# Thread: Domain and Range of Sine Function

1. ## Domain and Range of Sine Function

I need help finding the domain and range of one cycle of the sine function:

y=3sin(x+pi/4)-1

Can you explain it to me please?

I know for the domain is [-pi/4, ]

I just don't know the second part of it? how do I find this point on the graph exactly?

2. Hello, skeske1234!

Find the domain and range of one cycle of the function:

$y\:=\;3\sin\left(x+\tfrac{\pi}{4}\right)-1$
We are expected to know this . . .

For one cycle of the basic sine function, $y \:=\:\sin\theta,\quad \begin{array}{ccccc}\text{Domain:} & 0 \:\leq \:\theta \:\leq\: 2\pi \\ \text{Range:} & -1 \:\leq\:\sin\theta \:\leq\: 1 \end{array}$

In our problem, $\theta \,=\,x + \tfrac{\pi}{4}$

So . $0 \leq\theta\leq2\pi$ .becomes: . $0 \;\leq\; x+\tfrac{\pi}{4} \;\leq\; 2\pi$

. . Subtract $\tfrac{\pi}{4}$ from each side: . $-\tfrac{\pi}{4} \;\leq\; x \;\leq\; \tfrac{7\pi}{4}$

. . ${\color{blue}\text{Domain: }\;\left[-\frac{\pi}{4},\:\frac{7\pi}{4}\right]}$

And . $-1 \leq \sin\theta \leq 1$ . becomes: . $-1 \;\leq\;\sin(x + \tfrac{\pi}{4}) \;\leq\; 1$

. . Multiply by 3: . $-3 \;\leq\;3\sin(x+\tfrac{\pi}{4}) \;\leq\;3$

. . Subtract 1: . $-4 \;\leq\;3\sin(x + \tfrac{\pi}{4})-1 \;\leq\; 2$

. . ${\color{blue}\text{Range: }\;[-4,\:2]}$

3. For this sine function:

y=-2sin[0.5(x-5pi/6)]+4

Can you just check for me if I have the domain and range correct?

Domain:

[5pi/6, 17pi/6]

Range:

[6, 2]

another q:
is wave length the same thing as the period?

4. Originally Posted by skeske1234
For this sine function:

y=-2sin[0.5(x-5pi/6)]+4

Can you just check for me if I have the domain and range correct?

Domain:

[5pi/6, 17pi/6]

Mr F says: Since the period is 2pi/(1/2) = 4pi, you only have the domain for half a period ....

Range:

[6, 2]

Mr F says: Correct.

another q:
is wave length the same thing as the period?
No. For one thing, the dimension of wavelength is distance and the dimension of period is time.

But I suspect that's not really the question you need to ask, in which case that might not be the answer I want to give. Post the exact question that has led you to ask this. Mathematical period can be different from physical period .....

5. Originally Posted by mr fantastic
No. For one thing, the dimension of wavelength is distance and the dimension of period is time.

But I suspect that's not really the question you need to ask, in which case that might not be the answer I want to give. Post the exact question that has led you to ask this. Mathematical period can be different from physical period .....

My question is:

y=7cos[3pi(x-2)]+7

Wave length:
2/3

Period
2/3

Domain
[2, ] ????????????????!
What is the second part of it? Can you help?

Range
[0, 14]

Also, would you describe this cosine function as:

horizontally compressed by a factor of 1/3pi?

6. Originally Posted by skeske1234
My question is:

y=7cos[3pi(x-2)]+7

Are my answers correct? Mr F says: I don't know. What are y and x meant to represent? Do you mean wavelength and period in a physical sense? I suppose that in a mathematical sense wavelength and period are the same.

Wave length:
2/3

Period
2/3

Domain
[2, ] ????????????????!
What is the second part of it? Can you help? Mr F says: Assuming you want the dmain over one period, use the fact that the mathematical period is 2pi/3pi = 2/3 to get the other endpoint of the domain.

Range
[0, 14]

Also, would you describe this cosine function as:

horizontally compressed by a factor of 1/3pi? Mr F says: Horizontal diliation by factor of 1/(3pi).

..

7. so for the domain, would it be [2, 2/3] then?

or wouldn't it be
[2, 2pi-2]

8. Originally Posted by skeske1234
so for the domain, would it be [2, 2/3] then?

or wouldn't it be
[2, 2pi-2]
The period is 2/3 so a domain over one period is [2, 2 + 2/3].