Worked examples — Graphs of sin x, cos x, tan x — key features, period, amplitude

This page is a stress test. The parent note Graphs of sin, cos, tan built the three curves from the circle. Here we throw every kind of question at them — every sign, every quadrant, the degenerate "divide by zero" cases, the limiting "what happens near the cliff" cases, a word problem, and an exam-style twist — and solve each one slowly.

Before any example, we lay out the full battlefield so you can see that nothing is left uncovered.

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The scenario matrix

Every question this topic can ask lands in one of these cells. The right column names the example that kills it.

#	Case class	What makes it tricky	Killed by
C1	Read amplitude & period off $A\sin (Bx)$	mixing up multiply vs divide for $B$	Ex 1
C2	Negative $A$ (reflection)	does a minus change the amplitude?	Ex 2
C3	Full transform $A\sin (Bx+C)+D$	phase shift sign, midline	Ex 3
C4	Degenerate: $\cos x=0$ (asymptote of $\tan$)	tan is undefined, not $\infty$ as a value	Ex 4
C5	Limiting behaviour near an asymptote	sign of the blow-up ($+\infty$ vs $-\infty$)	Ex 4
C6	Signs across all four quadrants for $\tan$	where tan is $+$ / $-$	Ex 5
C7	Solve a trig equation using periodicity	listing all solutions in a window	Ex 6
C8	tan-period with a $B$ inside	$\pi/	B
C9	Word problem (Ferris wheel / SHM)	turning physics words into $A,B,C,D$	Ex 8
C10	Exam twist: two curves equal / intersection count	using period to count crossings	Ex 9

Prerequisites we lean on: Unit Circle and Radian Measure, Exact Values of Trig Ratios, Transformations of Graphs, Solving Trigonometric Equations, Simple Harmonic Motion.

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Ex 1 — Read amplitude & period (cell C1)

Setup. Compare with the template $y=A\sin (Bx+C)+D$. Here $A=4$, $B=3$, $C=0$, $D=0$.

1. Amplitude $=|A|=4$. Why this step? $A$ multiplies every height on the curve, so the biggest height becomes $4$ and the smallest $-4$. The word "amplitude" means the size of that swing, so we take the absolute value — size is never negative. In the figure this is the height of the red curve's peak.

2. Period $=\frac{2\pi }{|B|}=\frac{2\pi }{3}$. Why this step? One full wave happens when the thing inside the sine, namely $3x$, grows by a whole lap $2\pi$. Solving $3x=2\pi$ gives $x=\frac{2\pi }{3}$. A bigger $B$ crams the wiggle into a shorter $x$-distance — so the period shrinks. That's why we divide, never multiply. The figure marks this repeat-length on the axis.

3. Range $=[-4,4]$. Why this step? The range is the set of $y$-values the curve actually reaches. With midline $y=0$ (no $D$) and swing $\pm 4$ (the amplitude), the highest point is $+4$ and the lowest is $-4$, so every value between them is hit — that's the closed interval $[-4,4]$.

Verify: does the curve return to its start after $\frac{2\pi }{3}$? At $x=0$, $y=4\sin 0=0$. At $x=\frac{2\pi }{3}$, $y=4\sin (3\cdot \frac{2\pi }{3})=4\sin (2\pi )=0$. Same value, same rising slope — one full cycle confirmed. ✓

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Ex 2 — Negative amplitude coefficient (cell C2)

1. Amplitude $=|-2|=2$. Why this step? Amplitude is the size of the swing, so we take absolute value. A negative $A$ does not give a negative amplitude — amplitude is a distance, always $\ge 0$.

2. What the minus actually does: it flips the curve upside-down (reflection in the $x$-axis, shown by the red flipped curve in the figure). Ordinary $\cos x$ starts at its peak $+1$; multiply by $-2$ and $x=0$ now gives $-2$, its lowest point.

3. Where is the maximum $+2$? We need $-2\cos x=+2$, i.e. $\cos x=-1$. From the circle, the point sits at the far left $(-1,0)$, angle $x=\pi$. But cosine repeats every $2\pi$, so the maxima form a whole family: $x=\pi +2\pi n$ for any integer $n$ ($\dots ,-\pi ,\pi ,3\pi ,\dots$). Why this step? Flipping swaps peaks and troughs: the max of $-2\cos x$ sits where plain $\cos x$ had its min. And because the curve is periodic, that maximum recurs once every full period $2\pi$ — so we must list all of them, not just the first.

