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ACT Math Help: Trigonometry

Review real example questions for Trigonometry in ACT Math.

Question 1 / 10

0 of 10 answered

What is the period of the trigonometric function $f(x) = 3\sin\!\left(\dfrac{\pi}{2}x\right) - 4$ ?

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Question 1

What is the period of the trigonometric function $f(x) = 3\sin\!\left(\dfrac{\pi}{2}x\right) - 4$ ?

$\frac{\pi}{2}$
$2$
$3$
$4$ (correct answer)

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 $g(x)$ ?

4
5
7 (correct answer)
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

What is $\tan(\frac{\pi}{4})$ ?

$0$
$1$ (correct answer)
$\sqrt{3}$
$\frac{\sqrt{3}}{2}$

Explanation: In the unit circle, $\pi/4$ radians equals $45^\circ$ . Using SOH-CAH-TOA, tangent represents opposite over adjacent. For the special angle $\pi/4$ ( $45^\circ$ ), $\tan(\pi/4) = 1$ . Choice C ( $\sqrt{3}$ ) is actually $\tan(\pi/3)$ or $\tan(60^\circ)$ .

Question 4

What is $\sin(30^\circ)$ ?

$\frac{\sqrt{3}}{2}$
$\frac{1}{2}$ (correct answer)
$\frac{\sqrt{2}}{2}$
$0$

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 5

In a right triangle, if the opposite side to angle $\theta$ is 9 and the adjacent side is 12, what is $\tan(\theta)$ ?

12/7
4/3
3/4 (correct answer)
12/9

Explanation: By SOH-CAH-TOA, the tangent of an angle is the opposite side over the adjacent side. Here the opposite side is 9 and the adjacent side is 12, so $\tan(\theta) = 9/12$ , which reduces to 3/4. The value 4/3 is the reciprocal, produced by dividing adjacent by opposite, and 12/9 is that same reversed ratio left unsimplified. The value 12/7 pairs the adjacent side with a number that is not a side of this triangle at all, so it does not come from either trigonometric ratio. Write the ratio in words before plugging in numbers, then reduce it, since an unsimplified or flipped fraction is the most common way tangent problems go wrong.

Question 6

In a right triangle, if the opposite side to angle $\theta$ is 6 and the hypotenuse is 10, what is $\sin(\theta)$ ?

3/5 (correct answer)
4/5
5/6
6/5

Explanation: For angle θ, the opposite side is 6 and the hypotenuse is 10. Using SOH-CAH-TOA, sin(θ) = opposite/hypotenuse = 6/10. This simplifies to 3/5. Choice B (4/5) would be the cosine if the adjacent side were 8.

Question 7

What is $\tan(0^\circ)$ ?

√3
1
0 (correct answer)
√3/2

Explanation: tan(0°) is a standard unit circle value. Using SOH-CAH-TOA, tangent represents the opposite side over the adjacent side. For the angle 0°, tan(0°) = 0. Choice B (1) is actually tan(45°), not tan(0°).

Question 8

What angle $\theta$ satisfies $\sin(\theta)=\frac{\sqrt{2}}{2}$ , where $\theta$ is a common acute angle?

$30^\circ$
$45^\circ$ (correct answer)
$60^\circ$
$90^\circ$

Explanation: We need to find which common acute angle has sin(θ) = √2/2. From the unit circle and special triangles, sin(45°) = √2/2. This is the fundamental value for a 45° angle in a 45-45-90 triangle. Choice C gives 60°, but sin(60°) = √3/2, not √2/2.

Question 9

In the right triangle, if the opposite side is 6 and the adjacent side is 8, what is $\tan(\theta)$ ?

$\frac{8}{6}$
$\frac{4}{3}$
$\frac{6}{8}$
$\frac{3}{4}$ (correct answer)

Explanation: For angle θ, the opposite side is 6 and the adjacent side is 8. Using SOH-CAH-TOA, $tan(\theta) = opposite/adjacent$ . Therefore, $tan(\theta) = 6/8 = 3/4$ . Choice B ( $\frac{4}{3}$ ) would be adjacent/opposite, which is the reciprocal relationship (cotangent).

Question 10

Which equals $\sin(0^\circ)$ ?

$1$
$0$ (correct answer)
$\frac{1}{2}$
$\frac{\sqrt{2}}{2}$

Explanation: The angle 0° represents a position on the positive x-axis of the unit circle. At this position, the y-coordinate is 0, so sin(0°) = 0. This is a fundamental trigonometric value where the angle has no vertical component. Choice A gives 1, which is actually cos(0°), not sin(0°).