Graphs of Sine and Cosine

Study Sheet

Graphs of Sine and Cosine

Amplitude, period, phase shift, vertical shift, and modeling

The Basic Graphs y=sin⁡xy=\sin x and y=cos⁡xy=\cos x

Concept
The parent sine and cosine curves
xy−11π/2π3π/2y = sin xy = cos x
Sine (solid) starts on the midline; cosine (dashed) starts at its maximum.

Both y=sin⁡xy=\sin x and y=cos⁡xy=\cos x are periodic with period 2π2\pi: the graph repeats every 2π2\pi units. Each has amplitude 11, midline y=0y=0, maximum value 11, and minimum value −1-1.

  • y=sin⁡xy=\sin x starts at the midline: (0,0)(0,0), rises to a max, returns, falls to a min, returns.
  • y=cos⁡xy=\cos x starts at a maximum: (0,1)(0,1), falls to the midline, to a min, and back up.

Five-point method: divide one period into four equal pieces. For [0,2π][0,2\pi] the key xx-values are

0,π2,π,3π2,2π.0,\quad \dfrac{\pi}{2},\quad \pi,\quad \dfrac{3\pi}{2},\quad 2\pi.

Evaluate the function at these five points, then connect with a smooth wave.

The graphs of sine and cosine are waves: amplitude ∣A∣|A| (height), period 2πB\tfrac{2\pi}{B} (length of one cycle), phase shift CB\tfrac CB (horizontal slide) and midline DD in y=Asin⁡(Bx−C)+Dy = A\sin(Bx - C) + D. Cosine is sine shifted left by π2\tfrac\pi2. Sketch one period from the five key points and repeat.

Example
Worked example: graph y=sin⁡xy=\sin x with the five-point method
xy−11π/2π3π/2
The five key points of one period of y=sin⁡xy=\sin x.

Amplitude =1=1, period =2π=2\pi, midline y=0y=0. Split [0,2π][0,2\pi] into quarters and evaluate:

(0,0),(π2,1),(π,0),(3π2,−1),(2π,0).(0,0),\quad \left(\tfrac{\pi}{2},1\right),\quad (\pi,0),\quad \left(\tfrac{3\pi}{2},-1\right),\quad (2\pi,0).

The first quarter climbs from the midline to the maximum, the second returns to the midline, the third reaches the minimum, and the fourth comes back. Join the five points with a smooth wave, rounded at the top and bottom.

Side note
Where it goes wrong

Reading the phase shift as CC instead of CB\tfrac CB. Factor: sin⁡(2x−π)=sin⁡(2(x−π2))\sin(2x - \pi) = \sin\big(2(x - \tfrac\pi2)\big) shifts by π2\tfrac\pi2, not π\pi.

Tip

Tip: Cosine is just sine shifted π2\tfrac{\pi}{2} to the left: cos⁡x=sin⁡ ⁣(x+π2)\cos x=\sin\!\left(x+\tfrac{\pi}{2}\right). Remember “sine starts at the middle, cosine starts at the top.”

Try it
Quick check

Use the five-point method to list the key points of y=cos⁡xy=\cos x on [0,2π][0,2\pi].

Show the answer

Answer: (0,1)(0,1), (π2,0)\left(\tfrac{\pi}{2},0\right), (π,−1)(\pi,-1), (3π2,0)\left(\tfrac{3\pi}{2},0\right), (2π,1)(2\pi,1). The xx-values are the same quarter-points as for sine; only the yy-values change, because cosine starts at the top of its cycle instead of on the midline.

Amplitude ∣a∣|a| and Vertical Reflection

Concept
The role of aa in y=asin⁡xy=a\sin x and y=acos⁡xy=a\cos x

The number aa stretches the graph vertically.

