Practice questions

The key points of cosine on ${[0,2\pi]}$ are ${(0,1)}$, ${\left(\dfrac{\pi}{2},0\right)}$, ${(\pi,-1)}$, ${\left(\dfrac{3\pi}{2},0\right)}$, ${(2\pi,1)}$.

So the zeros are ${\dfrac{\pi}{2}}$ and ${\dfrac{3\pi}{2}}$. The high points are ${(0,1)}$ and ${(2\pi,1)}$. The low point is ${(\pi,-1)}$.

Write ${y=-\sin x+2}$. The graph of ${y=\sin x}$ is flipped over the ${x}$-axis, then moved up ${2}$. The key points are

${(0,2)}$, ${\left(\dfrac{\pi}{2},1\right)}$, ${(\pi,2)}$, ${\left(\dfrac{3\pi}{2},3\right)}$, ${(2\pi,2)}$

The values of ${-\sin x}$ go from ${-1}$ to ${1}$, so the values of ${2-\sin x}$ go from ${1}$ to ${3}$. The range is ${[1,3]}$.

The cosine graph is below the ${x}$-axis between its zeros ${\dfrac{\pi}{2}}$ and ${\dfrac{3\pi}{2}}$. So ${\cos x<0}$ when

${\dfrac{\pi}{2}<x<\dfrac{3\pi}{2}}$

The amplitude is ${|-4|=4}$. Here ${k=\dfrac{1}{3}}$, so the period is

${\dfrac{2\pi}{1/3}=6\pi}$

Factor ${2}$ out of the inside:

$\begin{align*}2x+\dfrac{\pi}{3}&=2\left(x+\dfrac{\pi}{6}\right)\\&=2\left(x-\left(-\dfrac{\pi}{6}\right)\right)\end{align*}$

So ${a=3}$, ${k=2}$, ${b=-\dfrac{\pi}{6}}$, and ${d=-1}$:

• Amplitude ${3}$, period ${\dfrac{2\pi}{2}=\pi}$.
• Phase shift ${-\dfrac{\pi}{6}}$: the graph moves ${\dfrac{\pi}{6}}$ to the left.
• Midline ${y=-1}$. The range is ${[-1-3,\,-1+3]=[-4,2]}$.

One period runs from ${-\dfrac{\pi}{6}}$ to ${-\dfrac{\pi}{6}+\pi=\dfrac{5\pi}{6}}$, in quarters of ${\dfrac{\pi}{4}}$. The ${x}$-values are

${-\dfrac{\pi}{6}}$, ${\dfrac{\pi}{12}}$, ${\dfrac{\pi}{3}}$, ${\dfrac{7\pi}{12}}$, ${\dfrac{5\pi}{6}}$

For cosine, the pattern is high, midline, low, midline, high, so the ${y}$-values are ${2}$, ${-1}$, ${-4}$, ${-1}$, ${2}$.

Midline and amplitude:

${d=\dfrac{7+1}{2}=4}$${a=\dfrac{7-1}{2}=3}$

The period is ${4\pi}$, so ${\dfrac{2\pi}{k}=4\pi}$ and ${k=\dfrac{1}{2}}$.

A sine curve crosses its midline going up at the start of a period, so ${b=\pi}$:

${y=3\sin\dfrac{1}{2}(x-\pi)+4}$

Here ${k=\dfrac{1}{2}}$, so the period is ${\dfrac{\pi}{1/2}=2\pi}$. For one branch, solve

$\begin{align*}-\dfrac{\pi}{2}&<\dfrac{x}{2}<\dfrac{\pi}{2}\\-\pi&<x<\pi\end{align*}$

So there are asymptotes at ${x=-\pi}$ and ${x=\pi}$. They repeat every ${2\pi}$: the asymptotes are ${x=\pi+2n\pi}$, where ${n}$ is any integer.

Cotangent has period ${\pi}$, so ${y=\cot 3x}$ has period ${\dfrac{\pi}{3}}$. Cotangent is undefined where its input is a multiple of ${\pi}$:

${3x=n\pi}$, so ${x=\dfrac{n\pi}{3}}$

The asymptotes are ${x=0}$, ${\pm\dfrac{\pi}{3}}$, ${\pm\dfrac{2\pi}{3}}$, and so on.

The matching cosine curve, ${y=3\cos 2x}$, has period ${\pi}$. So ${y=3\sec 2x}$ has period ${\pi}$ too.

The asymptotes are where ${\cos 2x=0}$:

${2x=\dfrac{\pi}{2}+n\pi}$, so ${x=\dfrac{\pi}{4}+\dfrac{n\pi}{2}}$

Secant values always have ${|\sec 2x|\ge 1}$. Multiplying by ${3}$ gives ${|3\sec 2x|\ge 3}$. So the range is ${(-\infty,-3]\cup[3,\infty)}$.

It starts at rest position moving up, so use sine with ${a=8}$. The period is ${\dfrac{2\pi}{\omega}=0.5}$, so ${\omega=4\pi}$:

${y=8\sin 4\pi t}$

The frequency is ${f=\dfrac{1}{0.5}=2}$ cycles per second (${2}$ Hz).

The radius is ${20}$ m, so the amplitude is ${20}$ and the midline is ${h=25}$. The period is ${10}$ minutes, so ${k=\dfrac{2\pi}{10}=\dfrac{\pi}{5}}$.

The rider starts at the lowest point. A cosine curve starts at its highest point, so use ${a=-20}$ to flip it:

${h=-20\cos\dfrac{\pi t}{5}+25}$

At ${t=4}$, with a calculator in radian mode:

$\begin{align*}h&=-20\cos\dfrac{4\pi}{5}+25\\&\approx -20(-0.809)+25\approx 41.2\end{align*}$

After ${4}$ minutes, the rider is about ${41.2}$ m above the ground.

Here ${\omega=4\pi}$, so ${f=\dfrac{4\pi}{2\pi}=2}$ cycles per second.

At ${t=5}$, ${\cos(20\pi)=1}$, so the spring is at the top of a swing:

${y=6e^{-0.2\cdot 5}=6e^{-1}\approx 2.21}$ cm