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

This page is a self-test. Each problem sits under a hidden solution — try it first, then reveal. Everything here builds on the parent note, and leans on Unit Circle and Radian Measure, Exact Values of Trig Ratios and Transformations of Graphs.

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Level 1 — Recognition

(Can you read a feature straight off the definition?)

Recall Solution L1.1

• Amplitude is half the total vertical swing. Here $A=4$ multiplies every height, so the wave rises to $+4$ and dips to $-4$. Amplitude $=|A|=4$.
• Period is how far in $x$ before the wave repeats. We did nothing to the inside ($x$ is bare, so $B=1$), so it repeats every full lap: $\frac{2\pi }{|B|}=\frac{2\pi }{1}=2\pi$.
• Answer: amplitude $4$, period $2\pi$.

Recall Solution L1.2

• $\cos x$ is the $x$-coordinate of a point on a circle of radius $1$. A coordinate on that circle can be as small as $-1$ (far left) and as large as $+1$ (far right), and everything between.
• Answer: range $[-1,1]$.

Recall Solution L1.3

• $\tan x=\frac{\sin x}{\cos x}$. A fraction is zero exactly when its top is zero (and its bottom is not).
• $\sin x=0$ at $x=0,\pi ,2\pi$. At each of these $\cos x\ne 0$, so all are valid.
• Answer: $x=0,\pi ,2\pi$.

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Level 2 — Application

(Plug into $y=A\sin (Bx+C)+D$ and read the machine.)

Recall Solution L2.1

• Amplitude $=|A|=3$: the $3$ multiplies every height.
• Period $=\frac{2\pi }{|B|}=\frac{2\pi }{2}=\pi$. Why divide? The inside $2x$ reaches $2\pi$ (one full cycle) as soon as $x$ reaches $\pi$. Bigger $B$ = faster wiggle = shorter period.
• Range $=[-3,3]$.
• Answer: amplitude $3$, period $\pi$, range $[-3,3]$.

Recall Solution L2.2

• $\tan$'s natural period is $\pi$ (its two sign-flips at $x+\pi$ cancel in the ratio). So $y=\tan (Bx)$ has period $\frac{\pi }{|B|}=\frac{\pi }{3}$.
• Asymptotes of $\tan (3x)$ occur where the inside $3x$ hits $\frac{\pi }{2}$ (where $\cos =0$): $3x=\frac{\pi }{2}\Rightarrow x=\frac{\pi }{6}$.
• Answer: period $\frac{\pi }{3}$; first positive asymptote at $x=\frac{\pi }{6}$.

Recall Solution L2.3

• $\cos x$ swings in $[-1,1]$. Multiply by $2$: swing $[-2,2]$. Add $1$ (the $D$ shift lifts everything): swing $[-1,3]$.
• Midline is the centre line the wave oscillates about: $y=D=1$.
• Answer: oscillates between $-1$ and $3$; midline $y=1$.

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Level 3 — Analysis

(Reason about why a value happens, using the circle.)

Recall Solution L3.1

• $\cos x$ is the horizontal position on the unit circle. It equals $-1$ only at the far left point $(-1,0)$.
• On the circle that point is reached at angle $x=\pi$ (half a lap). It is not reached again until $x=3\pi$, which is outside $[0,2\pi ]$.
• Answer: $x=\pi$ only.

Recall Solution L3.2

• Rearrange: $\sin x=\frac{1}{2}$. Why? Isolate the trig ratio first so we ask a clean question: "which angles have sine $\frac{1}{2}$?"
• The base angle with $\sin =\frac{1}{2}$ is $x=\frac{\pi }{6}$ (an exact value).
• $\sin$ is positive in both quadrant I and quadrant II (height above the axis is positive on the whole top half of the circle). The quadrant-II partner is $\pi -\frac{\pi }{6}=\frac{5\pi }{6}$.
• Answer: $x=\frac{\pi }{6},\frac{5\pi }{6}$. (See figure — both red points sit at the same height.)

Recall Solution L3.3

• Use the identity ${\sin }^{2}x=\frac{1-\cos 2x}{2}$ (a rearrangement of the double-angle identity, see Trigonometric Identities).
• The only $x$-dependence is $\cos 2x$. A cosine with inside $2x$ has period $\frac{2\pi }{2}=\pi$.
• Check directly: ${\sin }^2(x+\pi )=(\sin (x+\pi ){)}^2=(-\sin x{)}^2={\sin }^{2}x$. The minus square-away's, confirming repeat after $\pi$.
• Answer: period $=\pi$.

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Level 4 — Synthesis

(Combine transformation, periodicity and solving in one problem.)

Recall Solution L4.1

• Amplitude $=\frac{\max -\min }{2}=\frac{7-1}{2}=3$, so $A=3$.
• Midline / vertical shift $=\frac{\max +\min }{2}=\frac{7+1}{2}=4$, so $D=4$.
• Period $=\frac{2\pi }{B}=\pi \Rightarrow B=\frac{2\pi }{\pi }=2$.
• Answer: $y=3\sin (2x)+4$.

