Graphs of sine and cosine: amplitude, period and phase shift

Worked solutions

Problem 1

1. For $3\sin(4x) - 1$: $A = 3$, $B = 4$, $D = -1$Match it to $A\sin(B(x - C)) + D$ with $C = 0$.
2. Amplitude $3$, period $2\pi/4 = \tfrac\pi2$, midline $y = -1$The definitions.
3. Range $[-1 - 3, -1 + 3] = [-4, 2]$Maximum $D + |A|$, minimum $D - |A|$.
4. For $-2\cos(x/3) + 5$: $A = -2$, $B = \tfrac13$, $D = 5$$x/3 = \tfrac13x$.
5. Amplitude $|{-2}| = 2$, period $2\pi/\tfrac13 = 6\pi$, midline $y = 5$A small $B$ means a long period.
6. $3\sin(4x) - 1$: amplitude $3$, period $\tfrac\pi2$, midline $y = -1$, range $[-4, 2]$. $-2\cos(x/3) + 5$: amplitude $2$, period $6\pi$, midline $y = 5$, range $[3, 7]$The negative $A$ does not shrink anything: it starts the cosine at its minimum, $3$ at $x = 0$, instead of its maximum.

Problem 2

1. $2x - \tfrac\pi3 = 2\big(x - \tfrac\pi6\big)$Factor $B = 2$ out of the whole argument.
2. Amplitude $2$, period $2\pi/2 = \pi$, phase shift $\tfrac\pi6$Now it matches $A\sin(B(x - C)) + D$ with $C = \tfrac\pi6$.
3. The maximum is where $2\big(x - \tfrac\pi6\big) = \tfrac\pi2 + 2\pi k$$\sin$ peaks at $\tfrac\pi2$ plus whole periods.
4. $x = \tfrac\pi6 + \tfrac\pi4 + \pi k = \tfrac{5\pi}{12} + \pi k$Divide by $2$, then add $\tfrac\pi6$.
5. Amplitude $2$, period $\pi$, phase shift $\tfrac\pi6$ to the right; the first maximum for $x \ge 0$ is at $x = \tfrac{5\pi}{12} \approx 1.309$, where $y = 3$$k = -1$ gives $\tfrac{5\pi}{12} - \pi < 0$. Equivalently, the peak of $\sin(2x)$ at $\tfrac\pi4$ moved right by $\tfrac\pi6$.

Problem 3

1. $\tfrac x2 - \tfrac\pi4 = \tfrac12\big(x - \tfrac\pi2\big)$Factor out $B = \tfrac12$: phase shift $\tfrac\pi2$.
2. Period $2\pi/\tfrac12 = 4\pi$, so the key points are $\pi$ apartFive points split one period into four equal quarters.
3. $x = \tfrac\pi2, \tfrac{3\pi}2, \tfrac{5\pi}2, \tfrac{7\pi}2, \tfrac{9\pi}2$Start at the phase shift and add a quarter period each time.
4. $y = 2, 5, 2, -1, 2$Midline $2$, then $2 + 3$, midline, $2 - 3$, midline: the pattern of $\sin$.
5. $(\tfrac\pi2, 2)$, $(\tfrac{3\pi}2, 5)$, $(\tfrac{5\pi}2, 2)$, $(\tfrac{7\pi}2, -1)$, $(\tfrac{9\pi}2, 2)$At $x = \tfrac\pi2$ the argument is $0$, the start of a sine cycle, so the graph crosses the midline going up. Join the points with the smooth sine shape and repeat every $4\pi$.