  • Amplitude =∣a∣=max−min2=|a|=\dfrac{\text{max}-\text{min}}{2}: the distance from the midline to a peak.
  • Maximum value =∣a∣=|a|, minimum value =−∣a∣=-|a| (when midline is y=0y=0).
  • If a<0a<0, the graph is reflected across the xx-axis (flipped upside down).
  • aa does not change the period or the midline.

Multiplying by aa multiplies every output of sin⁡x\sin x by aa. The zeros stay zeros and the xx-values of the peaks do not move, so only the heights change: a peak of 11 becomes aa. When aa is negative, every positive output becomes negative and vice versa, which is a reflection across the xx-axis.

Example
Worked example: y=3cos⁡xy=3\cos x and y=−2sin⁡xy=-2\sin x
xy−3−2−1123π/2π3π/2y = 3cos xy = −2sin x
y=3cos⁡xy=3\cos x (solid) swings between ±3\pm3; y=−2sin⁡xy=-2\sin x (dashed) dips first.

y=3cos⁡xy=3\cos x: amplitude =∣3∣=3=|3|=3, max 33, min −3-3, period 2π2\pi, midline y=0y=0. y=−2sin⁡xy=-2\sin x: amplitude =∣−2∣=2=|-2|=2; because a=−2<0a=-2<0 the sine wave is reflected, so it goes down first from (0,0)(0,0).

Tip

Tip: A negative aa never makes the amplitude negative. Amplitude is always ∣a∣≥0|a|\ge 0; the sign only tells you whether the wave is flipped.

Try it
Quick check

For y=−12cos⁡xy=-\tfrac12\cos x, give the amplitude, the value at x=0x=0, and the xx-value in [0,2π][0,2\pi] where the maximum occurs.

Show the answer

Answer: amplitude 12\tfrac12; y(0)=−12y(0)=-\tfrac12; the maximum 12\tfrac12 is at x=πx=\pi. The negative aa flips cosine, so instead of starting at its top it starts at its bottom, −12cos⁡0=−12-\tfrac12\cos 0=-\tfrac12. Its top is where cos⁡x\cos x is lowest: cos⁡π=−1\cos\pi=-1 gives −12(−1)=12-\tfrac12(-1)=\tfrac12.

Period and the Coefficient bb

Concept
Period =2πb=\dfrac{2\pi}{b}

In y=asin⁡(bx)y=a\sin(bx) or y=acos⁡(bx)y=a\cos(bx) (with b>0b>0), the coefficient bb controls how fast the wave cycles:

 Period=2πb sob=2πPeriod.\boxed{\ \text{Period}=\dfrac{2\pi}{b}\ }\qquad\text{so}\qquad b=\dfrac{2\pi}{\text{Period}}.
  • b>1b>1 compresses the graph: more cycles, shorter period.
  • 0<b<10<b<1 stretches the graph: fewer cycles, longer period.

To find five key points, split one period into four equal steps of width 14⋅2πb=π2b\tfrac{1}{4}\cdot\tfrac{2\pi}{b}=\tfrac{\pi}{2b}.

Sine finishes one cycle when its input runs from 00 to 2π2\pi. With input bxbx, that happens when bxbx runs from 00 to 2π2\pi, i.e. when xx runs from 00 to 2πb\dfrac{2\pi}{b}. A larger bb feeds the sine its 2π2\pi sooner, so the cycle is shorter; b=2b=2 halves the period.

Example
Worked example: y=sin⁡2xy=\sin 2x
xy−11π/2π3π/2y = sin 2xone period = π
y=sin⁡2xy=\sin 2x (solid) fits two cycles where y=sin⁡xy=\sin x (dashed) fits one.

Here b=2b=2, so period =2π2=π=\dfrac{2\pi}{2}=\pi. The wave completes a full cycle in π\pi, i.e. two cycles on [0,2π][0,2\pi]. Key xx-values, stepping by π4\tfrac{\pi}{4}:  0, π4, π2, 3π4, π.\ 0,\ \tfrac{\pi}{4},\ \tfrac{\pi}{2},\ \tfrac{3\pi}{4},\ \pi.

