Graphing Sine & Cosine - Math 3

A cosine function is vertically stretched by a factor of 3, horizontally compressed by a factor of $\frac{1}{2}$ , and shifted down 4 units. If the original function was $y = \cos(x)$ , what is the midline of the transformed function?
Answer: $y = -4$ The transformed function is $y = 3\cos(2x) - 4$ . The vertical stretch affects amplitude (becomes 3), horizontal compression affects period ( $B = 2$ ), and the vertical shift down 4 units moves the midline from $y = 0$ to $y = -4$ . The midline is determined only by the vertical shift. Choice A gives the original midline before transformation. Choice C confuses the amplitude with the midline. Choice D represents an incorrect partial shift.
Two sinusoidal functions are given: $f(x) = 4\cos(2x)$ and $g(x) = 4\sin(2x + \frac{\pi}{2})$ . Which statement about these functions is correct?
Answer: The functions have identical graphs because $\sin(2x + \frac{\pi}{2}) = \cos(2x)$ by the cofunction identity Using the cofunction identity $\sin(\theta + \frac{\pi}{2}) = \cos(\theta)$ , we have $g(x) = 4\sin(2x + \frac{\pi}{2}) = 4\cos(2x) = f(x)$ . Therefore, the functions are identical. Choice A is wrong because they have the same period. Choice C misinterprets phase shift. Choice D incorrectly suggests that phase shift affects amplitude.
Consider the function $f(x) = A\sin(Bx)$ where $A > 0$ and $B > 0$ . If doubling the value of $A$ and halving the value of $B$ results in a new function $g(x)$ , how does the graph of $g(x)$ compare to the original graph of $f(x)$ ?
Answer: The amplitude doubles and the period doubles, resulting in a vertically and horizontally stretched graph Originally: $f(x) = A\sin(Bx)$ has amplitude $A$ and period $\frac{2\pi}{B}$ . After transformation: $g(x) = (2A)\sin(\frac{B}{2}x)$ has amplitude $2A$ (doubled) and period $\frac{2\pi}{B/2} = \frac{4\pi}{B}$ (doubled). So both amplitude and period double, creating vertical and horizontal stretching.
The function $h(x) = 3\sin(\frac{\pi x}{4}) - 2$ represents the height of a point on a rotating wheel above the ground, where $x$ is time in seconds. At what time $x$ (where $0 \leq x \leq 16$ ) does the point first reach its maximum height?
Answer: $x = 2$ seconds When analyzing sinusoidal functions like this one, you need to understand how transformations affect the basic sine curve. The function $h(x) = 3\sin(\frac{\pi x}{4}) - 2$ represents a transformed sine wave where the amplitude is 3, there's a vertical shift down by 2 units, and the period is affected by the coefficient $\frac{\pi}{4}$ . To find when the sine function reaches its maximum value, recall that $\sin(\theta) = 1$ when $\theta = \frac{\pi}{2} + 2\pi k$ (where k is any integer). For the first maximum, we need $\frac{\pi x}{4} = \frac{\pi}{2}$ . Solving this equation: multiply both sides by $\frac{4}{\pi}$ to get $x = 2$ . At this point, $h(2) = 3(1) - 2 = 1$ , which is indeed the maximum height. Looking at the incorrect answers: Choice A ( $x = 8$ ) occurs when $\frac{\pi x}{4} = 2\pi$ , where the sine function completes one full cycle and returns to its starting value, not the maximum. Choice B ( $x = 4$ ) corresponds to $\frac{\pi x}{4} = \pi$ , where $\sin(\pi) = 0$ , giving a middle value. Choice C ( $x = 6$ ) gives $\frac{\pi x}{4} = \frac{3\pi}{2}$ , where $\sin(\frac{3\pi}{2}) = -1$ , producing the minimum height. Remember: for sine functions, the first maximum always occurs at $\frac{\pi}{2}$ radians. Set the inside expression equal to $\frac{\pi}{2}$ and solve for your variable.
A function $f(x) = A\sin(Bx) + C$ has a maximum value of 7 and a minimum value of -1. If the function completes exactly 3 full cycles over the interval $[0, 2\pi]$ , what is the value of $B$ ?
Answer: $B = 3$ First, find the amplitude and midline. The amplitude is $A = \frac{7-(-1)}{2} = 4$ and midline is $C = \frac{7+(-1)}{2} = 3$ . For 3 complete cycles over $[0, 2\pi]$ , the period must be $\frac{2\pi}{3}$ . Since period = $\frac{2\pi}{B}$ , we have $\frac{2\pi}{B} = \frac{2\pi}{3}$ , so $B = 3$ . Choice B confuses cycles with the coefficient. Choice C inverts the relationship. Choice D uses the period value as B.
If $g(x) = 4\cos(\frac{x}{2}) - 3$ , what is the range of function $g$ ?
Answer: $[-7, 1]$ The cosine function has range $[-1, 1]$ . With amplitude 4, $4\cos(\frac{x}{2})$ has range $[-4, 4]$ . Subtracting 3 shifts everything down 3 units, giving range $[-4-3, 4-3] = [-7, 1]$ . Choice B incorrectly adds 3 instead of subtracting. Choice C ignores the vertical shift. Choice D uses the vertical shift value as the amplitude.
