Graphing Sine & Cosine - Algebra 3

1. The graph of a cosine function $f(x)=A\cos(Bx+C)+D$, with $A>0$ and $B>0$, has a local maximum at $\left(\frac{\pi}{4},1\right)$ and the next local minimum at $\left(\frac{3\pi}{4},-5\right)$. Which equation could define $f$?
   Answer: $f(x)=3\cos\left(2x-\frac{\pi}{2}\right)-2$ The maximum and minimum values give midline $D=\frac{1+(-5)}{2}=-2$ and amplitude $A=\frac{1-(-5)}{2}=3$. The horizontal gap from the maximum to the next minimum is $\frac{\pi}{2}$, which is half the period, so the period is $\pi$ and $B=2$. At a cosine maximum the argument is $0$; substituting $x=\frac{\pi}{4}$ gives $2\left(\frac{\pi}{4}\right)+C=0$, so $C=-\frac{\pi}{2}$. Thus $3\cos\left(2x-\frac{\pi}{2}\right)-2$ is correct. The choice with $+\frac{\pi}{2}$ makes the given point a minimum and the next point a maximum; the choice with $B=\frac{1}{2}$ gives period $4\pi$; the choice with amplitude $6$ has range from $4$ to $-8$, not from $1$ to $-5$.
2. The graph of $f(x)=4\sin(kx)+2$ has the property that the vertical distance between its maximum and minimum values equals the horizontal distance between an adjacent maximum and minimum. If $k>0$, what is $k$?
   Answer: $k=\frac{\pi}{8}$ The maximum value is $4+2=6$, and the minimum value is $-4+2=-2$, so the vertical distance between them is $6-(-2)=8$. For $y=4\sin(kx)+2$, an adjacent maximum and minimum are separated horizontally by half the period, $\pi/k$. Set $\pi/k=8$, giving $k=\pi/8$. The choice $8/\pi$ inverts the equation; $\pi/4$ treats the max-to-min gap as a full period; $4/\pi$ uses the amplitude incorrectly and also inverts the ratio.
3. What is the phase shift of $y=2\cos\left(\pi x+\frac{\pi}{2}\right)$ relative to $y=2\cos(\pi x)$?
   Answer: $\frac{1}{2}$ unit to the left Whenever you see a cosine function with a constant added inside the argument, your first instinct should be to factor out the coefficient of $x$. The phase shift is the horizontal translation of one graph relative to another, and the number $2$ out front only affects amplitude, not left/right movement. For $y=2\cos(\pi x+\frac{\pi}{2}$, factor the $\pi$: $$\cos\left(\pi x+\frac{\pi}{2}\right)=\cos\left(\pi\left(x+\frac{1}{2}\right)\right).$$ Comparing $\cos(\pi(x+\frac12))$ to $\cos(\pi x)$, the input $x$ has been replaced by $x+\frac12$. That replacement moves every point $\frac12$ unit to the left, because a point that used to occur at $x=0$ now occurs at $x=-\frac12$. So the phase shift is $\frac12$ unit left. The distractor "$\frac12$ unit to the right" recognizes the correct magnitude but forgets that $x+\frac12$ means left, not right; a right shift would require $x-\frac12$ inside. The choices with $\frac{\pi}{2}$ units come from grabbing the added constant $\frac{\pi}{2}$ directly as the shift without dividing by the frequency $\pi$. Since $\frac{\pi}{2}\div\pi=\frac12$, the magnitude cannot remain \frac{\pi}{2}}. And "$\frac{\pi}{2}$ units to the right" combines both the wrong magnitude and the wrong direction. Study takeaway: always rewrite $Bx+C$ as $B\left(x+\frac{C}{B}\right)$. The phase shift is $-\frac{C}{B}$, with the sign automatically telling you left or right.
