Equivalent Representations of Trigonometric Functions
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AP Precalculus › Equivalent Representations of Trigonometric Functions
A spring’s position is modeled by $f(t)=10\sin!\left(2t+\frac{\pi}{3}\right)$ (centimeters), where $t$ is seconds. Letting $t=\theta$ gives the polar form $r=10\sin!\left(2\theta+\frac{\pi}{3}\right)$. On the coordinate plane, the sinusoid has amplitude $10$, period $\pi$, and its first maximum occurs at $t=\frac{\pi}{12}$. This periodic motion matches simple harmonic oscillation. Based on the description, how does the phase shift change when converting the function to polar form?
It remains the same phase shift, since only $t$ is renamed $\theta$
It becomes a period change from $\pi$ to $2\pi$
It changes from $\frac{\pi}{3}$ to $\frac{\pi}{6}$
It disappears because polar form cannot include phase shifts
It becomes a vertical shift of $\frac{\pi}{3}$
Explanation
This question tests AP Precalculus understanding of equivalent representations of trigonometric functions, specifically how phase shifts behave when converting to polar form. When converting f(t) = 10sin(2t + π/3) to polar form by substituting t = θ, the result is r = 10sin(2θ + π/3). The phase shift, represented by the constant π/3 inside the sine function, remains unchanged because we're simply renaming the variable from t to θ. Choice C correctly identifies that the phase shift remains the same since only t is renamed θ. Choices A, B, D, and E incorrectly suggest the phase shift transforms into other parameters or disappears, misunderstanding that polar conversion is merely a coordinate system change. To help students: Emphasize that converting to polar form is a relabeling process, not a mathematical transformation. Practice identifying which parameters change (variable names) versus which remain constant (amplitude, frequency, phase) during conversion.
A vibration sensor reads $f(x)=3\sin!\left(\frac{x}{2}-\frac{\pi}{4}\right)$ (volts), where $x$ is time in seconds. Converting by setting $x=\theta$ gives $r=3\sin!\left(\frac{\theta}{2}-\frac{\pi}{4}\right)$. On the coordinate plane, the sinusoid has amplitude $3$, period $4\pi$, and crosses the midline at $x=\frac{\pi}{2}$. This periodic signal can represent a steady machine hum. Based on the description, what is the amplitude of the function given in polar form?
$\frac{\pi}{4}$
$\frac{1}{2}$
$\frac{3}{2}$
$4\pi$
$3$
Explanation
This question tests AP Precalculus understanding of equivalent representations of trigonometric functions, specifically identifying amplitude in polar form. The function f(x) = 3sin(x/2 - π/4) has an amplitude of 3, which is the coefficient multiplying the entire sine function. When converted to polar form r = 3sin(θ/2 - π/4), the amplitude remains 3 because it represents the maximum distance from the midline. Choice B correctly identifies the amplitude as 3. Choices A and E confuse amplitude with other parameters, while C and D incorrectly incorporate π or the frequency coefficient. To help students: Emphasize that amplitude is always the positive coefficient in front of the sine or cosine function. Practice identifying amplitude in various forms and stress that it remains constant through coordinate transformations.
A tide height model is $f(t)=5\sin!\left(\frac{\pi}{6}t\right)+2$ (meters), with $t$ in hours. Using $t=\theta$ gives the polar form $r=5\sin!\left(\frac{\pi}{6}\theta\right)+2$. On a coordinate plane, the midline is $y=2$ and peaks reach $7$ meters. This sinusoid represents periodic ocean tides. Refer to the passage above. Identify the period of the trigonometric function in the graphical representation.
$12$
$2\pi$
$\frac{\pi}{6}$
$6$
$\frac{12}{\pi}$
Explanation
This question tests AP Precalculus understanding of equivalent representations of trigonometric functions, specifically calculating the period from a given function. The function f(t) = 5sin(π/6·t) + 2 has a coefficient of π/6 multiplying t inside the sine function. Using the period formula P = 2π/B where B = π/6, we get P = 2π/(π/6) = 12 hours. Choice A correctly identifies the period as 12. The vertical shift of +2 affects the midline but not the period, and choices B through E represent common calculation errors or confusion between period and other function parameters. To help students: Reinforce that vertical shifts don't affect period, only horizontal behavior does. Practice extracting the coefficient of the variable from various function forms and applying the period formula systematically.