Problem 4

1. $\sin\big(x + \tfrac\pi2\big) = \sin x\cos\tfrac\pi2 + \cos x\sin\tfrac\pi2 = \cos x$The sum formula, with $\cos\tfrac\pi2 = 0$ and $\sin\tfrac\pi2 = 1$.
2. $5\cos(3x) = 5\sin\big(3x + \tfrac\pi2\big)$Step 1 with $3x$ in place of $x$.
3. $3x + \tfrac\pi2 = 3\big(x + \tfrac\pi6\big)$Factor out $B = 3$: $C = -\tfrac\pi6$.
4. $\cos\big(x + \tfrac\pi2\big) = \cos x\cos\tfrac\pi2 - \sin x\sin\tfrac\pi2 = -\sin x$The cosine sum formula.
5. $5\cos(3x) = 5\sin\big(3(x + \tfrac\pi6)\big)$, phase shift $-\tfrac\pi6$; $-2\sin x = 2\cos\big(x + \tfrac\pi2\big)$A cosine is a sine a quarter period earlier: here the period is $\tfrac{2\pi}3$, and a quarter of it is $\tfrac\pi6$. Any sinusoid can be written with either function; only $C$ changes. Combining a sine and a cosine of the same $B$ into one wave is the $a\sin x + b\cos x$ identity of the trig identities page.

Problem 5

1. $A = \tfrac{7 - (-1)}2 = 4$, $D = \tfrac{7 + (-1)}2 = 3$Half the distance from minimum to maximum, and the level halfway between them.
2. A maximum to the next minimum is half a period: $4 - 1 = 3$, so the period is $6$A cycle goes maximum, minimum, maximum.
3. $B = 2\pi/6 = \tfrac\pi3$Period $= 2\pi/B$.
4. $C = 1$$\cos$ peaks where its argument is $0$, so put that peak at $x = 1$.
5. As a sine, the peak sits a quarter period, $\tfrac64 = 1.5$, after the start of the cycle: $C = 1 - 1.5 = -\tfrac12$$\sin$ peaks a quarter period after its midline crossing.
6. $y = 4\cos\big(\tfrac\pi3(x - 1)\big) + 3 = 4\sin\big(\tfrac\pi3(x + \tfrac12)\big) + 3$Check: at $x = 1$ the cosine is $\cos 0 = 1$, so $y = 7$; at $x = 4$ it is $\cos\pi = -1$, so $y = -1$. In the sine form the arguments are $\tfrac\pi2$ and $\tfrac{3\pi}2$, with the same values.

Problem 6

1. Period $2\pi\big/\tfrac{2\pi}{365} = 365$ days, midline $12$ hours, amplitude $3$ hoursThe model repeats every year, between $9$ and $15$ hours.
2. At $t = 172$: the argument is $\tfrac{2\pi}{365} \cdot 92 \approx 1.5837$Close to $\tfrac\pi2 \approx 1.5708$.
3. $H(172) \approx 12 + 3(0.99992) \approx 15.00$Day $172$ is about June 21, the longest day: the peak is at $t = 80 + \tfrac{365}4 = 171.25$.
4. $H(t) > 14$ when $\sin\theta > \tfrac23$, with $\theta = \tfrac{2\pi}{365}(t - 80)$Subtract $12$, divide by $3$.
5. $\arcsin\tfrac23 \approx 0.7297 < \theta < \pi - 0.7297 \approx 2.4119$On one cycle the sine is above $\tfrac23$ between the two angles where it equals $\tfrac23$.
6. $t - 80 = \tfrac{365}{2\pi}\theta$, so $122.39 < t < 220.11$Undo the scaling and the shift.
7. Period $365$ days, midline $12$ hours, amplitude $3$ hours; $H(172) \approx 15.00$ hours; more than $14$ hours from day $123$ to day $220$That is $98$ whole days, from a stretch of about $97.7$ days: $H(122) \approx 13.98$ and $H(123) \approx 14.02$, $H(220) \approx 14.004$ and $H(221) \approx 13.97$.

Problem 7

1. Angular frequency $B = 120\pi \approx 376.99$ radians per secondThe coefficient of $t$.
2. Frequency $B/(2\pi) = 60$ cycles per second, $60$ HzOne cycle is $2\pi$ radians.
3. Period $1/60 \approx 0.01667$ sPeriod $= 2\pi/B$, the reciprocal of the frequency.
4. Mean of $\sin^2$ over a period is $\tfrac12$$\sin^2 u = \tfrac12(1 - \cos 2u)$, and $\cos 2u$ averages to $0$ over whole cycles.
5. $120\pi \approx 377$ rad/s, $60$ Hz, period $\tfrac1{60}$ s, amplitude $170$ V, RMS $170/\sqrt2 \approx 120.2$ VThe "$120$ volts" on a North American outlet is this RMS value; the peak is $170$. Hertz count cycles, radians per second count angle, and they differ by $2\pi$.

