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Reed College Department of Mathematics & Statistics

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Worksheet Sine and Cosine

2.

Use the trigonometric function \(f(x) = 3\sin(x)\) to answer the following questions.
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What is the amplitude of \(f(x)\text{?}\)
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What the period of \(f(x)\text{?}\)
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What is the maximum value of \(f(x)\text{?}\)
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What is the minimum value of \(f(x)\text{?}\)
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What is the midline of \(f(x)\text{?}\)
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Answer 1.
\(3\)
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Answer 2.
\(6.28319\)
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Answer 3.
\(3\)
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Answer 4.
\(-3\)
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Answer 5.
\(\text{y=}\)
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Answer 6.
\(0\)
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3.

Use the trigonometric function \(f(x) = -4\sin({{\frac{1}{3}}}x)\) to answer the following questions.
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What is the amplitude of \(f(x)\text{?}\)
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What is the period of \(f(x)\text{?}\)
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What is the maximum value of \(f(x)\text{?}\)
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What is the minimum value of \(f(x)\text{?}\)
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What is the midline of \(f(x)\text{?}\)
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Answer 1.
\(4\)
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Answer 2.
\(18.8496\)
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Answer 3.
\(4\)
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Answer 4.
\(-4\)
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Answer 5.
\(\text{y=}\)
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Answer 6.
\(0\)
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4.

Use the trigonometric function \(f(t) = -6\sin\left(t-\displaystyle\frac{4\pi}{8}\right)\) to answer the following questions.
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What is the amplitude of \(f(t)\text{?}\)
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What is the period of \(f(t)\) ?
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What is the maximum value of \(f(t)\) ?
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What is the minimum value of \(f(t)\text{?}\)
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What is the midline of \(f(t)\) ?
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The phase shift of \(f(t)\) is units
.
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If the video does not work, click here to watch on YouTube.
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Answer 1.
\(6\)
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Answer 2.
\(6.28319\)
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Answer 3.
\(6\)
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Answer 4.
\(-6\)
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Answer 5.
\(\text{y=}\)
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Answer 6.
\(0\)
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Answer 7.
\(1.5708\)
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Answer 8.
\(\text{right}\)
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5.

Use the trigonometric function \(f(t) = -\cos\left(t+\displaystyle\frac{\pi}{6}\right)+1\) to answer the following questions.
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What is the amplitude of \(f(t)\text{?}\)
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What is the period of \(f(t)\) ?
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What is the maximum value of \(f(t)\) ?
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What is the minimum value of \(f(t)\text{?}\)
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What is the midline of \(f(t)\) ?
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The phase shift of \(f(t)\) is units
.
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Answer 1.
\(1\)
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Answer 2.
\(6.28319\)
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Answer 3.
\(2\)
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Answer 4.
\(0\)
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Answer 5.
\(\text{y=}\)
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Answer 6.
\(1\)
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Answer 7.
\(0.523599\)
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Answer 8.
\(\text{left}\)
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6.

Use the trigonometric function \(f(x) = 6\sin\left(\displaystyle\frac{\pi}{3}(x-3)\right)+2\) to answer the following questions.
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What is the amplitude of \(f(x)\text{?}\)
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What is the period of \(f(x)\text{?}\)
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What is the maximum value of \(f(x)\text{?}\)
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What is the minimum value of \(f(x)\) ?
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What is the midline of \(f(x)\text{?}\)
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The phase shift of \(f(x)\) is units
.
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Answer 1.
\(6\)
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Answer 2.
\(6\)
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Answer 3.
\(8\)
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Answer 4.
\(-4\)
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Answer 5.
\(\text{y=}\)
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Answer 6.
\(2\)
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Answer 7.
\(3\)
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Answer 8.
\(\text{right}\)
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7.

The equation \(P=20\sin(2\pi t)+140\) models blood pressure, \(P\text{,}\) where \(t\) represents time in seconds. Find the blood pressure after \(28\) seconds. Also, give the maximum and minimum blood pressures.
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  1. Blood pressure after \(28\) seconds:
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  2. Maximum blood pressure:
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  3. Minimum blood pressure:
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Answer 1.
\(140\)
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Answer 2.
\(160\)
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Answer 3.
\(120\)
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8.