4. Which sine function has amplitude $5$, period $4\pi$, midline $y=-2$, and a minimum at $x=0$?
   Answer: $y=5\sin\left(\frac{1}{2}x-\frac{\pi}{2}\right)-2$ Amplitude $5$ and midline $-2$ give a range from $-7$ to $3$. Period $4\pi$ requires $B=\frac{1}{2}$. At $x=0$, $5\sin\left(-\frac{\pi}{2}\right)-2=-7$, which is the minimum. The $+\frac{\pi}{2}$ choice gives the maximum at $x=0$; the $B=\frac{1}{2}$ negative-sine choice gives the midline value $-2$ at $x=0$; and the $2x$ choice has period $\pi$.
5. Which function has a frequency of $\frac{1}{4}$ cycle per unit?
   Answer: $y=3\cos\left(\frac{\pi}{2}x\right)$ Whenever you see a question about trigonometric functions, the key is to connect the coefficient of $x$ to the period and frequency. For a function like $y = A\cos(Bx)$, the period is $\frac{2\pi}{B}$, and the frequency is the reciprocal: $\frac{B}{2\pi}$ cycles per unit. Here, you need that frequency to be $\frac{1}{4}$, so set $\frac{B}{2\pi} = \frac{1}{4}$, which gives $B = \frac{\pi}{2}$. Therefore, the function $y=3\cos\left(\frac{\pi}{2}x\right)$ has the correct $B$. The amplitude $3$ does not affect frequency. Now check the other choices. The function $y=3\cos(2x)$ has $B=2$, giving frequency $\frac{2}{2\pi} = \frac{1}{\pi}$, not $\frac{1}{4}$. Similarly, $y=3\cos(4x)$ has $B=4$, so its frequency is $\frac{4}{2\pi} = \frac{2}{\pi}$, far too large. The function $y=3\cos\left(\frac{\pi}{4}x\right)$ has $B=\frac{\pi}{4}$, yielding frequency $\frac{\pi/4}{2\pi} = \frac{1}{8}$, which is half of what you want. Each of these traps comes from confusing the coefficient $B$ with the frequency itself—remember that $B$ is related to the angular frequency, but the number of cycles per unit requires dividing by $2\pi$. A quick study tip: to avoid this mistake, always write the formula $\text{frequency} = \frac{B}{2\pi}$ and solve for $B$ given the desired frequency. Also, note that multiplying by a constant like $3$ only changes the amplitude, not the frequency or period. Practice recognizing that the argument inside cosine is $\text{frequency} \times 2\pi \times x$ when you need a specific cycle count.
6. Consider $y=4\sin(3x+\pi)-2$. Which statement correctly describes its graph?
   Answer: The period is $\frac{2\pi}{3}$, the phase shift is $\frac{\pi}{3}$ to the left, and the midline is $y=-2$. When you see $y=4\sin(3x+\pi)-2$, compare it to the standard form $y=A\sin(B(x-C))+D$. The value of $B$ controls the period, $C$ controls the phase shift, and $D$ controls the midline. The key move is to factor the coefficient of $x$ out of the argument first: $$3x+\pi = 3\left(x+\frac{\pi}{3}\right)$$ So $B=3$, which means the period is $\frac{2\pi}{3}$. The expression $x+\frac{\pi}{3}$ is $x-\left(-\frac{\pi}{3}\right)$, so the phase shift is $\frac{\pi}{3}$ to the left. The vertical shift is $-2$, so the midline is $y=-2$. Therefore the correct description is period $\frac{2\pi}{3}$, phase shift $\frac{\pi}{3}$ to the left, midline $y=-2$. The other choices each contain a common mistake. Saying the period is $2\pi$ forgets to divide by $B=3$. Saying the phase shift is $\frac{\pi}{3}$ to the right confuses the sign: a plus sign inside the argument shifts the graph left, not right. Saying the phase shift is $\pi$ to the left uses the unsimplified $+\pi$ instead of dividing by 3 after factoring. Saying the midline is $y=2$ drops the negative sign on the vertical shift. Strategy: for any sinusoidal function, rewrite the inside as $B(x-C)$ before reading period, phase shift, and midline. Factor first, then interpret.
