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anonymous

  • 5 years ago

How do you determine the domain of y=4cosx?

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  1. anonymous
    • 5 years ago
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    harpellet do u know this one The perimeter of square one is 36 cm greater than the perimeter of square two. The area of square one is 243 square cm more than the area of square two. Find the area of the larger square.

  2. anonymous
    • 5 years ago
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    domain is the values which can be assigned to x. since for the cos the value of y will repeat after an interval of pi, the domain is [2n*pi,(2n+1)pi] for n being an integer

  3. anonymous
    • 5 years ago
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    so woudl the anser be [-1,1] or [0,2pi]?

  4. amistre64
    • 5 years ago
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    the domain for cos(x) is: (-inf,inf). The definition of cosine is: x/r and r is always a positive number greater than 0, otherwise you simply have a "not a circle". since r is a constant, x can be any value between negative infinity and positive infinity

  5. amistre64
    • 5 years ago
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    streching the cosine function vertically does not change the domain....

  6. anonymous
    • 5 years ago
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    thank you very much. How about the phase shift of y=3cos(2x-1)?

  7. anonymous
    • 5 years ago
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    Think about your x- values if you move it to the right or left, it is the same domain. It won't change unless you restrict your x values. If that's confusing, then just know that phase shifts and stretches don't change your domain for a cosine and sine curve.

  8. amistre64
    • 5 years ago
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    is that cos(2(x-1))? or cos(2x-1)? to find the phase shift we got to get that "2" away from the x to study it. Lets assume its cos(2x-1) factor out the 2: 2(x - (1/2)) ; now we can see HOW the "x" all by its little lonesome is being harrased and pushed around. The phase shift tells us how far "x" is being moved. The x-(1/2) tells us that the phace shift is (+)1/2. The thing that modifies the x is the OPPOSITE of the phase shift.

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