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Math Help - Domain and Range of Sine Function

  1. #1
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    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?
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  2. #2
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    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]}

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  3. #3
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    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?
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  4. #4
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    Quote Originally Posted by skeske1234 View Post
    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 .....
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    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

    Are my answers correct?

    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?

    Ok thanks in advance!
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  6. #6
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    Quote Originally Posted by skeske1234 View Post
    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).

    Ok thanks in advance!
    ..
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  7. #7
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    so for the domain, would it be [2, 2/3] then?

    or wouldn't it be
    [2, 2pi-2]
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  8. #8
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    Quote Originally Posted by skeske1234 View Post
    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].
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