# 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

${y_3} = 2\sin \left( {t - \frac{\pi }{6}} \right).$

### 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.