Precalculus

Trigonometry

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Graphs of Sine and Cosine Functions

Solved Problems

Click or tap a problem to see the solution.

Example 1

Determine whether the function is even or odd: \[y = \frac{{{t^2} - \cos 2t}}{{\sin t}}.\]

Example 2

Determine whether the function is even or odd: \[y = \frac{{\sin 3t + {t^3}}}{{\cos t}}.\]

Example 3

Find the range of the function \[y = 5 + 4\sin x\cos x.\]

Example 4

Find the range of the function \[y = 2 + \sin 2x\cos 2x.\]

Example 5

Sketch a graph of the function \[y = 2\sin \left( {t - \frac{\pi }{6}} \right).\]

Example 6

Sketch a graph of the function \[y = - 2\cos 2t.\]

Example 1.

Determine whether the function is even or odd: \[y = \frac{{{t^2} - \cos 2t}}{{\sin t}}.\]

Solution.

We know that the sine function is odd, and the cosine function is even, that is

\[\cos \left( { - t} \right) = \cos t,\;\sin \left( { - t} \right) = - \sin t.\]

Then we have

\[y\left( { - t} \right) = \frac{{{t^2} - \cos 2t}}{{\sin t}} = \frac{{{{\left( { - t} \right)}^2} - \cos \left( {2\left( { - t} \right)} \right)}}{{\sin \left( { - t} \right)}} = \frac{{{{\left( { - t} \right)}^2} - \cos \left( { - 2t} \right)}}{{\sin \left( { - t} \right)}} = \frac{{{t^2} - \cos 2t}}{{ - \sin t}} = - \frac{{{t^2} - \cos 2t}}{{\sin t}} = - y\left( t \right).\]

Hence, the given function is odd.

Example 2.

Determine whether the function is even or odd: \[y = \frac{{\sin 3t + {t^3}}}{{\cos t}}.\]

Solution.

We use the even-odd properties of cosine and sine functions:

\[\cos \left( { - t} \right) = \cos t,\;\sin \left( { - t} \right) = - \sin t.\]

Then we can write:

\[y( - t) = \frac{{\sin \left( {3\left( { - t} \right)} \right) + {{\left( { - t} \right)}^3}}}{{\cos \left( { - t} \right)}} = \frac{{\sin \left( { - 3t} \right) + {{\left( { - t} \right)}^3}}}{{\cos \left( { - t} \right)}} = \frac{{ - \sin 3t - {t^3}}}{{\cos t}} = - \frac{{\sin 3t + {t^3}}}{{\cos t}} = - y\left( t \right).\]

Thus, the function \(y\left( t \right)\) is odd.

Example 3.

Find the range of the function \[y = 5 + 4\sin x\cos x.\]

Solution.

Using the double-angle identity we rewrite the given function in the form

\[y = 5 + 4\sin x\cos x = 5 + 2 \cdot 2\sin x\cos x = 5 + 2\sin 2x.\]

The function \(\sin 2x\) has the range \(\left[ { - 1,1} \right]\). Then the range of the function \(2\sin 2x\) is \(\left[ { - 2,2} \right]\). To determine the range of \(y\) we add \(5\) to the starting and ending values of the interval \(\left[ { - 2,2} \right].\) This yields:

\[\text{codom}\left( y \right) = \left[ { - 2 + 5,2 + 5} \right] = \left[ {3,7} \right].\]

Example 4.

Find the range of the function \[y = 2 + \sin 2x\cos 2x.\]

Solution.

By the double-angle formula, we can write:

\[y = 2 + \sin 2x\cos 2x = 2 + \frac{1}{2} \cdot 2\sin 2x\cos 2x = 2 + \frac{1}{2}\sin 4x.\]

The range of the sine function \(\sin 4x\) is \(\left[ { - 1,1} \right]\). Then the range of \(\frac{1}{2}\sin 4x\) is \(\left[ { - \frac{1}{2},\frac{1}{2}} \right].\) Adding \(2\) to the starting and ending points of the interval gives:

\[\text{codom}\left( y \right) = \left[ {2 + \left( { - \frac{1}{2}} \right),2 + \frac{1}{2}} \right] = \left[ {1.5,2.5} \right].\]

Example 5.

Sketch a graph of the function \[y = 2\sin \left( {t - \frac{\pi }{6}} \right).\]

Solution.

We begin with the basic sine function \({y_1} = \sin t\) and shift its graph by \({\frac{\pi }{6}}\) to the right to get the function

\[{y_2} = \sin \left( {t - \frac{\pi }{6}} \right).\]

Now we stretch the graphs of \(y_2\) vertically by multiplying the \(y-\)coordinate of each point by \(2.\) This gives us the resulting function

\[{y_3} = 2\sin \left( {t - \frac{\pi }{6}} \right).\]
Graph of the function y=2sin(t-pi/6)
Figure 4.

Example 6.

Sketch a graph of the function \[y = - 2\cos 2t.\]

Solution.

We will perform the following transformations successively:

\[\underbrace {\cos t}_{{y_1}} \to \underbrace {\cos 2t}_{{y_2}} \to \underbrace {2\cos 2t}_{{y_3}} \to \underbrace { - 2\cos 2t}_{{y_4}}.\]

The function \({y_2} = \cos 2t\) is obtained from function \({y_1} = \cos t\) by horizontal shrinking by a factor of \(2.\) The period of function \({y_2} = \cos 2t\) is equal to \(\pi.\)

The next transformation \(\underbrace {\cos 2t}_{{y_2}} \to \underbrace {2\cos 2t}_{{y_3}}\) stretches the graph of \({y_2} = \cos 2t\) vertically, so that the function \({y_3} = 2\cos 2t\) varies from \(-2\) to \(2.\)

The last transformation \(\underbrace {2\cos 2t}_{{y_3}} \to \underbrace { - 2\cos 2t}_{{y_4}}\) is a reflection about the \(x-\)axis.

These transformations are illustrated in Figure \(5\) below.

Graph of the function y=-2cos(2t)
Figure 5.
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