Consider the function $r(x) = 3\cos(4x) - 1$ . At what value of $x$ in the interval $[0, \frac{\pi}{2}]$ does this function first reach its minimum value?
Answer: $x = \frac{\pi}{4}$ The function $r(x) = 3\cos(4x) - 1$ reaches its minimum when $\cos(4x) = -1$ . This occurs when $4x = \pi + 2\pi k$ for integer $k$ . For $k = 0$ : $4x = \pi$ , so $x = \frac{\pi}{4}$ . At $x = 0$ : $r(0) = 3(1) - 1 = 2$ (maximum). At $x = \frac{\pi}{4}$ : $r(\frac{\pi}{4}) = 3(-1) - 1 = -4$ (minimum). Choice A gives the maximum. Choice B gives $r(\frac{\pi}{8}) = 3\cos(\frac{\pi}{2}) - 1 = -1$ (midline). Choice D gives the maximum again since $\cos(2\pi) = 1$ .
Two sinusoidal functions $f(x) = \sin(2x)$ and $g(x) = 2\sin(x)$ are compared on the interval $[0, 2\pi]$ . Which statement correctly describes their relationship?
Answer: Function $f(x)$ completes twice as many cycles as $g(x)$ and has the same amplitude throughout the interval When comparing sinusoidal functions, you need to analyze two key characteristics: amplitude (how high and low the function oscillates) and frequency (how often it completes a full cycle). For $f(x) = \sin(2x)$ , the coefficient 2 inside the sine function affects the period. The standard sine function $\sin(x)$ has period $2\pi$ , but $\sin(2x)$ has period $\frac{2\pi}{2} = \pi$ . This means $f(x)$ completes one full cycle every $\pi$ units, so over the interval $[0, 2\pi]$ , it completes 2 full cycles. The amplitude remains 1 since there's no coefficient in front of the sine. For $g(x) = 2\sin(x)$ , the coefficient 2 is outside the sine function, affecting amplitude but not period. This function has amplitude 2 and keeps the standard period of $2\pi$ , completing just 1 cycle over $[0, 2\pi]$ . Answer choice A is incorrect because the functions have different periods ( $\pi$ vs $2\pi$ ). Choice B is wrong because while $g(x)$ does have twice the amplitude, the functions don't complete the same number of cycles. Choice C incorrectly states that $f(x)$ has twice the amplitude when both functions actually have the same amplitude of 1 and 2 respectively, and mischaracterizes the amplitude relationship. Choice D correctly identifies that $f(x)$ completes twice as many cycles (2 vs 1) and recognizes that both functions have consistent amplitudes throughout the interval. Study tip: Remember that coefficients inside the function $\sin(bx)$ affect frequency/period, while coefficients outside $a\sin(x)$ affect amplitude.
A sinusoidal function has the form $y = A\cos(B(x - C)) + D$ . If this function has amplitude 4, period $3\pi$ , phase shift 2 units to the right, and midline $y = -1$ , what is the value of $A + B + C + D$ ?
Answer: $\frac{17}{3}$ From the given information: amplitude = 4, so $|A| = 4$ , thus $A = 4$ (assuming positive). Period = $3\pi$ , so $\frac{2\pi}{B} = 3\pi$ , giving $B = \frac{2}{3}$ . Phase shift = 2 units right, so $C = 2$ . Midline = -1, so $D = -1$ . Therefore: $A + B + C + D = 4 + \frac{2}{3} + 2 + (-1) = 5 + \frac{2}{3} = \frac{15}{3} + \frac{2}{3} = \frac{17}{3}$ .
A sinusoidal function has the form $y = A\cos(Bx) + C$ where $A > 0$ . If the function has a period of $\frac{4\pi}{3}$ , amplitude of 6, and passes through the point $(0, 2)$ , what is the value of $C$ ?
Answer: $C = -4$ When you encounter sinusoidal functions in the form $y = A\cos(Bx) + C$ , remember that each parameter has a specific role: $A$ controls amplitude, $B$ affects period, and $C$ represents the vertical shift (midline). Let's work through this systematically. First, we can find $A$ and $B$ from the given information. Since amplitude = 6 and $A > 0$ , we have $A = 6$ . The period formula is $\frac{2\pi}{B} = \frac{4\pi}{3}$ , so solving for $B$ : $B = \frac{2\pi}{\frac{4\pi}{3}} = \frac{3}{2}$ . Now our function is $y = 6\cos\left(\frac{3}{2}x\right) + C$ . To find $C$ , we use the given point $(0, 2)$ : $2 = 6\cos\left(\frac{3}{2} \cdot 0\right) + C$ $2 = 6\cos(0) + C$ $2 = 6(1) + C$ $2 = 6 + C$ $C = -4$ Looking at the wrong answers: Choice A ( $C = 8$ ) likely comes from adding the amplitude to the y-coordinate instead of recognizing the downward shift. Choice B ( $C = 2$ ) incorrectly assumes the y-intercept equals the vertical shift. Choice C ( $C = 6$ ) confuses the amplitude with the vertical shift parameter. The correct answer is D. Study tip: Always substitute known points into your function to find unknown parameters. Remember that $\cos(0) = 1$ , making $(0, y)$ points especially useful for finding the vertical shift in cosine functions.
