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    I know how to get max value but not min value q b i and ii
    R = SQUARE ROOT 10
    a is 1.25
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    (Original post by koolgurl14)
    I know how to get max value but not min value q b i and ii
    R = SQUARE ROOT 10
    a is 1.25
    The minimum value of \sin(x-\alpha) is -1, so the minimum value of R\sin(x-\alpha) = -R and this occurs precisely when \sin (x-\alpha) = -1 \Rightarrow x - \alpha = \frac{3\pi}{2} or some other reasonable value.
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    Easiest way to do this is to say what value of sin(.....) makes sqrt10 sin(.....) as small as possible. Given as you know that sin(...) ranges from -1 to 1, the smallest value occurs when sin(......) is -1, so the minimum value is -sqrt10. As for the second part, you know that sin(x-1.25)=-1, so just solve for x.
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    (Original post by Zacken)
    The minimum value of \sin(x-\alpha) is -1, so the minimum value of R\sin(x-\alpha) = -R and this occurs precisely when \sin (x-\alpha) = -1 \Rightarrow x - \alpha = \frac{3\pi}{2} or some other reasonable value.
    Hmm what about R>0 or does that not affect the results
    its the right answer btw im just asking why we should ignore the R>0
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    (Original post by koolgurl14)
    Hmm what about R>0 or does that not affect the results
    its the right answer btw im just asking why we should ignore the R>0
    What do you mean by ignoring R>0?

    We have an expression R\sin(x-\alpha). To minimise this, we look at the variable terms and notice that \sin is bounded between -1 (minimum) and 1 (maximum), so to maximise our entire function we pick the minimum sine value and hence say that the minimum of R\sin(x-\alpha) = R(-1) = -R.
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    (Original post by koolgurl14)
    Hmm what about R>0 or does that not affect the results
    its the right answer btw im just asking why we should ignore the R>0
    R>0 only applies to finding the value of R
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    Ahhh I see what you mean its just cause in the question it says the range of R is R>0 so I got confused but I guess its not actual R being less than 0?
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    (Original post by koolgurl14)
    Ahhh I see what you mean its just cause in the question it says the range of R is R>0 so I got confused but I guess its not actual R being less than 0?
    Ohh, it just means that when you do R^2 = 1^2 + (-3)^2 = 10 then you have R = \pm \sqrt{10} but since R > 0 you pick R = \sqrt{10} since -\sqrt{10} < 0
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    (Original post by Zacken)
    Ohh, it just means that when you do R^2 = 1^2 + (-3)^2 = 10 then you have R = \pm \sqrt{10} but since R > 0 you pick R = \sqrt{10} since -\sqrt{10} < 0

    ohhhhhh that makes a lot of sense thanks alot
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    (Original post by koolgurl14)
    ohhhhhh that makes a lot of sense thanks alot
    No problem!
 
 
 
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