, showing why the x-coordinate (cosine) stays the same while the y-coordinate (sine) changes sign. Choice B is correct because it applies the odd property correctly: sin(-θ) = -sin(θ). Choice A incorrectly treats sine as an even function, using sin(-θ) = sin(θ), when sine is actually odd with sin(-θ) = -sin(θ). The unit circle provides geometric intuition: reflecting across the x-axis (going from θ to -θ) keeps the x-coordinate (cosine) the same but flips the y-coordinate (sine), explaining why cosine is even and sine is odd. Key to symmetry and periodicity: remember that cosine is EVEN (cos(-x) = cos(x)), while sine and tangent are ODD (sin(-x) = -sin(x), tan(-x) = -tan(x)), and that sine and cosine repeat every 2π while tangent repeats every π.

Question 14

The graph of y=cos⁡(x)y=\cos(x) is symmetric about the yy-axis. Using this symmetry, what is cos⁡(−θ)\cos(-\theta) in terms of cos⁡(θ)\cos(\theta)?

  1. cos⁡(−θ)=−cos⁡(θ)\cos(-\theta)=-\cos(\theta)
  2. cos⁡(−θ)=sin⁡(θ)\cos(-\theta)=\sin(\theta)
  3. cos⁡(−θ)=cos⁡(θ)\cos(-\theta)=\cos(\theta) (correct answer)
  4. cos⁡(−θ)=1cos⁡(θ)\cos(-\theta)=\dfrac{1}{\cos(\theta)}
Explanation: This question tests understanding of the even/odd symmetry properties of trigonometric functions. Even functions have graphs symmetric about the y-axis (f(-x) = f(x)), while odd functions have graphs symmetric about the origin (f(-x) = -f(x)), and these symmetries are visible in the unit circle geometry. For angle θ, we use the even property of cosine: cos(-θ) = cos(θ). Choice C is correct because it applies the even property correctly. Choice A incorrectly treats cosine as an odd function, giving cos(-θ) = -cos(θ), when cosine is actually even with cos(-θ) = cos(θ). The unit circle provides geometric intuition: reflecting across the x-axis (going from θ to -θ) keeps the x-coordinate (cosine) the same but flips the y-coordinate (sine), explaining why cosine is even and sine is odd. Key to symmetry and periodicity: remember that cosine is EVEN (cos(-x) = cos(x)), while sine and tangent are ODD (sin(-x) = -sin(x), tan(-x) = -tan(x)), and that sine and cosine repeat every 2π while tangent repeats every π.

Question 15

On the unit circle, the point at angle θ\theta is (cos⁡(θ),sin⁡(θ))(\cos(\theta),\sin(\theta)), and the point at angle −θ-\theta reflects across the xx-axis. Based on this symmetry, what is sin⁡(−θ)\sin(-\theta) in terms of sin⁡(θ)\sin(\theta)?

  1. sin⁡(−θ)=sin⁡(θ)\sin(-\theta)=\sin(\theta)
  2. sin⁡(−θ)=−sin⁡(θ)\sin(-\theta)=-\sin(\theta) (correct answer)
  3. sin⁡(−θ)=cos⁡(θ)\sin(-\theta)=\cos(\theta)
  4. sin⁡(−θ)=∣sin⁡(θ)∣\sin(-\theta)=|\sin(\theta)|
Explanation: This question tests understanding of the even/odd symmetry properties of trigonometric functions. On the unit circle, a point at angle θ has coordinates (cos(θ), sin(θ)), and the point at angle -θ has coordinates (cos(-θ), sin(-θ)) = (cos(θ), -sin(θ)), reflecting across the x-axis. On the unit circle, point (cos(θ), sin(θ)) at angle θ reflects across the x-axis to point (cos(θ), -sin(θ)) at angle -θ, showing why the x-coordinate (cosine) stays the same while the y-coordinate (sine) changes sign. Choice B is correct because it applies the odd property correctly, recognizing that the reflection flips the sign of the sine value. Choice A incorrectly treats sine as an even function, using sin(-θ) = sin(θ), when sine is actually odd with sin(-θ) = -sin(θ). Key to symmetry and periodicity: remember that cosine is EVEN (cos(-x) = cos(x)), while sine and tangent are ODD (sin(-x) = -sin(x), tan(-x) = -tan(x)), and that sine and cosine repeat every 2π while tangent repeats every π. The unit circle provides geometric intuition: reflecting across the x-axis (going from θ to -θ) keeps the x-coordinate (cosine) the same but flips the y-coordinate (sine), explaining why cosine is even and sine is odd.

Question 16

The graph of y=cos⁡(x)y=\cos(x) is symmetric about the yy-axis. Using this even-function symmetry, what is cos⁡(−θ)\cos(-\theta) in terms of cos⁡(θ)\cos(\theta)?

  1. cos⁡(−θ)=cos⁡(θ)\cos(-\theta)=\cos(\theta) (correct answer)
  2. cos⁡(−θ)=−cos⁡(θ)\cos(-\theta)=-\cos(\theta)
  3. cos⁡(−θ)=sin⁡(θ)\cos(-\theta)=\sin(\theta)
  4. cos⁡(−θ)=cos⁡(θ+π)\cos(-\theta)=\cos(\theta+\pi)
Explanation: This question tests understanding of the even/odd symmetry properties of trigonometric functions. Even functions have graphs symmetric about the y-axis (f(-x) = f(x)), while odd functions have graphs symmetric about the origin (f(-x) = -f(x)), and these symmetries are visible in the unit circle geometry. For angle θ, we use the even property of cosine: cos(-θ) = cos(θ). Choice A is correct because it applies the even property correctly. Choice B incorrectly treats cosine as an odd function, giving cos(-θ) = -cos(θ), when cosine is actually even with cos(-θ) = cos(θ). Key to symmetry and periodicity: remember that cosine is EVEN (cos(-x) = cos(x)), while sine and tangent are ODD (sin(-x) = -sin(x), tan(-x) = -tan(x)), and that sine and cosine repeat every 2π while tangent repeats every π. The unit circle provides geometric intuition: reflecting across the x-axis (going from θ to -θ) keeps the x-coordinate (cosine) the same but flips the y-coordinate (sine), explaining why cosine is even and sine is odd.