A metronome’s side-to-side position is $f(t)=2\sin!\left(\pi t+\frac{\pi}{2}\right)$ (centimeters), with $t$ in seconds. Letting $t=\theta$ gives $r=2\sin!\left(\pi\theta+\frac{\pi}{2}\right)$. On the coordinate plane, the sinusoid has amplitude $2$, period $2$, and starts at a maximum when $t=0$. This matches a steady beat. Based on the description, identify the period of the trigonometric function in the graphical representation.
$\frac{1}{2}$
$2\pi$
$2$
$\pi$
$\frac{2}{\pi}$
Explanation
This question tests AP Precalculus understanding of equivalent representations of trigonometric functions, specifically determining period from a trigonometric equation. The function f(t) = 2sin(πt + π/2) has π as the coefficient of t, giving a period of 2π/π = 2 seconds using the formula P = 2π/B. The passage explicitly confirms the period is 2, making Choice B correct. Choice A confuses the coefficient π with the period, while choices C through E represent various calculation errors involving π. To help students: Reinforce the period formula P = 2π/B and practice extracting B from functions where it's multiplied by π. Use graphical representations to verify calculated periods match the visual cycle length.
A sound wave is modeled by $f(x)=7\sin!\left(4x-2\pi\right)$ (arbitrary units), where $x$ is time in seconds. Converting by setting $x=\theta$ gives $r=7\sin!\left(4\theta-2\pi\right)$. On the coordinate plane, the sinusoid has amplitude $7$, period $\frac{\pi}{2}$, and the phase shift is $\frac{\pi}{2}$ to the right. This periodic model matches a stable tone. Refer to the passage above. How does the phase shift change when converting the function to polar form?
It becomes a vertical shift of $2\pi$
It changes to a left shift of $\frac{\pi}{2}$
It stays the same, since $x$ is simply relabeled as $\theta$
It doubles because polar angles are measured differently
It becomes the amplitude because $-2\pi$ is outside the sine
Explanation
This question tests AP Precalculus understanding of equivalent representations of trigonometric functions, specifically phase shift behavior in polar conversion. The function f(x) = 7sin(4x - 2π) has a phase shift that can be found by solving 4x - 2π = 0, giving x = π/2 (shift right). When converting to polar form r = 7sin(4θ - 2π), the phase shift calculation remains identical: 4θ - 2π = 0 gives θ = π/2. Choice C correctly states the phase shift stays the same since x is simply relabeled as θ. Choices A, B, D, and E incorrectly suggest the phase shift transforms or relates to other parameters, misunderstanding the nature of coordinate conversion. To help students: Practice calculating phase shifts before and after conversion to verify they remain constant. Emphasize that relabeling variables doesn't change the function's behavior or characteristics.
A rotating beacon’s brightness is $f(t)=6\sin!\left(\frac{\pi}{4}t-\frac{\pi}{6}\right)$ (lumens), with $t$ in seconds. Using $t=\theta$, an equivalent polar form is $r=6\sin!\left(\frac{\pi}{4}\theta-\frac{\pi}{6}\right)$. On a coordinate plane, the sinusoid has amplitude $6$, period $8$, and is shifted right by $\frac{2}{3}$ seconds from $\sin!\left(\frac{\pi}{4}t\right)$. This periodicity matches a steady rotating light. Refer to the passage above. Identify the period of the trigonometric function in the graphical representation.
$8$
$\frac{8}{\pi}$
$\frac{4}{\pi}$
$\frac{\pi}{4}$
$\frac{2\pi}{8}$
Explanation
This question tests AP Precalculus understanding of equivalent representations of trigonometric functions, specifically identifying the period from a given trigonometric function. The function f(t) = 6sin(π/4·t - π/6) has a coefficient of π/4 multiplying t inside the sine function. The period of a sine function in the form sin(Bt) is calculated as 2π/B, so here the period is 2π/(π/4) = 8 seconds. The passage explicitly confirms the period is 8, making Choice A correct. Choices B through E represent common errors in period calculation, such as confusing the coefficient with the period or incorrectly manipulating π. To help students: Memorize the period formula P = 2π/B for sin(Bt) and cos(Bt). Practice extracting B from various function forms and emphasize that the period represents one complete cycle of the trigonometric function.