4. Range $=[-2,2]$. Why this step? The range collects every $y$-value the curve reaches. Midline is $y=0$ (no vertical shift), swing is $\pm 2$, so the curve sweeps from $-2$ up to $+2$ and hits everything between — the interval $[-2,2]$.

Verify: at $x=\pi$, $y=-2\cos \pi =-2(-1)=2$ (a maximum ✓). At $x=\pi +2\pi =3\pi$, $y=-2\cos 3\pi =-2(-1)=2$ (same max ✓). Range endpoints $\pm 2$ match the amplitude. ✓

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Ex 3 — Full transform, phase & vertical shift (cell C3)

Template: $y=A\sin (Bx+C)+D$ with $A=3$, $B=2$, $C=-\frac{\pi }{2}$, $D=1$.

1. Amplitude $=|3|=3$. Height swing is $\pm 3$. Why this step? $A$ scales every height, so the curve reaches $3$ above and $3$ below its midline.
2. Period $=\frac{2\pi }{|B|}=\frac{2\pi }{2}=\pi$. Inside grows by $2\pi$ when $x$ grows by $\pi$.
3. Phase shift $=-\frac{C}{B}=-\frac{-\pi /2}{2}=+\frac{\pi }{4}$. Why this step? A cycle starts when the inside equals zero: $2x-\frac{\pi }{2}=0\Rightarrow x=\frac{\pi }{4}$. A positive shift means the whole wave slides right by $\frac{\pi }{4}$ (see the red curve nudged rightward in the figure).
4. Midline $=D=1$. Why this step? $D$ is added to every $y$-value, lifting the whole wave up by $1$; the line it now oscillates about (the "average" level, halfway between peak and trough) is $y=1$, not $y=0$.
5. Range $=[-2,4]$. Why this step? Starting from the midline $y=1$ and swinging by the amplitude $3$ in each direction gives a top of $1+3=4$ and a bottom of $1-3=-2$; the curve hits everything between, so the range is $[-2,4]$.

Verify: the "start of a cycle" point should sit on the midline going up. At $x=\frac{\pi }{4}$: $y=3\sin (2\cdot \frac{\pi }{4}-\frac{\pi }{2})+1=3\sin (0)+1=1$. That equals the midline. ✓ And $\frac{\pi }{4}$ later, at $x=\frac{\pi }{4}+\frac{\pi }{4}=\frac{\pi }{2}$ (a quarter-period), we hit the peak: $y=3\sin (\frac{\pi }{2})+1=4$. ✓

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Ex 4 — Degenerate input & limiting behaviour of tan (cells C4, C5)

1. The value itself. $\tan \frac{\pi }{2}=\frac{\sin (\pi /2)}{\cos (\pi /2)}=\frac{1}{0}$. Why this step? $\tan$ is the ratio $\sin /\cos$. At the top of the circle $(0,1)$, the horizontal distance $\cos$ is exactly $0$. Dividing by $0$ has no answer, so $\tan \frac{\pi }{2}$ is undefined — not a number, not "$\infty$". There is a vertical asymptote here.

2. Approach from the left ($x$ just below $\frac{\pi }{2}$, red arrow in figure). Here $\sin x$ is close to $+1$ (positive) and $\cos x$ is a tiny positive number. Positive $÷$ tiny-positive $=$ huge positive: ${\lim }_{x\to (\pi /2{)}^{-}}\tan x=+\infty .$ Why this step? Just before the top of the circle the point is still in the right half, so $\cos >0$. Shrinking a positive denominator makes the ratio explode upward.

3. Approach from the right ($x$ just above $\frac{\pi }{2}$). Now the point has crossed to the left half: $\cos x$ is a tiny negative number, while $\sin x\approx +1$ still. Positive $÷$ tiny-negative $=$ huge negative: ${\lim }_{x\to (\pi /2{)}^{+}}\tan x=-\infty .$

Verify: plug in values straddling $\frac{\pi }{2}\approx 1.5708$. $\tan (1.5)\approx 14.10$ (large positive) and $\tan (1.65)\approx -12.60$ (large negative). The sign really does flip across the asymptote. ✓

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Ex 5 — Sign of tan in all four quadrants (cell C6)

$\tan x=\frac{\sin x}{\cos x}=\frac{y\text{-coordinate}}{x\text{-coordinate}}$. The sign is decided by the signs of the two coordinates.