7. A rider's height above the ground is modeled by $h(t)=-20\cos\left(\frac{\pi}{10}t\right)+25$, where $t$ is in seconds. At what positive time $t$ does the rider first reach the maximum height?
   Answer: $10$ seconds The maximum height is $45$, reached when $\cos\left(\frac{\pi}{10}t\right)=-1$. Then $\frac{\pi}{10}t=\pi+2\pi k$, so $t=10+20k$. The first positive time is $10$ seconds. At $5$ and $15$ seconds the rider is at the midline height $25$, and at $20$ seconds the rider is at the minimum height $5$.
8. Which equation is equivalent to $y=4\sin\left(2x+\frac{\pi}{2}\right)-3$?
   Answer: $y=4\cos(2x)-3$ When you see a sine expression with a phase shift of $\frac{\pi}{2}$, you're really being tested on cofunction identities. The key fact is $\sin(\theta+\frac{\pi}{2})=\cos\theta$. Applying it with $\theta=2x$ gives $$y=4\sin\left(2x+\frac{\pi}{2}\right)-3=4\cos(2x)-3,$$ so the equivalent equation is the one reading $y=4\cos(2x)-3$. The choice $y=-4\cos(2x)-3$ gets the sign wrong: $\sin(\theta+\frac{\pi}{2})$ is positive $\cos\theta$, not its negative. The choice $y=4\cos(2x)+3$ keeps the correct cosine term but shifts the graph in the wrong vertical direction; the original has $-3$, so the vertical shift must stay $-3$. The choice $y=4\cos\left(2x-\frac{\pi}{2}\right)-3$ is a different expression: $\cos(2x-\frac{\pi}{2})=\sin 2x$, so that option equals $4\sin 2x-3$, not the original $4\sin(2x+\frac{\pi}{2})-3$. It confuses which sign inside the argument produces the cofunction shift. Study tip: memorize the two core cofunction shifts: $\sin(x+\frac{\pi}{2})=\cos x$ and $\cos(x-\frac{\pi}{2})=\sin x$. On this exam, a $\frac{\pi}{2}$ phase shift is almost always a signal to swap sine and cosine — just check the sign and preserve the vertical shift exactly.
9. Which statement about $y=-2\cos(4x-\pi)+3$ is true?
   Answer: Amplitude $2$, period $\frac{\pi}{2}$, midline $y=3$, maximum value $5$. Whenever you see a transformed cosine like $y=-2\cos(4x-\pi)+3$, think of the standard form $y=A\cos(Bx-C)+D$. The amplitude is $|A|$, the period is $\frac{2\pi}{|B|}$, the midline is $y=D$, and the maximum and minimum values are $D+|A|$ and $D-|A|$. Here, $A=-2$, so the amplitude is $2$. Since $B=4$, the period is $\frac{2\pi}{4}=\frac{\pi}{2}$. The midline is $y=3$, because $D=3$. Therefore the maximum is $3+2=5$ and the minimum is $3-2=1$. So the true statement is: amplitude $2$, period $\frac{\pi}{2}$, midline $y=3$, maximum value $5$. Now look at the incorrect statements. The one with amplitude $2$, period $2\pi$, midline $y=3$, maximum value $5$ uses the wrong period: it ignores the horizontal compression caused by the coefficient $4$. The one with amplitude $-2$, period $\frac{\pi}{2}$, midline $y=-3$, minimum value $-5$ has two issues: amplitude is never negative, and the midline should be $y=3$, not $y=-3$. The statement with amplitude $4$, period $\frac{\pi}{2}$, midline $y=3$, maximum value $7$ mistakes the x-coefficient $4$ for the amplitude; the amplitude comes from the coefficient in front of cosine, which is $2$ after taking absolute value. Your takeaway: rewrite the function in standard form, then compute amplitude, period, and midline systematically. Always use $|A|$ for amplitude, and remember the maximum and minimum are just the midline plus or minus the amplitude.
10. A sinusoidal graph has amplitude $3$, period $4\pi$, midline $y=-1$, and is shifted $\frac{\pi}{2}$ to the right relative to $y=\cos x$. Which equation represents the graph?
