Pre-Calculus - Graphing the Sine and Cosine Functions

What is the period of this sine graph?

Trig period 1

$\frac{2}{3}\pi$

$\frac{3}{2}\pi$

$3\pi$

$\pi^2$

Explanation

The graph has 3 waves between 0 and $2\pi$ , meaning that the length of each of the waves is $2\pi$ divided by 3, or $\frac{2}{3}\pi$ .

What is the period of this sine graph?

Trig period 1

$\frac{2}{3}\pi$

$\frac{3}{2}\pi$

$3\pi$

$\pi^2$

Explanation

The graph has 3 waves between 0 and $2\pi$ , meaning that the length of each of the waves is $2\pi$ divided by 3, or $\frac{2}{3}\pi$ .

Which equation would produce this sine graph?

Phase shift 2

$y=2sin\left(x-\frac{\pi}{4}\right)-1$

$y=sin\left(x-\frac{3\pi}{4}\right)-1$

$y=sin\left(2x-\frac{\pi}{4}\right)$

$y=2sin\left(x-\frac{\pi}{2}\right)-1$

Explanation

The graph has an amplitude of 2 but has been shifted down 1:

Phase shift 2 dots

In terms of the equation, this puts a 2 in front of sin, and -1 at the end.

This makes it easier to see that the graph starts \[is at 0\] where $x=\frac{\pi}{4}$ .

The phase shift is $\frac{\pi}{4}$ to the right, or $x-\frac{\pi}{4}$ .

Which equation would produce this sine graph?

Phase shift 2

$y=2sin\left(x-\frac{\pi}{4}\right)-1$

$y=sin\left(x-\frac{3\pi}{4}\right)-1$

$y=sin\left(2x-\frac{\pi}{4}\right)$

$y=2sin\left(x-\frac{\pi}{2}\right)-1$

Explanation

The graph has an amplitude of 2 but has been shifted down 1:

Phase shift 2 dots

In terms of the equation, this puts a 2 in front of sin, and -1 at the end.

This makes it easier to see that the graph starts \[is at 0\] where $x=\frac{\pi}{4}$ .

The phase shift is $\frac{\pi}{4}$ to the right, or $x-\frac{\pi}{4}$ .

Write the equation for a sine graph with a maximum at $\left(\frac{ \pi }{8}, 3\right)$ and a minimum at $\left(\frac{ 9 \pi }{8 } , -1 \right)$ .

$y = 2 \sin (x + \frac{ 3 \pi }{ 8 } ) + 1$

$y = 2 \sin (x - \frac{ 3 \pi }{8 } ) - 1$

$y = 2 \sin ( \frac{1}{2}( x - \frac{ \pi }{8})) + 1$

$y = 3 \sin (\frac{ 1}{2} x )$

$y = 2 \sin (x - \frac{ \pi }{ 8 }) + 1$

Explanation

To write this equation, it is helpful to sketch a graph:

Trig graph 1

Indicating the maximum and minimum points, we can see that this graph has been shifted up 1, and it has an amplitude of 2.

The distance from the maximum to the minimum point is half the wavelength. In this case, the wavelength is $\frac{ 9 \pi }{ 8 } - \frac{ \pi }{8 } = \frac{ 8 \pi }{8 } = \pi$ . That means the full wavelength is $2 \pi$ , and the frequency is 1.

This sketch shows that the graph starts to the left of the y-axis. To figure out exactly where, subtract $\frac{ \pi }{2}$ from the maximum x-coordinate, $\frac{ \pi }{ 8 }$ :

$\frac{ \pi }{ 8 } - \frac{ \pi }{2} = \frac{ \pi }{8 } - \frac{ 4 \pi }{ 8 } = \frac{ - 3 \pi }{ 8 }$ .

Our equation will be in the form $y = A \cdot \sin ( f (x - h )) + k$ where A is the amplitude, f is the frequency, h is the horizontal shift, and k is the vertical shift.

This graph has an equation of

$y = 2 \sin ( x + \frac{ 3 \pi }{ 8 } ) + 1$ .

Which of the given functions has the greatest amplitude?

$2\sin(x)$

$\sin(4x)$

$\sin(x+3)$

$\sin\frac{2\pi }{3}(x+2)$

$\frac{1}{2}\sin(4x)$

Explanation

The amplitude of a function is the amount by which the graph of the function travels above and below its midline. When graphing a sine function, the value of the amplitude is equivalent to the value of the coefficient of the sine. Similarly, the coefficient associated with the x-value is related to the function's period. The largest coefficient associated with the sine in the provided functions is 2; therefore the correct answer is $2\sin(x)$ .

The amplitude is dictated by the coefficient of the trigonometric function. In this case, all of the other functions have a coefficient of one or one-half.

Which of the given functions has the greatest amplitude?