1. Quadrant I ($0<x<\frac{\pi }{2}$): point is top-right, $x>0,y>0$. So $\frac{+}{+}=+$. Test: $\tan \frac{\pi }{4}=\frac{\sqrt{2}/2}{\sqrt{2}/2}=1>0$.
2. Quadrant II ($\frac{\pi }{2}<x<\pi$): top-left, $x<0,y>0$. So $\frac{+}{-}=-$. Test: $\tan \frac{3\pi }{4}=-1<0$.
3. Quadrant III ($\pi <x<\frac{3\pi }{2}$): bottom-left, $x<0,y<0$. So $\frac{-}{-}=+$. Test: $\tan \frac{5\pi }{4}=1>0$.
4. Quadrant IV ($\frac{3\pi }{2}<x<2\pi$): bottom-right, $x>0,y<0$. So $\frac{-}{+}=-$. Test: $\tan \frac{7\pi }{4}=-1<0$.

Why this pattern matters: the sign is positive exactly in quadrants I and III — the two "diagonally opposite" ones. That is a picture of tan's period being $\pi$: quadrant III is quadrant I shifted by half a turn, and $\tan$ repeats every half turn, so it must carry the same sign.

Verify: $\tan \frac{\pi }{4}=1$, $\tan \frac{3\pi }{4}=-1$, $\tan \frac{5\pi }{4}=1$, $\tan \frac{7\pi }{4}=-1$ — signs $+,-,+,-$. ✓

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Ex 6 — Solve an equation using periodicity (cell C7)

1. Which point has $x$-coordinate $\frac{1}{2}$? On the unit circle, $\cos =\frac{1}{2}$ at the exact angle $x=\frac{\pi }{3}$ (from Exact Values of Trig Ratios). Why this step? $\cos x$ is the horizontal coordinate; we want where it equals $\frac{1}{2}$.

2. But cosine is even ($\cos (-x)=\cos x$), and the circle is symmetric top-to-bottom. So the point at angle $-\frac{\pi }{3}$, equivalently $2\pi -\frac{\pi }{3}=\frac{5\pi }{3}$, has the same $x$-coordinate $\frac{1}{2}$. Why this step? The vertical line $x=\frac{1}{2}$ cuts the circle in two places (one above, one below the axis, marked red in the figure). Two intersection points ⇒ two angles.

3. Solutions in $[0,2\pi ]$: $x=\frac{\pi }{3}$ and $x=\frac{5\pi }{3}$.

Verify: $\cos \frac{\pi }{3}=0.5$ ✓ and $\cos \frac{5\pi }{3}=0.5$ ✓. Both lie inside $[0,2\pi ]$. ✓

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Ex 7 — tan-period with a coefficient (cell C8)

1. Period $=\frac{\pi }{|B|}$ with $B=\frac{1}{2}$, giving $\frac{\pi }{1/2}=2\pi$. Why $\pi$ on top, not $2\pi$? Tan's natural period is already $\pi$ (its two sign-flips at a half-turn cancel — see parent note). So we scale $\pi$, not $2\pi$. Here $B=\frac{1}{2}<1$ stretches the graph, making the period larger.

2. Asymptotes occur where the inside equals $\frac{\pi }{2}+n\pi$ (where cosine of the inside is zero): $\frac{x}{2}=\frac{\pi }{2}+n\pi ⟹x=\pi +2\pi n.$ Why this step? $\tan (\text{stuff})$ blows up when $\text{stuff}=\frac{\pi }{2}+n\pi$; solve for $x$. First two for $x>0$: $x=\pi$ (with $n=0$) and $x=3\pi$ (with $n=1$), the red dashed walls in the figure.

Verify: the gap between consecutive asymptotes must equal the period: $3\pi -\pi =2\pi$. ✓ Matches the period found in step 1. ✓

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Ex 8 — Word problem: Ferris wheel / SHM (cell C9)

This is Simple Harmonic Motion dressed as words. We map words → the four constants of $h(t)=A\cos (Bt+C)+D$ (or a sine).