    Answer: $y=3\cos\left(\frac{1}{2}x-\frac{\pi}{4}\right)-1$ Whenever you see a sinusoidal graph question, start by translating the key features into the general form $y=A\cos(B(x-C))+D$. Amplitude is $|A|=3$; midline $y=-1$ gives $D=-1$; period $4\pi$ means $B=\frac{2\pi}{4\pi}=\frac12$. The graph is shifted $\frac{\pi}{2}$ right relative to cosine, so inside the parentheses you need $x-\frac{\pi}{2}$. Since $B=\frac12$, the argument becomes $\frac12\left(x-\frac{\pi}{2}\right)=\frac12 x-\frac{\pi}{4}$. Therefore the correct equation is $y=3\cos\left(\frac12 x-\frac{\pi}{4}\right)-1$. The choice with $y=3\cos\left(\frac12 x-\frac{\pi}{2}\right)-1$ has the right period and midline, but it equals $3\cos\left(\frac12(x-\pi)\right)-1$, a right shift of $\pi$, not $\frac{\pi}{2}$. The choice $y=3\cos\left(2x-\frac{\pi}{2}\right)+1$ uses $B=2$, giving period $\pi$, and has midline $+1$ instead of $-1$. The choice $y=3\cos\left(\frac12 x+\frac{\pi}{4}\right)-1$ equals $3\cos\left(\frac12(x+\frac{\pi}{2})\right)-1$, which shifts left, not right. A solid strategy: always factor out $B$ before reading the phase shift. Writing $\cos(Bx-C)$ as $\cos\left(B\left(x-\frac CB\right)\right)$ reveals the true horizontal shift and helps you avoid sign and scaling traps.
11. A cosine curve has a maximum at $x=1$, a minimum at $x=4$, midline $y=2$, and amplitude $3$. Which equation could represent it?
    Answer: $y=3\cos\left(\frac{\pi}{3}(x-1)\right)+2$ When you see a cosine curve described by its maximum, minimum, midline, and amplitude, connect each feature to the general form $y=A\cos(B(x-h))+k$. The midline $y=2$ gives $k=2$. The amplitude $3$ gives $A=3$. A positive cosine curve reaches its maximum when $\cos=1$, so the location of the maximum tells you the horizontal shift: maximum at $x=1$, so $h=1$. That means the correct form must be $y=3\cos(B(x-1))+2$. Now find $B$: the distance from a maximum to the next minimum is half a period. Here that distance is $4-1=3$, so the full period is $6$. Since $B=\frac{2\pi}{\text{period}}$, $B=\frac{2\pi}{6}=\frac{\pi}{3}$. Therefore the equation is $y=3\cos\left(\frac{\pi}{3}(x-1)\right)+2$. The choice with $\frac{\pi}{6}(x-1)$ has the right shift and midline, but its period is $12$, not $6$, so the minimum would occur at $x=7$, not $x=4$. The choice with $\frac{\pi}{3}(x-4)$ has the correct period but shifts the maximum to $x=4$. The sine version $y=3\sin\left(\frac{\pi}{3}(x-1)\right)+2$ gives the midline at $x=1$, not a maximum. For any trig graph question, use the maximum and minimum x-values to find both the horizontal shift and half the period. That single step usually determines the correct equation.
