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.

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.

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°).

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.

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.

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, $\pi/6$ 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 $sin(60^\circ$).

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°).

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°).

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.