Let \(f(x)\) be the trigonometric function \(f(x) = \sin(x)\text{.}\)
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On the interval \([0, 2\pi)\text{,}\) the maximum value(s) of \(f(x)\) occur(s) at what `x`-value(s)?
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If there is more than one answer, enter as a comma separated list.
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\(x=\)
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Answer.
\(\frac{\pi }{2}\)
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9.

Let \(f(x)\) be the trigonometric function \(f(x) = \sin(x)\text{.}\)
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On the interval \([0, 2\pi)\text{,}\) the minimum value(s) of the function occur(s) at what `x`-value(s)?
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If there is more than one answer, enter as a comma separated list.
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\(x=\)
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Answer.
\(\frac{3\pi }{2}\)
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10.

\(\displaystyle{ y = {-5\cos\mathopen{}\left(2x\right)} }\)
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What is the amplitude of the given function?
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What is the period of the given function?
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A B
C D
Hint.
First, remember that the amplitude is given by the coefficient \(A\) of \(A \sin(x)\) or \(A \cos(x)\text{.}\)
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Then, remember that from a function \(\sin( B x )\) or \(\cos( B x )\text{,}\) the period \(p\) is given by the equation \(\displaystyle{ p = \frac{ 2 \pi }{ B } }\text{.}\)
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Answer 1.
\(5\)
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Answer 2.
\(\frac{2\pi }{2}\)
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Answer 3.
\(\text{D}\)
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11.

Does this graph have the shape of \(\sin(x)\) or \(\cos(x)\text{?}\)
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What is the amplitude of the graphed function?
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State the formula for the function that is graphed above. \(y =\)
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Hint.
Do you recall the difference between the graphs of sine and cosine?\(\sin(x)\) passes through the origin, and \(\cos(x)\) does not.
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Amplitude is used to describe the height of the wave.How high does this wave get?
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Remember that the formula for a sine or cosine wave looks like:\(y = A \sin(x)\) or \(y = A \cos(x)\)
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where \(A\) represents the amplitude.
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Answer 1.
\(\cos\mathopen{}\left(x\right)\)
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Answer 2.
\(3\)
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Answer 3.
\(3\cos\mathopen{}\left(x\right)\)
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Solution.
We can tell that this is a graph of \(\cos(x)\) because we don’t pass through the origin.
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Our amplitude must be \(3\) because our graph extends up to \(3\) and down to \(-3\) on the \(y\)-axis.
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Therefore, our formula for this graph is \(y={3\cos\mathopen{}\left(x\right)}\text{.}\)
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12.

Practice
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Match the equations given below with their corresponding graphs:
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Answer 1.
\(\text{C}\)
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Answer 2.
\(\text{F}\)
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Answer 3.
\(\text{D}\)
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Answer 4.
\(\text{A}\)
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Solution.
  1. \(y = {3\sin\mathopen{}\left(x\right)}\) is the graph of sine with positive coefficient and amplitude 3; this corresponds to graph C.
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  2. \(y = {-2\sin\mathopen{}\left(x\right)}\) is the graph of sine with negative coefficient (so it is the vertically-reflected, or ’upside-down’, version of sine) and amplitude 2; this corresponds to graph F.
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  3. \(y = {2\cos\mathopen{}\left(x\right)}\) is the graph of cosine with positive coefficient and amplitude 2; this corresponds to graph D.
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  4. \(y = {2\sin\mathopen{}\left(x\right)}\) is the graph of sine with positive coefficient and amplitude 2; this corresponds to graph A.
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13.