12. Which sequence of transformations maps $y=\cos x$ to $y=3\cos(2x-\pi)-1$?
    Answer: Vertically stretch by $3$, horizontally compress by a factor of $\frac{1}{2}$, shift right $\frac{\pi}{2}$, shift down $1$. Whenever you are asked to describe transformations from a base trig function, rewrite the equation as $y=a\cos(b(x-c))+d$. Then each parameter maps directly: amplitude $|a|$, period $\frac{2\pi}{|b|}$, horizontal shift $c$, vertical shift $d$. For $y=3\cos(2x-\pi)-1$, factor the inside: $2x-\pi=2(x-\frac{\pi}{2})$, so the function is $3\cos[2(x-\frac{\pi}{2})]-1$. Thus $a=3$ means vertical stretch by $3$; $b=2$ means horizontal compression by a factor of $\frac12$; $c=\frac{\pi}{2}$ means shift right $\frac{\pi}{2}$; $d=-1$ means shift down $1$. That matches the choice with vertical stretch by $3$, horizontal compression by $\frac12$, shift right $\frac{\pi}{2}$, and shift down $1$. The distractor with "shift right $\pi$" comes from forgetting to factor out the $2$: the phase shift is $\frac{\pi}{2}$, not $\pi$. The one with "horizontally stretch by $2$" reverses the effect of $b>1$, which compresses the graph horizontally. The one with "shift left $\frac{\pi}{2}$" confuses the sign inside: $x-\frac{\pi}{2}$ moves the graph right, not left. Study tip: always rewrite as $y=a\cos(b(x-c))+d$ first. If $b$ is not factored out, you cannot correctly read the phase shift.
13. In $y=a\cos(bx-c)+d$, assume $a>0$ and $b>0$. If $a$ is halved and $b$ is doubled while $c$ and $d$ stay the same, which statement about the new graph is true?
    Answer: Amplitude is halved, period is halved, and phase shift is halved. Whenever you see transformations of $y=a\cos(bx-c)+d$, identify each parameter's job: amplitude is $a$, period is $2\pi/b$, and phase shift is $c/b$. Here $a>0$ and $b>0$, with $c$ and $d$ fixed. If $a$ is halved, the amplitude becomes half of its original value. If $b$ is doubled, the period becomes $2\pi/(2b)=\pi/b$, which is half the original period $2\pi/b$. The phase shift becomes $c/(2b)$, which is also half the original $c/b$. The vertical shift $d$ stays the same. So the true statement is: amplitude is halved, period is halved, and phase shift is halved. The choice saying "amplitude is halved, period is doubled, and phase shift is halved" gets the period backwards: doubling $b$ shrinks the period, not lengthens it. The choice saying "amplitude is halved, period is halved, and phase shift is doubled" treats phase shift as if it moved with $b$, but since phase shift is $c/b$, doubling $b$ actually halves it. The choice saying "amplitude is unchanged, period is unchanged, and phase shift is unchanged" ignores that changing $a$ and $b$ directly changes amplitude and period. A reliable strategy is to rewrite the function as $y=a\cos\left(b\left(x-\frac{c}{b}\right)\right)+d$ before comparing. That instantly shows how amplitude, period, and phase shift each respond to changes in $a$ and $b$.
14. A sine graph has midline $y=-1$ and amplitude $3$. It increases through its midline at $x=1$ and reaches its first maximum at $x=3$. Which equation represents the graph?
    Answer: $y=3\sin\left(\frac{\pi}{4}(x-1)\right)-1$ Whenever you see a sine graph described in words, translate the clues into the standard form $y=A\sin(B(x-h)))+k$. The midline $y=-1$ tells you $k=-1$; the amplitude $3$ tells you the vertical coefficient is $3$. Now locate horizontal behavior: a sine curve that increases through its midline acts like $\sin x$ at $x=0$. Since this graph increases through midline at $x=1$, the phase shift is $x-1$. From an upward midline crossing to the first maximum is one-quarter cycle. Here that distance is $3-1=2$, so one-quarter period $=2$ and the full period $=8$. For $y=3\sin(B(x-1))-1$, $\frac{2\pi}{B}=8$, so $B=\frac{\pi}{4}$. Thus the matching equation is $y=3\sin\left(\frac{\pi}{4}(x-1)\right)-1$. The choice using $\frac{\pi}{2}(x-1)$ has the right midline, amplitude, and shift, but its period is only $4$, so by $x=3$ it has already cycle back to midline rather than reaching a maximum. The cosine choice $3\cos\left(\frac{\pi}{4}(x-1)\right)-1$ has the correct period, but a cosine curve starts at a maximum at $x=1$, not at the midline going up. The choice with $x+1$ shifts the curve in the wrong direction: it crosses midline at $x=-1$ and reaches its maximum at $x=1$, opposite the given condition. A useful check: test the two points $x=1$ and $x=3$ in each equation. The correct graph must give $-1$ at $x=1$ and $2$ at $x=3$; this quickly eliminates the incorrect periods, signs, and cosine setup.