Tip

Tip: Amplitude and period are independent. In y=4sin⁡2xy=4\sin 2x the amplitude is 44 and the period is still 2π2=π\dfrac{2\pi}{2}=\pi — the 44 has no effect on the period.

Try it
Quick check

Find the period of (a) y=3sin⁡x2y=3\sin\dfrac{x}{2} and (b) y=cos⁡4xy=\cos 4x. For (b), list the five key xx-values of the first cycle.

Show the answer

(a) 4π4\pi. b=12b=\tfrac12, so the period is 2π1/2=4π\dfrac{2\pi}{1/2}=4\pi; the 33 affects only the height. (b) π2\dfrac{\pi}{2}. 2π4=π2\dfrac{2\pi}{4}=\dfrac{\pi}{2}, and the quarter-period is π8\dfrac{\pi}{8}, so the key xx-values are 0, π8, π4, 3π8, π20,\ \dfrac{\pi}{8},\ \dfrac{\pi}{4},\ \dfrac{3\pi}{8},\ \dfrac{\pi}{2}.

Phase (Horizontal) Shift and Vertical Shift

Concept
Shifting the graph: y=asin⁡(bx−c)+dy=a\sin(bx-c)+d

Two more numbers slide the whole graph:

  • Vertical shift dd: raises (d>0d>0) or lowers (d<0d<0) the graph. The new midline is y=dy=d; max =d+∣a∣=d+|a|, min =d−∣a∣=d-|a|.
  • Phase (horizontal) shift =cb=\dfrac{c}{b}: found by setting the inside equal to zero, bx−c=0⇒x=cbbx-c=0\Rightarrow x=\dfrac{c}{b}. A positive result shifts right, negative shifts left.

A cycle of sine begins where its input is 00. For sin⁡(bx−c)\sin(bx-c) the input is 00 at x=cbx=\dfrac{c}{b}, not at x=0x=0, so the whole picture has been slid until its starting point sits at cb\dfrac{c}{b}. Adding dd outside adds the same amount to every output, lifting every point, including the midline, by dd.

Example
Worked example: y=sin⁡ ⁣(x−π2)y=\sin\!\left(x-\frac{\pi}{2}\right) and y=cos⁡x+2y=\cos x+2
xy−1123π/2π3π/22πy = sin(x − π/2)y = cos x + 2
The solid sine cycle now starts at x=π2x=\tfrac{\pi}{2} (dot); the dashed cosine rides on the midline y=2y=2.

y=sin⁡ ⁣(x−π2)y=\sin\!\left(x-\tfrac{\pi}{2}\right): set x−π2=0⇒x=π2x-\tfrac{\pi}{2}=0\Rightarrow x=\tfrac{\pi}{2}, so shift π2\tfrac{\pi}{2} to the right. (It coincides with −cos⁡x-\cos x.) y=cos⁡x+2y=\cos x+2: no horizontal shift; d=2d=2 moves the midline up to y=2y=2, so it swings between 11 and 33.

Tip

Tip: Always factor out bb before reading the phase shift: sin⁡(2x−π)=sin⁡ ⁣(2(x−π2))\sin(2x-\pi)=\sin\!\big(2(x-\tfrac{\pi}{2})\big) shows the shift is π2\tfrac{\pi}{2}, not π\pi. Equivalently, phase shift =cb=\dfrac{c}{b}.

Try it
Quick check

For y=cos⁡(3x+π)−2y=\cos(3x+\pi)-2, find the phase shift (with direction), the midline, and the range.

Show the answer

Answer: shift π3\dfrac{\pi}{3} to the left; midline y=−2y=-2; range [−3,−1][-3,-1]. Factor out b=3b=3: cos⁡(3x+π)=cos⁡ ⁣(3(x+π3))\cos(3x+\pi)=\cos\!\big(3(x+\tfrac{\pi}{3})\big), so the cycle starts where x+π3=0x+\tfrac{\pi}{3}=0, at x=−π3x=-\tfrac{\pi}{3}. The −2-2 lowers the midline to y=−2y=-2, and the amplitude 11 gives −2±1-2\pm1.