Practice
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Match the equations given below with their corresponding graphs:
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Answer 1.
\(\text{E}\)
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Answer 2.
\(\text{B}\)
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Answer 3.
\(\text{F}\)
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Answer 4.
\(\text{D}\)
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Solution.
  1. \(y = {\cos\mathopen{}\left(2x\right)}\) is the graph of \(\cos(\theta)\) with \(\theta = {2x}\text{;}\)
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    A complete oscillation (period) for cosine happens at \(\theta = 2\pi\text{,}\) which corresponds to \({2x} = 2\pi\) (or \(x = {\pi }\));
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    With a period of \({\pi }\text{,}\) we get \({2}\) oscillations between \(0\) and \(2\pi\text{;}\) this corresponds to graph E.
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  2. \(y = {\sin\mathopen{}\left(\frac{x}{2}\right)}\) is the graph of \(\sin(\theta)\) with \(\theta = {\frac{x}{2}}\text{;}\)
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    A complete oscillation (period) for sine happens at \(\theta = 2\pi\text{,}\) which corresponds to \({\frac{x}{2}} = 2\pi\) (or \(x = {4\pi }\));
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    With a period of \({4\pi }\text{,}\) we get \({\frac{1}{2}}\) an oscillation between \(0\) and \(2\pi\text{;}\) this corresponds to graph B.
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  3. \(y = {\sin\mathopen{}\left(x\right)}\) is the graph of \(\sin(\theta)\) with \(\theta = {x}\text{;}\)
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    A complete oscillation (period) for sine happens at \(\theta = 2\pi\text{,}\) which corresponds to \({x} = 2\pi\) (or \(x = {2\pi }\));
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    With a period of \({2\pi }\text{,}\) we get \({1}\) oscillation between \(0\) and \(2\pi\text{;}\) this corresponds to graph F.
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  4. \(y = {\sin\mathopen{}\left(2x\right)}\) is the graph of \(\sin(\theta)\) with \(\theta = {2x}\text{;}\)
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    A complete oscillation (period) for sine happens at \(\theta = 2\pi\text{,}\) which corresponds to \({2x} = 2\pi\) (or \(x = {\pi }\));
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    With a period of \({\pi }\text{,}\) we get \({2}\) oscillations between \(0\) and \(2\pi\text{;}\) this corresponds to graph D.
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14.