15. A cosine graph has range $[-5,3]$ and period $\pi$. Which equation could represent it?
    Answer: $y=4\cos(2x)-1$ Whenever you see a cosine graph described by its range and period, think of the general form $y=A\cos(Bx)+D$. Your job is to extract the amplitude $|A|$, the midline $D$, and the frequency factor $B$ from those two facts. The midpoint of the range $[-5,3]$ is $D=\frac{-5+3}{2}=-1$, so the graph is shifted down 1. The distance from midline to either endpoint is $4$, so $A=4$. And for cosine, the period is $\frac{2\pi}{B}$; a period of $\pi$ means $B=2$. Therefore the equation is $y=4\cos(2x)-1$. Each wrong answer misses one of these pieces: $y=4\cos(2x)+1$ has the right amplitude and period but shifts up, giving range $[-3,5]$. $y=8\cos(2x)-1$ has the right shift and period but amplitude 8, giving range $[-9,7]$. $y=4\cos(x)-1$ has the right amplitude and shift but period $2\pi$, since $B=1$, not $\pi$. These distractors test whether you can confidently connect range to midline/amplitude and period to $B$. When you see a trig graph question, first find midline: it is the average of max and min. Then find amplitude: half the distance between max and min. Finally get $B$ from period using $$B=\frac{2\pi}{\text{period}}
16. Which statement correctly describes the graph of $f(x)=-2\cos(3x+\pi)-4$ when it is simplified to the equivalent positive-amplitude cosine form?
    Answer: Amplitude $2$, period $\frac{2\pi}{3}$, midline $y=-4$, phase shift $0$. Using $\cos(3x+\pi)=-\cos(3x)$, the function becomes $f(x)=2\cos(3x)-4$. Therefore its amplitude is $2$, its period is $\frac{2\pi}{3}$, its midline is $y=-4$, and its horizontal shift is $0$. The choice with midline $y=4$ has the wrong sign; the choice with amplitude $-2$ confuses amplitude with a vertical reflection; the choice with period $2\pi$ ignores the horizontal compression by $3$.
17. For $y=3\cos(2x-\pi)+1$, what is the $x$-coordinate of the first maximum to the right of the $y$-axis?
    Answer: $x=\frac{\pi}{2}$ Whenever you see a cosine function like $y=3\cos(2x-\pi)+1$, remember that a maximum occurs when the cosine argument equals a multiple of $2\pi$, because $\cos(0)=1$ is the largest cosine value. Set the inside equal to $2\pi k$: $2x-\pi = 2\pi k \quad \Rightarrow \quad x=\frac{\pi}{2}+\pi k$ For the first maximum to the right of the $y$-axis, choose the smallest nonnegative solution. With $k=0$, you get $x=\frac{\pi}{2}$, so that is the correct coordinate. The choice $x=\pi$ is a trap: substituting gives $2\pi-\pi=\pi$, and $\cos\pi=-1$, so that point is actually a minimum, not a maximum. Similarly, $x=0$ gives $2(0)-\pi=-\pi$, so $\cos(-\pi)=-1$, also a minimum. The choice $x=\frac{\pi}{4}$ makes the argument $2(\frac{\pi}{4})-\pi=-\frac{\pi}{2}$, where $\cos=0$, so that point is on the midline, not at a peak. The key takeaway: for maximums of cosine, solve $\text{argument}=2\pi k$, not zero generically. Here, solving $2x-\pi=0$ directly already gives $x=\frac{\pi}{2}$, but if there were several peaks, you would add multiples of the period, $\pi$, and then choose the first one to the right of the $y$-axis.

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