Graphing the Full Sinusoid y=asin⁡(bx−c)+dy=a\sin(bx-c)+d

Concept
Five steps for any sinusoid
  • Amplitude =∣a∣=|a|.   2. Period =2πb=\dfrac{2\pi}{b}.   3. Phase shift =cb=\dfrac{c}{b} (right if ++).
  • [4.] Midline y=dy=d; max =d+∣a∣=d+|a|, min =d−∣a∣=d-|a|.
  • [5.] Start the cycle at x=cbx=\dfrac{c}{b}, then plot five points a quarter-period 2π4b\dfrac{2\pi}{4b} apart.

The five numbers are independent because each acts on a different stage of the formula: bb and cc change which input the sine sees, aa scales the output, and dd shifts it. So you can read them off in any order, but when plotting it is easiest to find the starting xx and the quarter-period first (the horizontal skeleton), then the midline and amplitude (the vertical one).

Example
Worked example: y=2sin⁡ ⁣(2x−π2)+1y=2\sin\!\left(2x-\frac{\pi}{2}\right)+1
xy−1123π/4π/23π/4π5π/43π/2midline y = 1
One cycle from x=π4x=\tfrac{\pi}{4} to x=5π4x=\tfrac{5\pi}{4}, between −1-1 and 33.

a=2, b=2, c=π2, d=1a=2,\ b=2,\ c=\tfrac{\pi}{2},\ d=1.

  • Amplitude =∣2∣=2=|2|=2;   Period =2π2=π=\dfrac{2\pi}{2}=\pi.
  • Phase shift =cb=π/22=π4=\dfrac{c}{b}=\dfrac{\pi/2}{2}=\dfrac{\pi}{4} to the right.
  • Midline y=1y=1; max =1+2=3=1+2=3; min =1−2=−1=1-2=-1.
  • Cycle starts at x=π4x=\tfrac{\pi}{4}; quarter-period =π4=\dfrac{\pi}{4}. Key points at x=π4,π2,3π4,π,5π4x=\tfrac{\pi}{4},\tfrac{\pi}{2},\tfrac{3\pi}{4},\pi,\tfrac{5\pi}{4} give y=1,3,1,−1,1y=1,3,1,-1,1.
Tip

Tip: The vertical shift dd and midline are the “sea level” of the wave. Find max/min by adding and subtracting the amplitude from dd: max=d+∣a∣\text{max}=d+|a|, min=d−∣a∣\text{min}=d-|a|.

Try it
Quick check

For y=3cos⁡ ⁣(π4x−π2)−1y=3\cos\!\left(\dfrac{\pi}{4}x-\dfrac{\pi}{2}\right)-1, find the amplitude, period, phase shift, midline, maximum, and minimum.

Show the answer

Answer: amplitude 33; period 88; shift 22 to the right; midline y=−1y=-1; max 22; min −4-4. Period =2ππ/4=8=\dfrac{2\pi}{\pi/4}=8. Phase shift =cb=π/2π/4=2=\dfrac{c}{b}=\dfrac{\pi/2}{\pi/4}=2. Max and min are −1±3-1\pm 3. The cycle starts at its maximum at x=2x=2, since there the input is 00 and cos⁡0=1\cos 0=1.

Writing an Equation from a Graph or Description

Concept
Reverse-engineering a sinusoid

Given a graph or a word description, recover a,b,c,da,b,c,d:

  • d=max+min2d=\dfrac{\text{max}+\text{min}}{2} (midline).   2. ∣a∣=max−min2|a|=\dfrac{\text{max}-\text{min}}{2} (amplitude).
  • [3.] b=2πPeriodb=\dfrac{2\pi}{\text{Period}}.
  • [4.] Choose sine or cosine to match the starting behavior, then set cc from the horizontal shift. A maximum at x=x0x=x_0 suggests y=acos⁡ ⁣(b(x−x0))+dy=a\cos\!\big(b(x-x_0)\big)+d. A midline point rising suggests sine.