Problem 8

1. $\sin(2x)$ has period $\pi$ and $\cos(3x)$ has period $\tfrac{2\pi}3$$2\pi/B$ for each.
2. If $T$ is a period of $f$, it is one of $f'' = -4\sin(2x) - 9\cos(3x)$ tooDifferentiate $f(x + T) = f(x)$ twice.
3. So $T$ is a period of $9f + f'' = 5\sin(2x)$ and of $4f + f'' = -5\cos(3x)$Each combination cancels one term, so each term repeats on its own.
4. $T = \pi j = \tfrac{2\pi}3 k$ for positive integers $j$, $k$, so $3j = 2k$; the smallest is $j = 2$, $k = 3$, $T = 2\pi$A sinusoid repeats only after whole periods: $2\pi$ is the least common multiple of $\pi$ and $\tfrac{2\pi}3$, and $f(x + 2\pi) = f(x)$.
5. For $g$: a period would need $T = 2\pi j = \tfrac{2\pi}{\sqrt2}k$, so $\sqrt2 = k/j$The same argument, with $g'' = -\sin x - 2\sin(\sqrt2\,x)$, makes $T$ a period of each term; but $\sqrt2$ is irrational, so no such $j$, $k$ exist.
6. $f$ has period $2\pi$; $g$ is not periodicA sum of sinusoids is periodic exactly when the ratio of their frequencies is rational, and its period is the least common multiple of theirs. $g$ comes close to repeating, but never exactly.

Problem 9

1. A horizontal stretch by $4$ replaces $x$ with $\tfrac x4$; a shift right by $\pi$ replaces $x$ with $x - \pi$Each moves the point that used to be at $x$.
2. (a) Stretch: $\sin\tfrac x4$. Then shift: $\sin\tfrac{x - \pi}4$The shift replaces $x$ in the current equation.
3. (b) Shift: $\sin(x - \pi)$. Then stretch: $\sin\big(\tfrac x4 - \pi\big) = \sin\big(\tfrac14(x - 4\pi)\big)$The stretch also stretches the shift that was already applied.
4. (a) $y = \sin\big(\tfrac14(x - \pi)\big)$, phase shift $\pi$; (b) $y = \sin\big(\tfrac14(x - 4\pi)\big)$, phase shift $4\pi$; both have period $8\pi$The order matters for the shift but not the period. In $A\sin(B(x - C)) + D$ the $C$ is the shift applied last, after the stretch, which is why it is read with $B$ factored out.

Problem 10

1. Pair $i$ has period $2\pi/\omega_i = 2\pi \cdot 10000^{2i/d}$Period $= 2\pi/B$ with $B = \omega_i$.
2. $i = 0$: $2\pi \approx 6.28$; $i = 128$: $2\pi \cdot 10000^{1/2} = 200\pi \approx 628.3$$2i/d = 0$ and $256/512 = \tfrac12$.
3. $i = 255$: $2\pi \cdot 10000^{510/512} \approx 2\pi \cdot 9646.6 \approx 60611$The periods grow geometrically from $2\pi$ to almost $20000\pi$.
4. With $a = \omega_i p$ and $b = \omega_i k$: $\sin(a + b) = \cos b\,\sin a + \sin b\,\cos a$The sine sum formula.
5. $\cos(a + b) = -\sin b\,\sin a + \cos b\,\cos a$The cosine sum formula.
6. Periods $2\pi \approx 6.28$, $200\pi \approx 628.3$ and about $60611$ positions; $R_k = \begin{bmatrix}\cos\omega_i k & \sin\omega_i k\\ -\sin\omega_i k & \cos\omega_i k\end{bmatrix}$ sends $(\sin\omega_i p, \cos\omega_i p)$ to $(\sin\omega_i(p + k), \cos\omega_i(p + k))$Steps 4 and 5 are the two rows of $R_k$ applied to $(\sin a, \cos a)$, and only $b = \omega_i k$ appears in $R_k$. It is a rotation, so an offset of $k$ positions is the same linear map wherever it happens. Fast pairs separate nearby positions; slow pairs tell far-apart positions apart.