Practice
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Enter an equation for each graph:
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Graph A: \(y =\)
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Graph B: \(y =\)
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Graph C: \(y =\)
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Graph D: \(y =\)
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Graph E: \(y =\)
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Graph F: \(y =\)
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Answer 1.
\(\cos\mathopen{}\left(4x\right)\)
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Answer 2.
\(\cos\mathopen{}\left(x\right)\)
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Answer 3.
\(\sin\mathopen{}\left(x\right)\)
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Answer 4.
\(\sin\mathopen{}\left(4x\right)\)
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Answer 5.
\(\cos\mathopen{}\left(2x\right)\)
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Answer 6.
\(\sin\mathopen{}\left(2x\right)\)
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Solution.
  1. Graph A is a sine wave with \({2}\) oscillations between \(0\) and \(2\pi\text{;}\)
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    This means that each period has length \(\frac{2\pi}{{2}}\text{,}\) or \({\pi }\text{.}\)
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    In order to write an equation for \(\sin(Bx)\) with period \({\pi }\text{,}\) we need to use the fact that \(\sin(Bx)\) has period \(\frac{2\pi}{B}\text{.}\)
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    If \(\frac{2\pi}{B}\) is going to be equal to \({\pi }\text{,}\) that means we need to have \(B = {2}\text{,}\) giving us \(y = {\cos\mathopen{}\left(4x\right)}\) as the equation for this graph.
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    🔗
  2. Graph B is a cosine wave with \({2}\) oscillations between \(0\) and \(2\pi\text{;}\)
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    This means that each period has length \(\frac{2\pi}{{2}}\text{,}\) or \({\pi }\text{.}\)
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    In order to write an equation for \(\cos(Bx)\) with period \({\pi }\text{,}\) we need to use the fact that \(\cos(Bx)\) has period \(\frac{2\pi}{B}\text{.}\)
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    If \(\frac{2\pi}{B}\) is going to be equal to \({\pi }\text{,}\) that means we need to have \(B = {2}\text{,}\) giving us \(y = {\cos\mathopen{}\left(x\right)}\) as the equation for this graph.
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    🔗
  3. Graph C is a cosine wave with \({4}\) oscillations between \(0\) and \(2\pi\text{;}\)
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    This means that each period has length \(\frac{2\pi}{{4}}\text{,}\) or \({\frac{\pi }{2}}\text{.}\)
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    In order to write an equation for \(\cos(Bx)\) with period \({\frac{\pi }{2}}\text{,}\) we need to use the fact that \(\cos(Bx)\) has period \(\frac{2\pi}{B}\text{.}\)
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    If \(\frac{2\pi}{B}\) is going to be equal to \({\frac{\pi }{2}}\text{,}\) that means we need to have \(B = {4}\text{,}\) giving us \(y = {\sin\mathopen{}\left(x\right)}\) as the equation for this graph.
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    🔗
  4. Graph D is a sine wave with \({4}\) oscillations between \(0\) and \(2\pi\text{;}\)
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    This means that each period has length \(\frac{2\pi}{{4}}\text{,}\) or \({\frac{\pi }{2}}\text{.}\)
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    In order to write an equation for \(\sin(Bx)\) with period \({\frac{\pi }{2}}\text{,}\) we need to use the fact that \(\sin(Bx)\) has period \(\frac{2\pi}{B}\text{.}\)
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    If \(\frac{2\pi}{B}\) is going to be equal to \({\frac{\pi }{2}}\text{,}\) that means we need to have \(B = {4}\text{,}\) giving us \(y = {\sin\mathopen{}\left(4x\right)}\) as the equation for this graph.
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    🔗
  5. Graph E is a cosine wave with \({1}\) oscillation between \(0\) and \(2\pi\text{;}\)
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    This means that each period has length \(\frac{2\pi}{{1}}\text{,}\) or \({2\pi }\text{.}\)
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    In order to write an equation for \(\cos(Bx)\) with period \({2\pi }\text{,}\) we need to use the fact that \(\cos(Bx)\) has period \(\frac{2\pi}{B}\text{.}\)
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    If \(\frac{2\pi}{B}\) is going to be equal to \({2\pi }\text{,}\) that means we need to have \(B = {1}\text{,}\) giving us \(y = {\cos\mathopen{}\left(2x\right)}\) as the equation for this graph.
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    🔗
  6. Graph F is a sine wave with \({1}\) oscillation between \(0\) and \(2\pi\text{;}\)
    🔗
    This means that each period has length \(\frac{2\pi}{{1}}\text{,}\) or \({2\pi }\text{.}\)
    🔗
    In order to write an equation for \(\sin(Bx)\) with period \({2\pi }\text{,}\) we need to use the fact that \(\sin(Bx)\) has period \(\frac{2\pi}{B}\text{.}\)
    🔗
    If \(\frac{2\pi}{B}\) is going to be equal to \({2\pi }\text{,}\) that means we need to have \(B = {1}\text{,}\) giving us \(y = {\sin\mathopen{}\left(2x\right)}\) as the equation for this graph.
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15.