The midline sits halfway between the highest and lowest values, so it is their average; the amplitude is the distance from the midline to either extreme, half the total swing. Those two facts fix dd and ∣a∣|a| without any formula for the curve. The period fixes bb, and a single labelled point, best a maximum, fixes the shift.

Example
Worked example: write an equation
xy246πy = 3max 7min −1
y=4cos⁡2x+3y=4\cos 2x+3: maxima π\pi apart, midline halfway between 77 and −1-1.

A wave has maximum 77, minimum −1-1, period π\pi, and a maximum at x=0x=0. d=7+(−1)2=3d=\dfrac{7+(-1)}{2}=3;   ∣a∣=7−(−1)2=4|a|=\dfrac{7-(-1)}{2}=4;   b=2ππ=2b=\dfrac{2\pi}{\pi}=2. A maximum at x=0x=0 with no shift matches cosine:  y=4cos⁡(2x)+3 \boxed{\,y=4\cos(2x)+3\,}.

Example
Worked example: modeling with a Ferris wheel
th10203010203040center 22 m
Height in metres against time in seconds: boarding at the bottom (22 m), top at t=20t=20.

A Ferris wheel has radius 2020 m; its center is 2222 m above the ground. It makes one revolution every 4040 s and a rider boards at the bottom at t=0t=0. Model the height h(t)h(t). Amplitude =20=20 (the radius); midline d=22d=22 (center height); period =40⇒b=2π40=π20=40\Rightarrow b=\dfrac{2\pi}{40}=\dfrac{\pi}{20}. Starting at the bottom (a minimum) matches −cos⁡-\cos:

 h(t)=−20cos⁡ ⁣(π20 t)+22 \boxed{\,h(t)=-20\cos\!\left(\dfrac{\pi}{20}\,t\right)+22\,}

Check: h(0)=−20+22=2h(0)=-20+22=2 m (bottom); h(20)=20+22=42h(20)=20+22=42 m (top). ✓

Tip

Modeling tip: In real problems, amplitude =12(high−low)=\tfrac12(\text{high}-\text{low}), midline =12(high+low)=\tfrac12(\text{high}+\text{low}), and the period is the time for one full cycle. Use +cos⁡+\cos if it starts at the high point, −cos⁡-\cos if it starts at the low point, and sin⁡\sin if it starts at the midline.

Try it
Quick check

A sinusoid has a maximum of 1010 at t=3t=3 and the next minimum, 44, at t=7t=7. Write an equation using cosine.

Show the answer

Answer: y=3cos⁡ ⁣(π4(t−3))+7y=3\cos\!\left(\dfrac{\pi}{4}(t-3)\right)+7. Midline d=10+42=7d=\tfrac{10+4}{2}=7 and amplitude 10−42=3\tfrac{10-4}{2}=3. A maximum to the next minimum is half a period, so the period is 2(7−3)=82(7-3)=8 and b=2π8=π4b=\dfrac{2\pi}{8}=\dfrac{\pi}{4}. A cosine starts at its maximum, so shift it to start at t=3t=3. Check: at t=7t=7 the input is π\pi and 3cos⁡π+7=43\cos\pi+7=4.

