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ACT Math

ACT Math Help: Trigonometry

Review real example questions for Trigonometry in ACT Math.

Question 1

What is the period of the trigonometric function f(x)=3sin⁡ ⁣(π2x)−4f(x) = 3\sin\!\left(\dfrac{\pi}{2}x\right) - 4f(x)=3sin(2π​x)−4?

  1. π2\frac{\pi}{2}2π​
  2. 222
  3. 333
  4. 444
Explanation: The correct answer is D (4). The period of f(x) = A sin(bx) + c is given by 2π/b. Here b = π/2. Period = 2π ÷ (π/2) = 2π × (2/π) = 4. A (π/2) reports the b-value itself as the period rather than computing 2π/b. B (2) results from computing 2π/b with b = π (misreading the coefficient as π instead of π/2): 2π/π = 2. C (3) reports the amplitude coefficient rather than the period — confusing the A and b parameters. Pro tip: the period formula is 2π/b where b is the coefficient multiplying x, not x itself.

Question 2

The function g(x) \= -3\cos\\!\left(\dfrac{\pi}{2}x\right) + k has a minimum value of 1. What is the maximum value of g(x)g(x)g(x)?

  1. 4
  2. 5
  3. 7
  4. 10
Explanation: This is a trigonometric graph analysis question testing how a negative coefficient interacts with the vertical shift to determine maximum value. Choice C (7) is correct — identify the parameters of g(x) = −3cos(πx/2) + k. Amplitude = 3, vertical shift = k. The minimum of −3cos(πx/2) occurs when cos = +1 (giving −3), so the minimum of g is −3 + k = 1 → k = 4. The maximum occurs when cos = −1 (giving +3), so the maximum of g = 3 + k = 3 + 4 = 7. Choice A (4) reports k itself — finding the vertical shift but confusing it with the maximum value. Choice B (5) adds the amplitude to the minimum: 1 + |−3| = 1 + 4 = 5, incorrectly treating the amplitude as k. Actually B = min + amplitude = 1 + 4 = 5 if student thinks range = 2×amplitude centered at min. Choice D (10) adds k and the amplitude twice: 4 + 3 + 3 = 10 or similar double-counting. Pro tip: When a cosine function has a NEGATIVE leading coefficient, the function reaches its maximum when cosine is at its MINIMUM (−1), not its maximum. Always think: "What value of the trig function makes the WHOLE expression as large as possible?" Here, −3(−1) = +3 is the largest the trig part can be, giving max = 3 + k.

Question 3

Which equals sin⁡(60∘)\sin(60^\circ)sin(60∘)?

  1. 12\frac{1}{2}21​
  2. 22\frac{\sqrt{2}}{2}22​​
  3. 32\frac{\sqrt{3}}{2}23​​
  4. 3\sqrt{3}3​
Explanation: The angle 60° is a special angle from the unit circle and 30-60-90 triangles. Using these fundamental trigonometric values, sin(60°) = √3/2. This is a standard result that should be memorized along with other special angle values. Choice A gives 1/2, which is actually sin(30°), not sin(60°).

Question 4

In a right triangle, angle θ\thetaθ has adjacent side length 121212 and hypotenuse length 131313. What is cos⁡(θ)\cos(\theta)cos(θ)?​​

  1. 1312\frac{13}{12}1213​
  2. 513\frac{5}{13}135​
  3. 1213\frac{12}{13}1312​
  4. 125\frac{12}{5}512​
Explanation: For angle θ, the adjacent side = 12 and the hypotenuse = 13. Using CAH: cos = adjacent/hypotenuse, we get cos(θ) = 12/13. Choice B shows 5/13, which would be sin(θ) using the opposite side length of 5.

Question 5

In a right triangle, the side opposite angle θ\thetaθ is 888 and the hypotenuse is 171717. What is sin⁡(θ)\sin(\theta)sin(θ)?​​

  1. 1517\frac{15}{17}1715​
  2. 817\frac{8}{17}178​
  3. 178\frac{17}{8}817​
  4. 815\frac{8}{15}158​
Explanation: For angle θ, the opposite side = 8 and the hypotenuse = 17. Using SOH: sin = opposite/hypotenuse, we get sin(θ) = 8/17. Choice A shows 15/17, which would be cos(θ) using the adjacent side length of 15.

Question 6

What is tan⁡(π4)\tan(\frac{\pi}{4})tan(4π​)?

  1. 000
  2. 111
  3. 3\sqrt{3}3​
  4. 32\frac{\sqrt{3}}{2}23​​
Explanation: In the unit circle, π/4\pi/4π/4 radians equals 45∘45^\circ45∘. Using SOH-CAH-TOA, tangent represents opposite over adjacent. For the special angle π/4\pi/4π/4 (45∘45^\circ45∘), tan⁡(π/4)=1\tan(\pi/4) = 1tan(π/4)=1. Choice C (3\sqrt{3}3​) is actually tan⁡(π/3)\tan(\pi/3)tan(π/3) or tan⁡(60∘)\tan(60^\circ)tan(60∘).

Question 7

In the unit circle, what is sin⁡(π6)\sin(\frac{\pi}{6})sin(6π​)?

  1. 1/2
  2. √3/2
  3. √2/2
  4. 0
Explanation: In the unit circle, π/6\pi/6π/6 radians equals 30∘30^\circ30∘. Using SOH-CAH-TOA, sine represents the y-coordinate (or opposite/hypotenuse). For the special angle π/6\pi/6π/6 (30∘30^\circ30∘), sin(π/6)=1/2sin(\pi/6) = 1/2sin(π/6)=1/2. Choice B (3/2\sqrt{3}/23​/2) is actually sin(π/3)sin(\pi/3)sin(π/3) or sin(60∘sin(60^\circsin(60∘).

Question 8

What is sin⁡(90∘)\sin(90^\circ)sin(90∘)?

  1. −1-1−1
  2. 0
  3. 12\frac{1}{2}21​
  4. 1
Explanation: 90° is where the unit circle intersects the positive y-axis. Using SOH-CAH-TOA, sin(90°) = opposite/hypotenuse = 1. At 90°, the y-coordinate reaches its maximum value. Choice B gives 0, which is sin(0°), not sin(90°).

Question 9

What is sin⁡(30∘)\sin(30^\circ)sin(30∘)?​​

  1. 32\frac{\sqrt{3}}{2}23​​
  2. 12\frac{1}{2}21​
  3. 22\frac{\sqrt{2}}{2}22​​
  4. 000
Explanation: The angle 30° is a special angle on the unit circle. sin(30°) = 1/2, which is a standard value to memorize. Choice A shows √3/2, which is actually cos(30°), not sin(30°).

Question 10

In a right triangle, if the opposite side to angle θ\thetaθ is 5 and the adjacent side is 12, what is tan⁡(θ)\tan(\theta)tan(θ)?

  1. 5/12
  2. 12/5
  3. 5/13
  4. 13/5
Explanation: For angle θ\thetaθ, the opposite side is 5 and the adjacent side is 12. Using SOH-CAH-TOA, tan⁡(θ)=oppositeadjacent\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}tan(θ)=adjacentopposite​. Therefore, tan⁡(θ)=512\tan(\theta) = \frac{5}{12}tan(θ)=125​. Choice B (125\frac{12}{5}512​) incorrectly flipped the ratio, using adjacent/opposite instead.