Practice
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Enter an equation for each graph:
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Graph A: \(y =\)
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Graph B: \(y =\)
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Graph C: \(y =\)
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Graph D: \(y =\)
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Graph E: \(y =\)
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Graph F: \(y =\)
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Answer 1.
\(\cos\mathopen{}\left(x-\frac{\pi }{2}\right)\)
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Answer 2.
\(\cos\mathopen{}\left(x+\frac{\pi }{4}\right)\)
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Answer 3.
\(\cos\mathopen{}\left(x+\frac{5\pi }{6}\right)\)
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Answer 4.
\(\cos\mathopen{}\left(x+\frac{\pi }{3}\right)\)
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Answer 5.
\(\cos\mathopen{}\left(x-\frac{3\pi }{4}\right)\)
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Answer 6.
\(\cos\mathopen{}\left(x-\frac{\pi }{3}\right)\)
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Solution.
  1. Graph A is a cosine wave that has been shifted to the right;
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    Looking at the grid lines, we see that the interval between 0 and \(\pi\) has been split into halves, and our graph has been shifted right by one of those halves.
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    If we want an equation for the cosine wave that has been shifted right by one-halves of \(\pi\text{,}\) we need to take \(y = \cos(x)\) and replace \(x\) by \(\left( {x-\frac{\pi }{2}} \right)\text{.}\)
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    Therefore, our equation for Graph A is \(y = {\cos\mathopen{}\left(x-\frac{\pi }{2}\right)}\text{.}\)
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    🔗
  2. Graph B is a cosine wave that has been shifted to the left;
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    Looking at the grid lines, we see that the interval between 0 and \(\pi\) has been split into quarters, and our graph has been shifted left by one of those quarters.
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    If we want an equation for the cosine wave that has been shifted left by one-quarters of \(\pi\text{,}\) we need to take \(y = \cos(x)\) and replace \(x\) by \(\left( {x+\frac{\pi }{4}} \right)\text{.}\)
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    Therefore, our equation for Graph B is \(y = {\cos\mathopen{}\left(x+\frac{\pi }{4}\right)}\text{.}\)
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  3. Graph C is a cosine wave that has been shifted to the left;
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    Looking at the grid lines, we see that the interval between 0 and \(\pi\) has been split into sixths, and our graph has been shifted left by five of those sixths.
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    If we want an equation for the cosine wave that has been shifted left by five-sixths of \(\pi\text{,}\) we need to take \(y = \cos(x)\) and replace \(x\) by \(\left( {x+\frac{5\pi }{6}} \right)\text{.}\)
    🔗
    Therefore, our equation for Graph C is \(y = {\cos\mathopen{}\left(x+\frac{5\pi }{6}\right)}\text{.}\)
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    🔗
  4. Graph D is a cosine wave that has been shifted to the left;
    🔗
    Looking at the grid lines, we see that the interval between 0 and \(\pi\) has been split into thirds, and our graph has been shifted left by one of those thirds.
    🔗
    If we want an equation for the cosine wave that has been shifted left by one-thirds of \(\pi\text{,}\) we need to take \(y = \cos(x)\) and replace \(x\) by \(\left( {x+\frac{\pi }{3}} \right)\text{.}\)
    🔗
    Therefore, our equation for Graph D is \(y = {\cos\mathopen{}\left(x+\frac{\pi }{3}\right)}\text{.}\)
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    🔗
  5. Graph E is a cosine wave that has been shifted to the right;
    🔗
    Looking at the grid lines, we see that the interval between 0 and \(\pi\) has been split into quarters, and our graph has been shifted right by three of those quarters.
    🔗
    If we want an equation for the cosine wave that has been shifted right by three-quarters of \(\pi\text{,}\) we need to take \(y = \cos(x)\) and replace \(x\) by \(\left( {x-\frac{3\pi }{4}} \right)\text{.}\)
    🔗
    Therefore, our equation for Graph E is \(y = {\cos\mathopen{}\left(x-\frac{3\pi }{4}\right)}\text{.}\)
    🔗
    🔗
  6. Graph F is a cosine wave that has been shifted to the right;
    🔗
    Looking at the grid lines, we see that the interval between 0 and \(\pi\) has been split into thirds, and our graph has been shifted right by one of those thirds.
    🔗
    If we want an equation for the cosine wave that has been shifted right by one-thirds of \(\pi\text{,}\) we need to take \(y = \cos(x)\) and replace \(x\) by \(\left( {x-\frac{\pi }{3}} \right)\text{.}\)
    🔗
    Therefore, our equation for Graph F is \(y = {\cos\mathopen{}\left(x-\frac{\pi }{3}\right)}\text{.}\)
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16.

Decide whether the following graph appears to be a periodic function. If so enter the value of its period in the blank. If the graph does not appear to be periodic enter NONE.
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The period is (Enter NONE if not periodic.)
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(Click on graph to enlarge)
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Answer.
\(4\)
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Solution.
SOLUTION
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17.