$2\sin(x)$

$\sin(4x)$

$\sin(x+3)$

$\sin\frac{2\pi }{3}(x+2)$

$\frac{1}{2}\sin(4x)$

Explanation

The amplitude is dictated by the coefficient of the trigonometric function. In this case, all of the other functions have a coefficient of one or one-half.

Write the equation for a sine graph with a maximum at $\left(\frac{ \pi }{8}, 3\right)$ and a minimum at $\left(\frac{ 9 \pi }{8 } , -1 \right)$ .

$y = 2 \sin (x + \frac{ 3 \pi }{ 8 } ) + 1$

$y = 2 \sin (x - \frac{ 3 \pi }{8 } ) - 1$

$y = 2 \sin ( \frac{1}{2}( x - \frac{ \pi }{8})) + 1$

$y = 3 \sin (\frac{ 1}{2} x )$

$y = 2 \sin (x - \frac{ \pi }{ 8 }) + 1$

Explanation

To write this equation, it is helpful to sketch a graph:

Trig graph 1

Indicating the maximum and minimum points, we can see that this graph has been shifted up 1, and it has an amplitude of 2.

This sketch shows that the graph starts to the left of the y-axis. To figure out exactly where, subtract $\frac{ \pi }{2}$ from the maximum x-coordinate, $\frac{ \pi }{ 8 }$ :

$\frac{ \pi }{ 8 } - \frac{ \pi }{2} = \frac{ \pi }{8 } - \frac{ 4 \pi }{ 8 } = \frac{ - 3 \pi }{ 8 }$ .

Our equation will be in the form $y = A \cdot \sin ( f (x - h )) + k$ where A is the amplitude, f is the frequency, h is the horizontal shift, and k is the vertical shift.

This graph has an equation of

$y = 2 \sin ( x + \frac{ 3 \pi }{ 8 } ) + 1$ .

Write the equation for a cosine graph with a maximum at $\left( \frac{ \pi }{3} , 3 \right)$ and a minimum at $\left(\frac{ 4 \pi }{3} , -3 \right)$ .

$y = 3 \cos (x - \frac{ \pi }{3} )$

$y = 2 \cos (x + \frac{ \pi }{3} )+ 1$

$y = 3 \cos (\frac{ 1}{2} ( x - \frac{ \pi }{3} ))$

$y = -3 \cos (x - \frac{ \pi } {3 })$

$y = 2 \cos (\frac{ 1}{2} ( x + \frac{ \pi }{3} )) + 1$

Explanation

In order to write this equation, it is helpful to sketch a graph:

Trig graph 2

The dotted line is at $x = \frac{ \pi }{3}$ , where the maximum occurs and therefore where the graph starts. This means that the graph is shifted to the right $\frac{ \pi }{3}$ .

The distance from the maximum to the minimum is half the entire wavelength. Here it is $\frac{ 4 \pi }{3} - \frac{ \pi }{3} = \frac{ 3 \pi }{3} = \pi$ .

Since half the wavelength is $\pi$ , that means the full wavelength is $2 \pi$ so the frequency is just 1.

The amplitude is 3 because the graph goes symmetrically from -3 to 3.

The equation will be in the form $y = A \cdot \cos (f (x - h)) + k$ where A is the amplitude, f is the frequency, h is the horizontal shift, and k is the vertical shift.

This equation is

$y = 3 \cos (x - \frac{ \pi } {3 })$ .

Write the equation for a cosine graph with a maximum at $\left( \frac{ \pi }{3} , 3 \right)$ and a minimum at $\left(\frac{ 4 \pi }{3} , -3 \right)$ .

$y = 3 \cos (x - \frac{ \pi }{3} )$

$y = 2 \cos (x + \frac{ \pi }{3} )+ 1$

$y = 3 \cos (\frac{ 1}{2} ( x - \frac{ \pi }{3} ))$

$y = -3 \cos (x - \frac{ \pi } {3 })$

$y = 2 \cos (\frac{ 1}{2} ( x + \frac{ \pi }{3} )) + 1$

Explanation

In order to write this equation, it is helpful to sketch a graph:

Trig graph 2

The dotted line is at $x = \frac{ \pi }{3}$ , where the maximum occurs and therefore where the graph starts. This means that the graph is shifted to the right $\frac{ \pi }{3}$ .

The distance from the maximum to the minimum is half the entire wavelength. Here it is $\frac{ 4 \pi }{3} - \frac{ \pi }{3} = \frac{ 3 \pi }{3} = \pi$ .

Since half the wavelength is $\pi$ , that means the full wavelength is $2 \pi$ so the frequency is just 1.

The amplitude is 3 because the graph goes symmetrically from -3 to 3.

The equation will be in the form $y = A \cdot \cos (f (x - h)) + k$ where A is the amplitude, f is the frequency, h is the horizontal shift, and k is the vertical shift.

This equation is

$y = 3 \cos (x - \frac{ \pi } {3 })$ .

Graphing the Sine and Cosine Functions

Help Questions

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