1. Midline $D=12$. Why? The height oscillates about the centre of the wheel, $12\text{ m}$ up.
2. Amplitude $A=10$. Why? The passenger swings $10\text{ m}$ above and below the centre — the radius.
3. Period $=40\text{ s}$, so $B=\frac{2\pi }{40}=\frac{\pi }{20}$. Why? One full circle every $40$ s; the inside must grow by $2\pi$ over $40$ s.
4. Start at the lowest point. At $t=0$ we want $h=12-10=2$. Using $-\cos$ gives $-\cos (0)=-1$, the minimum. So: $h(t)=12-10\cos \left(\frac{\pi }{20}t\right).$ Why the minus? Plain $\cos$ starts at its max; we need to start at the min, so we flip it.
5. Height at $t=10$: $h(10)=12-10\cos \left(\frac{\pi }{20}\cdot 10\right)=12-10\cos \left(\frac{\pi }{2}\right)=12-10(0)=12\text{ m}.$

Verify (units & sanity): at $t=0$, $h=12-10\cos 0=12-10=2\text{ m}$ (lowest, ✓ correct radius below centre). At the quarter-turn $t=10\text{ s}$, height $=12\text{ m}$ = centre height, exactly where a rising passenger should be a quarter of the way up. At $t=20$ (half turn): $h=12-10\cos \pi =12+10=22\text{ m}$ = top. ✓ All in metres. ✓

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Ex 9 — Exam twist: count intersections (cell C10)

1. Rearrange to a single tan. Where $\cos x\ne 0$, divide both sides by $\cos x$: $\frac{\sin x}{\cos x}=1⟹\tan x=1.$ Why this step? Turning "two curves equal" into "$\tan x=$ constant" lets us use tan's tidy period of $\pi$ to count solutions.

2. Solve $\tan x=1$. The first solution is $x=\frac{\pi }{4}$. Because $\tan$ repeats every $\pi$, the next is $\frac{\pi }{4}+\pi =\frac{5\pi }{4}$ (the two red crossing points in the figure). Why this step? The period being $\pi$ (not $2\pi$) means solutions of $\tan x=c$ come once per $\pi$-block, so two of them fit inside a $2\pi$ window.

3. Check we didn't lose any at $\cos x=0$. At $x=\frac{\pi }{2},\frac{3\pi }{2}$ we have $\cos =0$ but $\sin =\pm 1\ne 0$, so $\sin x=\cos x$ fails there — no lost solutions. Count = 2.

Verify: $\sin \frac{\pi }{4}=\cos \frac{\pi }{4}=\frac{\sqrt{2}}{2}$ ✓ and $\sin \frac{5\pi }{4}=\cos \frac{5\pi }{4}=-\frac{\sqrt{2}}{2}$ ✓. Both inside $[0,2\pi ]$; total $2$. ✓

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Active recall

Recall Which matrix cell does each trap belong to?

• "Amplitude of $-5\sin x$ is $-5$" ::: C2 — amplitude is a size, answer is $5$
• "$\tan (\pi /2)=\infty$" ::: C4 — it is undefined; $\infty$ is not a value
• "$\tan x=1$ has one solution in $[0,2\pi ]$" ::: C10 — tan's period is $\pi$, so there are two ($\frac{\pi }{4}$ and $\frac{5\pi }{4}$)
• "Period of $\tan (x/2)$ is $\pi /2$" ::: C8 — it is $\pi ÷\frac{1}{2}=2\pi$

Recall Quick numeric self-test

• Period of $4\sin (3x)$? ::: $2\pi /3$
• Sign of $\tan$ in quadrant III? ::: positive
• Solutions of $\cos x=\frac{1}{2}$ in $[0,2\pi ]$? ::: $\pi /3$ and $5\pi /3$
• Ferris-wheel height at half-turn (Ex 8)? ::: $22\text{ m}$

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Connections

• Parent: the three graphs — the features stress-tested here
• Unit Circle and Radian Measure — every sign argument came from coordinates on the circle
• Exact Values of Trig Ratios — supplied $\frac{\pi }{3},\frac{\pi }{4}$ values
• Transformations of Graphs — $A,B,C,D$ decoding in Ex 1–3
• Solving Trigonometric Equations — periodicity used in Ex 6, 9
• Simple Harmonic Motion — the Ferris-wheel model in Ex 8
• Trigonometric Identities — $\tan =\sin /\cos$ underlies Ex 4, 5, 9