The table below gives the height \(h=f(t)\) in feet of a weight on a spring where \(t\) is time in seconds.
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\(t\) (sec) 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30
\(h\) (feet) 4.2 3 2 1.8 2 3 4.2 5.4 6.4 6.6 6.4 5.4 4.2 3 2 1.8
(a) What is the period of \(\ f(t)\text{?}\) (include )
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(b) What is the the midline of \(\ f(t)\text{?}\) (include )
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(c) What is the the amplitude of \(\ f(t)\text{?}\) (include )
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Answer 1.
\(24\ {\rm s}\)
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Answer 2.
\(4.2\ {\rm ft}\)
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Answer 3.
\(2.4\ {\rm ft}\)
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Solution.
SOLUTION\(h\)\(\frac{ 6.6 + 1.8}{2} = 4.2\)\(6.6 - 4.2 = 2.4\)
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18.

(a) Find another angle \(\phi\) between \(0^\circ\) and \(360^\circ\) that has the same cosine as \(55^{\circ}\text{.}\) (That is, find \(\phi\) satisfying \(\cos(\phi) = \cos(55^\circ)\text{.}\))
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\(\phi =\) degrees.
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(b) Find another angle \(\phi\) between \(0^\circ\) and \(360^\circ\) that has the same sine as \(55^{\circ}\text{.}\) (That is, find \(\phi\) satisfying \(\sin(\phi) = \sin(55^\circ)\text{.}\))
🔗
\(\phi =\) degrees.
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Answer 1.
\(305\)
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Answer 2.
\(125\)
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Solution.
SOLUTION
\(55^{\circ}\)\(P\)\(a\)\(55^{\circ}\text{,}\)\(Q\)\(x\)\(P\text{,}\)\(Q\)\(\phi = 360^{\circ} - 55^{\circ} = 305^{\circ}\text{.}\)\(R\)\(y\)\(P\text{,}\)\(R\)\(\phi = 180^{\circ} - 55^{\circ} = 125^{\circ}\text{.}\)
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19.

Below is the graph of the function \(f(x) = 10 \sin{ \left( \frac{\pi}{5} x \right)}\) in blue, and a second sinusoidal function \(y = g(x)\) in red, which is a horizontal shift of \(y = f(x)\text{.}\) Find a formula for the function \(g(x)\text{.}\)
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\(g(x) =\)
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(Click on graph to enlarge)
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Answer.
\(10\sin\mathopen{}\left(\frac{\pi }{5}x-\frac{2\pi }{5}\right)\)
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Solution.
SOLUTION\(\frac{ 2 }{10}\)\(\phi = (2 \pi ) \cdot \left( \frac{ 2 }{10} \right) = \frac{2 \pi}{5}\text{.}\)\(g(x)\)
\begin{equation*} g(x) = 10 \sin{ \left( \frac{\pi}{5} \ x - \phi \right) } = 10 \sin{ \left( \frac{\pi}{5} \ x - \frac{2 \pi}{5} \right) }. \end{equation*}
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20.

Find a possible formula for the trigonometric function whose values are in the following table (note the values in the table may be slightly rounded):
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\(x\) 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00
\(g(x)\) 4.00 5.00 5.73 6.00 5.73 5.00 4.00 3.00 2.27 2.00 2.27 3.00 4.00
\(g(x) =\)
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Answer.
\(2\sin\mathopen{}\left(\frac{2\pi }{3}x\right)+4\)
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Solution.
SOLUTION\(g(x)\)\(x = 0\)\(g(x)=A \sin{(Bx)}+k\text{,}\)\(A\)\(k\)\(\frac{2 \pi}{B}\text{.}\)\(k \approx 4.00\text{.}\)\(A = \mbox{max} - k = 6.00 - 4.00 = 2\text{.}\)\(\mbox{period} = 3 = \frac{2 \pi}{B}\text{,}\)\(B = \frac{ 2 \pi}{3}\text{.}\)\(g(x) = 2 \sin{\left( \frac{2 \pi}{3} x \right)}+4\text{.}\)
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