A sinusoidal function has amplitude 5, period $\frac{2\pi}{3}$ , and midline $y = -2$ . If the function equals its midline value when $x = 0$ and is increasing at that point, which equation represents this function?
Answer: $y = 5\sin(3x) - 2$ Period $\frac{2\pi}{3}$ means $B = \frac{2\pi}{\frac{2\pi}{3}} = 3$ . At $x = 0$ , the function equals its midline and is increasing. For sine: $\sin(0) = 0$ (midline) and sine is increasing at $x = 0$ . For cosine: $\cos(0) = 1$ (maximum), not midline. Choice B starts at maximum, not midline. Choice C has wrong B value ( $\frac{3}{2}$ instead of 3). Choice D starts at minimum due to negative amplitude.
The graph of $y = \cos(x)$ is transformed to create the function $g(x) = 2\cos(3x - \pi) + 1$ . Which statement correctly describes the relationship between the period of $g(x)$ and the period of the parent cosine function?
Answer: The period of $g(x)$ is one-third the period of $\cos(x)$ because the coefficient 3 compresses horizontally The period of $\cos(x)$ is $2\pi$ . For $g(x) = 2\cos(3x - \pi) + 1$ , the coefficient 3 multiplying $x$ creates a horizontal compression, making the period $\frac{2\pi}{3}$ . This is indeed one-third of the original period. The amplitude (2) and vertical shift (+1) do not affect the period, and the phase shift ( $-\pi$ ) only translates the graph horizontally without changing the period.
If $p(x) = 2\sin(3x) + 5$ and $q(x) = -2\cos(3x) + 5$ , which statement about these functions is correct?
Answer: Both functions have the same amplitude and the same range of values When you encounter trigonometric functions in the form $a\sin(bx) + c$ or $a\cos(bx) + c$ , you need to identify three key characteristics: amplitude ( $|a|$ ), period ( $\frac{2\pi}{|b|}$ ), and midline ( $y = c$ ). For $p(x) = 2\sin(3x) + 5$ : The amplitude is $|2| = 2$ , the period is $\frac{2\pi}{3}$ , and the midline is $y = 5$ . Since the function oscillates 2 units above and below the midline, the range is $[3, 7]$ . For $q(x) = -2\cos(3x) + 5$ : The amplitude is $|-2| = 2$ (amplitude is always positive), the period is $\frac{2\pi}{3}$ , and the midline is $y = 5$ . The negative sign flips the cosine function vertically but doesn't change the amplitude. The range is also $[3, 7]$ . Both functions have identical amplitude (2) and range ([3, 7]), making C correct. A is wrong because both functions have the same midline at $y = 5$ , not different y-values. B incorrectly claims $p(x)$ has a larger amplitude than $q(x)$ —they're both 2. D makes the common error of thinking the negative sign affects the period, but it only reflects the function across the midline without changing the period. Remember: The coefficient's absolute value determines amplitude, and negative signs only affect vertical reflection, not amplitude or period. Focus on the structure $a\sin(bx) + c$ to quickly identify these properties.
The function $y = A\cos(Bx) + C$ passes through the points $(0, 7)$ , $(\frac{\pi}{4}, 2)$ , and $(\frac{\pi}{2}, -3)$ . What is the value of the amplitude $A$ ?
Answer: $A = 5$ From $(0, 7)$ : $7 = A\cos(0) + C = A + C$ . From $(\frac{\pi}{2}, -3)$ : $-3 = A\cos(\frac{B\pi}{2}) + C$ . If $B = 2$ , then $\cos(\pi) = -1$ , so $-3 = -A + C$ . Solving: $A + C = 7$ and $-A + C = -3$ gives $2A = 10$ , so $A = 5$ and $C = 2$ . Checking with middle point: $2 = 5\cos(\frac{\pi}{2}) + 2 = 5(0) + 2 = 2$ ✓. Choice A results from calculation error. Choice C confuses amplitude with the y-value at x = 0. Choice D uses $2A$ instead of $A$ .
The graph of $y = \cos(x)$ is transformed to create $y = -2\cos(3x) + 1$ . Which statement correctly describes the transformation?
Answer: The amplitude is doubled, period is divided by 3, graph is reflected over x-axis, then shifted up 1 unit From $y = -2\cos(3x) + 1$ : amplitude = $|-2| = 2$ (doubled), period = $\frac{2\pi}{3}$ (divided by 3), the negative sign reflects over the x-axis, and +1 shifts up 1 unit. Choice A incorrectly states period is tripled instead of divided by 3. Choice C confuses x-axis and y-axis reflection. Choice D incorrectly identifies amplitude as halved when $|-2| = 2$ means doubled.

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