Trigonometry
Help Questions
ACT Math › Trigonometry
What is the period of the trigonometric function $f(x) = 3\sin!\left(\dfrac{\pi}{2}x\right) - 4$?
$4$
$3$
$\frac{\pi}{2}$
$2$
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.
Which equals $\sin(60^\circ)$?
$\frac{\sqrt{2}}{2}$
$\frac{1}{2}$
$\sqrt{3}$
$\frac{\sqrt{3}}{2}$
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°).
In a right triangle, angle $\theta$ has adjacent side length $12$ and hypotenuse length $13$. What is $\cos(\theta)$?
$\frac{12}{5}$
$\frac{5}{13}$
$\frac{12}{13}$
$\frac{13}{12}$
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.
In a right triangle, the side opposite angle $\theta$ is $8$ and the hypotenuse is $17$. What is $\sin(\theta)$?
$\frac{15}{17}$
$\frac{17}{8}$
$\frac{8}{15}$
$\frac{8}{17}$
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.
What is $\tan(\frac{\pi}{4})$?
$0$
$\frac{\sqrt{3}}{2}$
$\sqrt{3}$
$1$
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 $\tan(60^\circ)$.
In the unit circle, what is $\sin(\frac{\pi}{6})$?
√2/2
√3/2
1/2
0
Explanation
In the unit circle, $\pi/6$ radians equals $30^\circ$. Using SOH-CAH-TOA, sine represents the y-coordinate (or opposite/hypotenuse). For the special angle $\pi/6$ ($30^\circ$), $sin(\pi/6) = 1/2$. Choice B ($\sqrt{3}/2$) is actually $sin(\pi/3)$ or $sin(60^\circ$).
What is $\sin(90^\circ)$?
0
$-1$
$\frac{1}{2}$
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°).
What is $\sin(30^\circ)$?
$\frac{1}{2}$
$0$
$\frac{\sqrt{2}}{2}$
$\frac{\sqrt{3}}{2}$
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°).
In a right triangle, if the opposite side to angle $\theta$ is 5 and the adjacent side is 12, what is $\tan(\theta)$?
5/13
5/12
12/5
13/5
Explanation
For angle $\theta$, the opposite side is 5 and the adjacent side is 12. Using SOH-CAH-TOA, $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$. Therefore, $\tan(\theta) = \frac{5}{12}$. Choice B ($\frac{12}{5}$) incorrectly flipped the ratio, using adjacent/opposite instead.
In a right triangle, the hypotenuse is $10$ and the side adjacent to angle $\theta$ is $6$. What is $\sin(\theta)$?
$\frac{3}{5}$
$\frac{5}{3}$
$\frac{4}{5}$
$\frac{6}{10}$
Explanation
Given hypotenuse = 10 and adjacent = 6, we need to find the opposite side using the Pythagorean theorem: opposite = √(10² - 6²) = √64 = 8. Using SOH: sin = opposite/hypotenuse, we get sin(θ) = 8/10 = 4/5. Choice A shows 3/5, which incorrectly assumes the opposite side is 6.