Going Deeper: Advanced Sine & Cosine Graphs

Concept
Combining two sinusoids: asin⁡x+bcos⁡x=Rsin⁡(x+ϕ)a\sin x+b\cos x=R\sin(x+\phi)

A sum of a sine and a cosine of the same frequency is itself a single sinusoid. For any constants a,ba,b,

 asin⁡x+bcos⁡x=Rsin⁡(x+ϕ) R=a2+b2,tan⁡ϕ=ba.\boxed{\ a\sin x+b\cos x=R\sin(x+\phi)\ }\qquad R=\sqrt{a^2+b^2},\quad \tan\phi=\dfrac{b}{a}.
  • R=a2+b2R=\sqrt{a^2+b^2} is the amplitude of the combined wave (always ≥0\ge 0).
  • The phase angle ϕ\phi satisfies cos⁡ϕ=aR\cos\phi=\dfrac{a}{R} and sin⁡ϕ=bR\sin\phi=\dfrac{b}{R}; use both signs to place ϕ\phi in the correct quadrant — do not trust arctan⁡\arctan alone.
  • You may also write it as Rcos⁡(x−ψ)R\cos(x-\psi) with tan⁡ψ=ab\tan\psi=\dfrac{a}{b}. Same curve, different bookkeeping.

This is why superposition of two oscillations (of equal period) never produces a new shape — only a shifted, rescaled sine.

Example
Worked example: rewrite 3sin⁡x+4cos⁡x3\sin x+4\cos x as a single sine
xyπ/2π3π/25−5
The sum peaks at 55, at x=π2−0.927≈0.64x=\tfrac{\pi}{2}-0.927\approx 0.64, not at 3+4=73+4=7.

Here a=3, b=4a=3,\ b=4. Then R=32+42=25=5R=\sqrt{3^2+4^2}=\sqrt{25}=5, and cos⁡ϕ=35\cos\phi=\tfrac{3}{5}, sin⁡ϕ=45\sin\phi=\tfrac{4}{5} (both positive, so ϕ\phi is in the first quadrant):

ϕ=arctan⁡ ⁣43≈0.927 rad⟹3sin⁡x+4cos⁡x=5sin⁡(x+0.927).\phi=\arctan\!\tfrac{4}{3}\approx 0.927\ \text{rad}\quad\Longrightarrow\quad 3\sin x+4\cos x=5\sin(x+0.927).

So the sum has amplitude 55, period 2π2\pi, and peaks a little before x=π2x=\tfrac{\pi}{2} (shifted left by 0.9270.927). In the figure the solid curve is the combined sinusoid and the dashed curves are its ingredients.

Tip

Tip: The combined amplitude R=a2+b2R=\sqrt{a^2+b^2} is never simply a+ba+b. For 3sin⁡x+4cos⁡x3\sin x+4\cos x the peaks of the two pieces occur at different xx-values, so they never add to 3+4=73+4=7; the true maximum is R=5R=5.

Concept
Max, min, and the period of a sum

Once a sinusoid is written as y=Asin⁡(bx−c)+dy=A\sin(bx-c)+d (or found via RR above), reading its extremes is mechanical:

  • Maximum =d+∣A∣=d+|A|, reached when the inside angle =π2=\tfrac{\pi}{2}; minimum =d−∣A∣=d-|A|, when the inside angle =−π2=-\tfrac{\pi}{2} (or 3π2\tfrac{3\pi}{2}).
  • Solve for the location: set bx−c=π2+2πkbx-c=\tfrac{\pi}{2}+2\pi k and solve for xx to find where the max occurs.

Period of a sum of different frequencies. y=sin⁡(b1x)+sin⁡(b2x)y=\sin(b_1 x)+\sin(b_2 x) is generally not a simple sinusoid. Its period is the least common multiple of the two periods 2πb1\tfrac{2\pi}{b_1} and 2πb2\tfrac{2\pi}{b_2} (when that LCM exists):

Period(sin⁡2x+sin⁡3x)=lcm⁡ ⁣(2π2,2π3)=lcm⁡(π,2π3)=2π.\text{Period}\big(\sin 2x+\sin 3x\big)=\operatorname{lcm}\!\left(\tfrac{2\pi}{2},\tfrac{2\pi}{3}\right)=\operatorname{lcm}(\pi,\tfrac{2\pi}{3})=2\pi.

If the ratio b1/b2b_1/b_2 is irrational, the sum is not periodic at all.