Question 17

For the tangent function, the smallest positive period pp satisfies tan⁡(θ+p)=tan⁡(θ)\tan(\theta+p)=\tan(\theta) for all θ\theta where tangent is defined. What is the period of tan⁡(θ)\tan(\theta)?

  1. π2\frac{\pi}{2}
  2. 2π2\pi
  3. π\pi (correct answer)
  4. 4π4\pi
Explanation: This question tests understanding of the periodicity of trigonometric functions. The period of a function is the smallest positive value p such that f(θ + p) = f(θ) for all θ in the domain. The tangent function has period π because this is the smallest positive value p such that tan(θ + p) = tan(θ) for all θ where tangent is defined; tangent's vertical asymptotes repeat every π radians. Choice C is correct because tangent has period π, not 2π like sine and cosine. Choice B incorrectly claims the period of tangent is 2π when it's actually π. For tangent, remember its period is π (not 2π) because tan(θ) = sin(θ)/cos(θ), and both sine and cosine change sign when moving π radians, making their ratio the same.

Question 18

Based on symmetry of the unit circle, the point at angle θ\theta is (cos⁡(θ),sin⁡(θ))(\cos(\theta),\sin(\theta)) and the point at angle −θ-\theta is (cos⁡(θ),−sin⁡(θ))(\cos(\theta),-\sin(\theta)). Using this, what is tan⁡(−θ)\tan(-\theta) in terms of tan⁡(θ)\tan(\theta)?

  1. tan⁡(−θ)=tan⁡(θ)\tan(-\theta)=\tan(\theta)
  2. tan⁡(−θ)=−tan⁡(θ)\tan(-\theta)=-\tan(\theta) (correct answer)
  3. tan⁡(−θ)=1tan⁡(θ)\tan(-\theta)=\frac{1}{\tan(\theta)}
  4. tan⁡(−θ)=cos⁡(θ)\tan(-\theta)=\cos(\theta)
Explanation: This question tests understanding of the odd symmetry property of the tangent function. On the unit circle, point (cos(θ), sin(θ)) at angle θ reflects across the x-axis to point (cos(θ), -sin(θ)) at angle -θ, showing why the x-coordinate (cosine) stays the same while the y-coordinate (sine) changes sign. Since tan(θ) = sin(θ)/cos(θ), we have tan(-θ) = sin(-θ)/cos(-θ) = -sin(θ)/cos(θ) = -tan(θ), proving tangent is an odd function. Choice B is correct because it applies the odd property correctly, recognizing that tan(-θ) = -tan(θ). Choice A incorrectly treats tangent as an even function, using tan(-θ) = tan(θ), when tangent is actually odd with tan(-θ) = -tan(θ). Key to symmetry and periodicity: remember that cosine is EVEN (cos(-x) = cos(x)), while sine and tangent are ODD (sin(-x) = -sin(x), tan(-x) = -tan(x)), and that sine and cosine repeat every 2π while tangent repeats every π.

Question 19

Using the periodicity of tangent, tan⁡(θ+π)=tan⁡(θ)\tan(\theta+\pi)=\tan(\theta) for all θ\theta where tangent is defined. What is tan⁡(5π4)\tan\left(\frac{5\pi}{4}\right)?

  1. −1-1
  2. 11 (correct answer)
  3. 00
  4. undefined
Explanation: This question tests understanding of the periodicity of trigonometric functions. The tangent function has period π, meaning tan(θ + π) = tan(θ) for all θ where tangent is defined. Since 5π/4 = π/4 + π, we can write tan(5π/4) = tan(π/4 + π) = tan(π/4) = 1. Choice B is correct because it uses the period π correctly to simplify 5π/4 to π/4, then evaluates tan(π/4) = 1. Choice A incorrectly gives -1, perhaps confusing the quadrant or not properly applying the periodicity of tangent. For tangent, remember its period is π (not 2π) because tan(θ) = sin(θ)/cos(θ), and both sine and cosine change sign when moving π radians, making their ratio the same.

Question 20

Based on the unit circle, examine the graph shown. The graph shows the relationship between angle measures and the corresponding y-coordinates of points on the unit circle. What can be concluded about the symmetry of this function?

  1. The function is even because f(−x)=f(x)f(-x) = f(x) for all values
  2. The function is odd because f(−x)=−f(x)f(-x) = -f(x) for all values (correct answer)
  3. The function has rotational symmetry about the point (π,0)(\pi, 0)
  4. The function has both even and odd properties in different intervals
Explanation: The graph shows y=sin⁡(x)y = \sin(x), which represents the y-coordinate of points on the unit circle. The sine function is odd, meaning sin⁡(−x)=−sin⁡(x)\sin(-x) = -\sin(x). This is evident from the unit circle: if a point (cos⁡(θ),sin⁡(θ))(\cos(\theta), \sin(\theta)) corresponds to angle θ\theta, then the point for angle −θ-\theta is (cos⁡(θ),−sin⁡(θ))(\cos(\theta), -\sin(\theta)). The graph demonstrates this symmetry by being symmetric about the origin - if you rotate the graph 180¬∞ about the origin, it maps onto itself.