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    So, I need help on not a particular question but on understanding max and min values for trig and what value of theta these max and min values occur out, after the rcos(x-a) or rsin(x-a) questions...

    I know that:
    - max value at which sin(x-a) = 1
    - min value at which sin (x-a) = 0
    I don't understand why...

    Also, what would:
    - max value at which cos(x-a) = ?
    - min value at which cos(x-a) = ?

    Thank you.
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    (Original post by ReallyWigga)
    So, I need help on not a particular question but on understanding max and min values for trig and what value of theta these max and min values occur out, after the rcos(x-a) or rsin(x-a) questions...

    I know that:
    - max value at which sin(x-a) = 1
    - min value at which sin (x-a) = 0
    I don't understand why...

    Also, what would:
    - max value at which cos(x-a) = ?
    - min value at which cos(x-a) = ?

    Thank you.
    No the minimum of sin(anything) is -1. I have a feeling I know why you're confused but you'll need to post the whole question.
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    (Original post by notnek)
    No the minimum of sin(anything) is -1. I have a feeling I know why you're confused but you'll need to post the whole question.
    Ok, I'm very confused now...
    Name:  Screen Shot 2017-01-22 at 14.50.42.png
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    So this is the question(part c) by the way) which gave the answers as sin(x-a) = 0 for the minimum... I don't understand why though...

    Also, I was doing another question and it gave cos(x-a) = -1 as the minimum so I'm definitely cofuggled.
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    (Original post by ReallyWigga)
    Ok, I'm very confused now...
    Name:  Screen Shot 2017-01-22 at 14.50.42.png
Views: 374
Size:  190.7 KB
    So this is the question(part c) by the way) which gave the answers as sin(x-a) = 0 for the minimum... I don't understand why though...

    Also, I was doing another question and it gave cos(x-a) = -1 as the minimum so I'm definitely cofuggled.
    The value of [sin(3\theta - \alpha)]^2 is always positive, so what is always greater than or equal to 0, as it is squared, so what is the smallest value it can be?
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    (Original post by ReallyWigga)
    Ok, I'm very confused now...

    So this is the question(part c) by the way) which gave the answers as sin(x-a) = 0 for the minimum... I don't understand why though...

    Also, I was doing another question and it gave cos(x-a) = -1 as the minimum so I'm definitely cofuggled.
    I thought it might be this question that confused you.

    Notice that the trig expression is squared in this function H. So replacing it with R\sin ... you get something of the form

    \displaystyle H(\theta) = 4 +5\left(R\sin(3\theta - \alpha)\right)^2

    The minimum value of \sin(3\theta - \alpha) is -1 so the minimum of the stuff in the brackets is -R. This is what you're used to.

    But if you square -R you get R^2 and it's no longer small anymore. So if you want to find the minimum of H(\theta) then you want the squared expression to be as small as possible. This happens when the stuff in the brackets is equal to 0.

    Does that make sense?
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    (Original post by notnek)
    I thought it might be this question that confused you.

    Notice that the trig expression is squared in this function H. So replacing it with R\sin ... you get something of the form

    \displaystyle H(\theta) = 4 +5\left(R\sin(3\theta - \alpha)\right)^2

    The minimum value of \sin(3\theta - \alpha) is -1 so the minimum of the stuff in the brackets is -R. This is what you're used to.

    But if you square -R you get R^2 and it's no longer small anymore. So if you want to find the minimum of H(\theta) then you want the squared expression to be as small as possible. This happens when the stuff in the brackets is equal to 0.

    Does that make sense?
    (Original post by NotNotBatman)
    The value of [sin(3\theta - \alpha)]^2 is always positive, so what is always greater than or equal to 0, as it is squared, so what is the smallest value it can be?
    I see that, since the bracket is squared, it could never be a negative. But I don't understand the terms used when you want the expression to be as small as possible?
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    (Original post by ReallyWigga)
    I see that, since the bracket is squared, it could never be a negative. But I don't understand the terms used when you want the expression to be as small as possible?
    Part ci) asks you to find the minimum value of H(\theta). In other words it's asking you to find the smallest that H(\theta) could possibly be. Since there is a squared term in the expression, the smallest that H(\theta) could be is 4+5\times 0 = 4 and this occurs when R\sin(3\theta - \alpha) is equal to 0.

    If that doesn't help then can you please explain in full what you do / do not understand? The more you tell us, the more chance there is that we will be able to help you.
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    (Original post by notnek)
    Part ci) asks you to find the minimum value of H(\theta). In other words it's asking you to find the smallest that H(\theta) could possibly be.
    I don't really understand what you mean by the smallest possible value? I see how you have worked it out, but that statement I don't get.

    Would that mean, to find the maximum, we have to use the greatest possible value, i.e = 1?

    So, in a normal context where nothing is squared...
    Rsin(x-a) = 1 -- maximum
    Rsin(x-a) = -1 -- minimum??
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    (Original post by ReallyWigga)
    I don't really understand what you mean by the smallest possible value? I see how you have worked it out, but that statement I don't get.

    Would that mean, to find the maximum, we have to use the greatest possible value, i.e = 1?

    So, in a normal context where nothing is squared...
    Rsin(x-a) = 1 -- maximum
    Rsin(x-a) = -1 -- minimum??
    What do you think when you see the word "minimum"? "minimum" and "smallest possible value" should mean the same thing to you.

    I think that there's some understanding that you're missing but I don't know what it is. You'll need to go more in depth to explain your issue.

    Would that mean, to find the maximum, we have to use the greatest possible value, i.e = 1?

    So, in a normal context where nothing is squared...
    Rsin(x-a) = 1 -- maximum
    Rsin(x-a) = -1 -- minimum??
    Yes that's all correct.
 
 
 
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