Recall Solution L4.2

• $\cos (\text{anything})=0$ when that "anything" equals $\frac{\pi }{2}+n\pi$ (every place $\cos$ crosses zero).
• So $2x=\frac{\pi }{2}+n\pi \Rightarrow x=\frac{\pi }{4}+\frac{n\pi }{2}$.
• March $n=0,1,2,3$: $x=\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$. Next ($n=4$) gives $\frac{9\pi }{4}>2\pi$: stop.
• Answer: $4$ solutions. (Sense check: inside $2x$ runs $0\to 4\pi$, i.e. two full laps, and $\cos$ crosses zero twice per lap $=4$.)

Recall Solution L4.3

• They cross where $\sin x=\cos x$. Divide both sides by $\cos x$ (valid where $\cos x\ne 0$): $\tan x=1$.
• $\tan x=1$ at $x=\frac{\pi }{4}$ (base) and repeats every $\pi$ (tan's period): so $x=\frac{\pi }{4}$ and $\frac{\pi }{4}+\pi =\frac{5\pi }{4}$.
• Check we lost nothing by dividing: at the excluded points $\cos x=0$, $\sin x=\pm 1\ne 0$, so no crossing hides there.
• Answer: $2$ crossings, at $x=\frac{\pi }{4}$ and $\frac{5\pi }{4}$. (See figure.)

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Level 5 — Mastery

(Multi-step, unfamiliar packaging — no formula to copy.)

Recall Solution L5.1

• (i) Maximum height. $\cos$ ranges over $[-1,1]$. The term $-8\cos (\cdot )$ is largest when $\cos (\cdot )=-1$ (the minus flips it), giving $-8(-1)=8$. Max height $=10+8=18$ m.
• (ii) One revolution = period. Period $=\frac{2\pi }{|B|}$ with $B=\frac{\pi }{15}$: $\frac{2\pi }{\pi /15}=2\pi \cdot \frac{15}{\pi }=30$ s.
• (iii) Height $10$. Set $10-8\cos (\frac{\pi }{15}t)=10\Rightarrow \cos (\frac{\pi }{15}t)=0$. This happens when $\frac{\pi }{15}t=\frac{\pi }{2}$ (first positive zero of cos), so $t=\frac{15}{2}=7.5$ s.
• Answer: (i) $18$ m, (ii) $30$ s, (iii) $7.5$ s.

Recall Solution L5.2

• Period. No $B$ multiplies $x$ (coefficient is $1$), so period $=\frac{\pi }{1}=\pi$. The $-\frac{\pi }{4}$ only slides the graph right, it does not change spacing.
• Asymptotes happen where the inside equals $\frac{\pi }{2}+n\pi$: $x-\frac{\pi }{4}=\frac{\pi }{2}+n\pi \Rightarrow x=\frac{3\pi }{4}+n\pi$.
• The smallest positive value is at $n=0$: $x=\frac{3\pi }{4}$. (Check $n=-1$: $x=-\frac{\pi }{4}<0$, rejected.)
• Answer: period $\pi$; first positive asymptote $x=\frac{3\pi }{4}$.

Recall Solution L5.3

• Let $u=3x$. As $x$ runs $0\to 2\pi$, $u$ runs $0\to 6\pi$ (three laps). Solve $\sin u=\frac{\sqrt{3}}{2}$ over $[0,6\pi ]$.
• Base angle: $\sin u=\frac{\sqrt{3}}{2}\Rightarrow u=\frac{\pi }{3}$. Quadrant-II partner: $\pi -\frac{\pi }{3}=\frac{2\pi }{3}$. These are the two per lap.
• Add $2\pi$ for each further lap: $u=\frac{\pi }{3},\frac{2\pi }{3},\frac{7\pi }{3},\frac{8\pi }{3},\frac{13\pi }{3},\frac{14\pi }{3}$ (all $\le 6\pi$).
• Divide each by $3$ (since $x=u/3$): $x=\frac{\pi }{9},\frac{2\pi }{9},\frac{7\pi }{9},\frac{8\pi }{9},\frac{13\pi }{9},\frac{14\pi }{9}$.
• Answer: $6$ solutions: $x=\frac{\pi }{9},\frac{2\pi }{9},\frac{7\pi }{9},\frac{8\pi }{9},\frac{13\pi }{9},\frac{14\pi }{9}$.

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Connections

• Solving Trigonometric Equations — periodicity gives the full solution list (L3–L5)
• Transformations of Graphs — amplitude/period/shift machinery (L2, L4)
• Exact Values of Trig Ratios — the $\frac{1}{2},\frac{\sqrt{3}}{2}$ base angles used above
• Unit Circle and Radian Measure — why sine repeats a value each lap
• Simple Harmonic Motion — the Ferris-wheel model in L5.1
• Trigonometric Identities — ${\sin }^{2}x$ period proof (L3.3)