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    After expressing an equation in the form RSin(x-alpha), how do i find the 'greatest and least possible values'

    I have 2sinx-3cosx = rt13(x-0.983)

    Write down the greatest and least possible values of 1 + 2sinx-3cosx

    The write down bit signifies to me that this shouldnt take too long.

    im not sure what to do...???
    thanks
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    (Original post by starwarsjedi123)
    Hi
    After expressing an equation in the form RSin(x-alpha), how do i find the 'greatest and least possible values'

    I have 2sinx-3cosx = rt13(x-0.983)

    Write down the greatest and least possible values of 1 + 2sinx-3cosx

    The write down bit signifies to me that this shouldnt take too long.

    im not sure what to do...???
    thanks
    Given something of the form RSin(x-alpha):

    What is the maximum value that sin(x) can take for any value of x?

    Hence what is the maximum value that sin(x-alpha) can take for any values of x or alpha?

    Hence what is the maximum value of..
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    The maximum value is 1? and the minimum is -1?

    So maximum of sin(x-alpha) would be x-alpha=1 ??
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    (Original post by starwarsjedi123)
    The maximum value is 1? and the minimum is -1?

    So maximum of sin(x-alpha) would be x-alpha=1 ??
    If you quote people like so, they can see that you've replied

    Correct, max is 1 and min is -1 for sinx.

    For sin(x-alpha), this is also the same. For any values of x and alpha, we can define some value c = x - alpha, giving sin(x-alpha) = sin(c), which looks just like sinx so the maximum is still 1 and -1.

    With that in mind, you have the R to worry about. So the max of Rsin(x-alpha) is..
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    its simply rt13 is it not??? rt13 stretches sinx parallel to y axis??


    so the answer to my original question is simply 1+rt13 and 1-rt13??

    (Original post by SeanFM)
    If you quote people like so, they can see that you've replied

    Correct, max is 1 and min is -1 for sinx.

    For sin(x-alpha), this is also the same. For any values of x and alpha, we can define some value c = x - alpha, giving sin(x-alpha) = sin(c), which looks just like sinx so the maximum is still 1 and -1.

    With that in mind, you have the R to worry about. So the max of Rsin(x-alpha) is..
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    (Original post by starwarsjedi123)
    its simply rt13 is it not??? rt13 stretches sinx parallel to y axis??


    so the answer to my original question is simply 1+rt13 and 1-rt13??
    Correct :borat: well done.
 
 
 
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