Concept
Intersections and counting solutions on an interval

To find where two graphs meet, or how many times an equation is satisfied on an interval, think geometrically:

  • Intersection of two curves y=f(x)y=f(x) and y=g(x)y=g(x): set f(x)=g(x)f(x)=g(x) and solve. Each solution is one crossing point.
  • Number of solutions of sin⁡x=k\sin x=k on [0,2π)[0,2\pi): draw the horizontal line y=ky=k. If −1<k<1-1<k<1 there are 2 solutions; if k=±1k=\pm 1 there is exactly 1 (a tangent touch at the peak/trough); if ∣k∣>1|k|>1 there are 0.
  • For a compressed wave sin⁡(bx)=k\sin(bx)=k on [0,2π)[0,2\pi), the line crosses roughly 2b2b times because the wave completes bb full cycles. Count crossings, do not just solve once.
Example
Worked example: count solutions of cos⁡x=12\cos x=\tfrac12 vs. cos⁡2x=12\cos 2x=\tfrac12 on [0,2π)[0,2\pi)
xy−11π/2π3π/2y = 1/2
The line y=12y=\tfrac12 meets cos⁡2x\cos 2x (solid) four times and cos⁡x\cos x (dashed) twice.

The line y=12y=\tfrac12 meets y=cos⁡xy=\cos x where x=π3x=\tfrac{\pi}{3} and x=5π3x=\tfrac{5\pi}{3} — 2 solutions. But y=cos⁡2xy=\cos 2x completes two full cycles on [0,2π)[0,2\pi), so the same line cuts it 4 times: 2x=π3,5π3,7π3,11π32x=\tfrac{\pi}{3},\tfrac{5\pi}{3},\tfrac{7\pi}{3},\tfrac{11\pi}{3}, giving x=π6,5π6,7π6,11π6x=\tfrac{\pi}{6},\tfrac{5\pi}{6},\tfrac{7\pi}{6},\tfrac{11\pi}{6}.

Doubling the frequency doubled the number of solutions.

Example
Worked example: tide model with a solve-for-time step
th24681012123456h = 5
Depth in metres over 1212 hours; the 55 m line is crossed at t=2t=2 and t=10t=10.

At a harbor the water is 66 m deep at high tide and 22 m at low tide; high tide occurs at t=0t=0 hours and the cycle repeats every 1212 hours. Model the depth and find when the depth first reaches 55 m.

Build the model. Midline d=6+22=4d=\tfrac{6+2}{2}=4; amplitude ∣A∣=6−22=2|A|=\tfrac{6-2}{2}=2; period 12⇒b=2π12=π612\Rightarrow b=\tfrac{2\pi}{12}=\tfrac{\pi}{6}. It starts at a maximum, so use +cos⁡+\cos:

 h(t)=2cos⁡ ⁣(π6t)+4 \boxed{\,h(t)=2\cos\!\left(\tfrac{\pi}{6}t\right)+4\,}

Solve for time. Set h(t)=5h(t)=5:

2cos⁡ ⁣(π6t)+4=5⟹cos⁡ ⁣(π6t)=12π6t=π3 or 5π3⟹t=2 h or t=10 h.\begin{aligned} 2\cos\!\left(\tfrac{\pi}{6}t\right)+4&=5 &&\Longrightarrow&& \cos\!\left(\tfrac{\pi}{6}t\right)=\tfrac12\\ \tfrac{\pi}{6}t&=\tfrac{\pi}{3}\ \text{or}\ \tfrac{5\pi}{3} &&\Longrightarrow&& t=2\ \text{h}\ \text{or}\ t=10\ \text{h}. \end{aligned}

The depth first hits 55 m at t=2t=2 hours (falling), and again at t=10t=10 hours (rising back up).

Concept
Writing a sinusoid from scattered conditions, and identical graphs

From scattered data. You do not need a clean starting point. Given a few facts:

  • Get the midline and amplitude from any known high and low: d=12(high+low)d=\tfrac12(\text{high}+\text{low}), ∣A∣=12(high−low)|A|=\tfrac12(\text{high}-\text{low}).
  • Get bb from the period, or from the distance between a max and the next min (that gap is a half-period).
  • Pin the horizontal shift with one labeled point — ideally a max, min, or midline crossing — by forcing the model to pass through it.

Different formulas, identical graphs. Because sine and cosine are shifts of each other, many equations describe the same curve. All four below are one graph:

sin⁡ ⁣(x+π2)=cos⁡x=−cos⁡(x+π)=sin⁡ ⁣(π2−x).\sin\!\left(x+\tfrac{\pi}{2}\right)=\cos x=-\cos(x+\pi)=\sin\!\left(\tfrac{\pi}{2}-x\right).

A reflection − ⁣sin⁡x-\!\sin x equals a half-period shift sin⁡(x−π)\sin(x-\pi); adding 2π2\pi to a phase changes nothing. When two answers look different, test a few points (or compare A,b,dA,b,d and one peak location) to confirm they match.

Tip

Big-picture tip: Every equal-frequency combination, reflection, and phase shift of sine and cosine is still a single sinusoid — same midline spacing, same period, just relocated and rescaled. Reach for Rsin⁡(x+ϕ)R\sin(x+\phi) to unify them, and always finish a modeling problem with an explicit solve-for-tt step.

Try it
Quick check

Write 5sin⁡x−12cos⁡x5\sin x-12\cos x in the form Rsin⁡(x+ϕ)R\sin(x+\phi), and state its maximum value.

Show the answer

Answer: 13sin⁡(x+ϕ)13\sin(x+\phi) with ϕ=−arctan⁡125≈−1.176\phi=-\arctan\tfrac{12}{5}\approx -1.176; maximum 1313. R=52+122=13R=\sqrt{5^2+12^2}=13. Then cos⁡ϕ=513>0\cos\phi=\tfrac{5}{13}>0 and sin⁡ϕ=−1213<0\sin\phi=-\tfrac{12}{13}<0, so ϕ\phi is in Quadrant IV: ϕ=−arctan⁡125≈−1.176\phi=-\arctan\tfrac{12}{5}\approx -1.176 rad. A sinusoid of amplitude 1313 with midline 00 has maximum 1313.

Why It Works

The graph is the unit circle unrolled: as the angle grows, the yy-coordinate of the moving point rises to 11, falls to −1-1 and returns — sine. The period of sin⁡(Bx)\sin(Bx) is 2πB\tfrac{2\pi}{B} because BxBx completes one full cycle of 2π2\pi when xx advances by 2πB\tfrac{2\pi}{B}. The cosine graph is the sine graph shifted because cos⁡x=sin⁡(x+π2)\cos x = \sin(x + \tfrac\pi2), which is the cofunction identity on the unit circle.

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

Formulas, Proofs & Tips

Tip
Amplitude, period and shifts
y=asin⁡(b(x−c))+d:amplitude=∣a∣,period=2π∣b∣,shift=c,midline=dy=a\sin\big(b(x-c)\big)+d:\quad \text{amplitude}=|a|,\quad \text{period}=\frac{2\pi}{|b|},\quad \text{shift}=c,\quad \text{midline}=d

What it means. aa stretches the wave, bb squeezes it, cc slides it sideways, dd moves the centre line.

Example. y=3sin⁡(2x)y=3\sin(2x) has amplitude 33 and period 2π2=π\tfrac{2\pi}{2}=\pi.

Why it works. sin⁡\sin repeats when its input advances by 2π2\pi. Here the input is b(x−c)b(x-c), so xx only needs to advance by 2πb\tfrac{2\pi}{b} to complete a cycle — the period shrinks as bb grows.

Tip. Amplitude uses ∣a∣|a| — a negative aa flips the wave but does not make the amplitude negative.