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Maths C3 - Harmonic Identities - Finding Max and Min Values... HELP???

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    (Original post by RDKGames)
    Seems right.
    Sorry I know you're against the use of quadrant diagrams but I seem to make less mistakes this way even though it takes a little longer. The good thing is I think I'm starting to understand Harmonic Identity questions involving Min or Max values
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    Just want to thank you guys (Kevin De Bruyne, notnek and RDKGames) for helping me out and being patient with me with a topic that I always seemed to struggle with! Hopefully it makes more sense to me from now on.
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    (Original post by notnek)
    So do you now know where \sin(3\theta-1.1071...)=0 comes from? I'm asking because this hasn't been explained to you since you said that you don't know where it comes from.
    (Original post by Philip-flop)
    Just want to thank you guys (Kevin De Bruyne, notnek and RDKGames) for helping me out and being patient with me with a topic that I always seemed to struggle with! Hopefully it makes more sense to me from now on.
    Just gonna say it in case he doesn't...

    Your expression is H(\theta)=4+5(\sqrt{20}\sin(3 \theta +\alpha))^2=4+100\sin^2(3 \theta+ \alpha)

    To minimise H, you need to minimise \sin^2(3\theta+\alpha). Clearly the minimum value of this is 0, and it is achieved when \sin(3\theta+\alpha)=0 (because if it were equal anything other than 0, the square of it would not be 0)
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    (Original post by RDKGames)
    Just gonna say it in case he doesn't...

    Your expression is H(\theta)=4+5(\sqrt{20}\sin(3 \theta +\alpha))^2=4+100\sin^2(3 \theta+ \alpha)

    To minimise H, you need to minimise \sin^2(3\theta+\alpha). Clearly the minimum value of this is 0, and it is achieved when \sin(3\theta+\alpha)=0 (because if it were equal anything other than 0, the square of it would not be 0)
    Yeah so the expression is...
    H(\theta)=4+5[\sqrt{20}\sin(3 \theta +1.10714...)]^2

    And to find it's minimum value we let the  sin(3 \theta - 1.10714...) = 0 because we want it to give the smallest possible value for this minimum, thus giving...
     H(\theta) = 4+5 [\sqrt{20} (0) ]^2 = 4

    Right?
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    (Original post by Philip-flop)
    Yeah so the expression is...
    H(\theta)=4+5[\sqrt{20}\sin(3 \theta +\alpha)]^2

    And to find it's minimum value we let the  sin(3 \theta - 1.10714...) = 0 because we want it to give the smallest possible value for this minimum, thus giving...
     H(\theta) = 4+5 [\sqrt{20} (0) ]^2 = 4

    Right?
    You're not wrong, but I don't like the way your argument goes from \sin to \sin^2, rather than the other way round.

    The minimum value of \sin is clearly -1 when minimised but that would mean \sin^2 is 1.

    But if you start your argument by minimising \sin^2, so it is 0, then \sin is 0.
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    (Original post by RDKGames)
    You're not wrong, but I don't like the way your argument goes from \sin to \sin^2, rather than the other way round.

    The minimum value of \sin is clearly -1 when minimised but that would mean \sin^2 is 1.

    But if you start your argument by minimising \sin^2, so it is 0, then \sin is 0.
    Oh I see, so it's better to expand the bracket first so...
    H(\theta)=4+5[\sqrt{20}\sin(3 \theta +1.10714...)]^2

    Becomes...
    H(\theta)=4+100sin^2(3 \theta +1.10714...)

    H(\theta)=4+100sin(3 \theta +1.10714...)^2

    So for the minimum value  sin(3 \theta + 1.10714...) = 0

    Therefore min value of  H(\theta) = 4 + 100(0)^2 = 4

    Is that the way I should do it?
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    (Original post by Philip-flop)
    Oh I see, so it's better to expand the bracket first so...
    H(\theta)=4+5[\sqrt{20}\sin(3 \theta +1.10714...)]^2

    Becomes...
    H(\theta)=4+100sin^2(3 \theta +1.10714...)

    H(\theta)=4+100sin(3 \theta +1.10714...)^2

    So for the minimum value  sin(3 \theta + 1.10714...) = 0

    Therefore min value of  H(\theta) = 4 + 100(0)^2 = 4

    Is that the way I should do it?
    I guess, there's nothing wrong with it... but again, you go from \sin to \sin^2 when it makes more sense to do it the other way around since H(\theta) is dependent on \sin^2, so you want THIS minimised which means \sin^2 =0 and this IMPLIES \sin =0

    Maybe I'm just being pedantic here and I doubt you would lose ANY marks for this, but to minimise H you should notice that first saying "min of \sin" then going onto "min of \sin^2" doesn't flow very nicely because the min of \sin is -1 (not 0 as you claim) and this value does NOT minimise H. Whereas if you went from saying first "min of \sin^2" (which is 0 and minimised H first of all) IMPLIES "min of \sin" (which is also 0 as a result) then it makes sense. It's more logical this way.

    All in all, this bit just lacks argument as to WHY \sin(3 \theta + 1.10714...) is picked as 0 and not its minimum which would be -1.

    (Original post by Philip-flop)

    H(\theta)=4+100sin(3 \theta +1.10714...)^2

    So for the minimum value  sin(3 \theta + 1.10714...) = 0
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    Philip-flop Just adding to what RDKGames said here:
    All in all, this bit just lacks argument as to WHY \sin(3 \theta + 1.10714...) is picked as 0 and not its minimum which would be -1.
    A large percentage of students would think that \sin(3 \theta + 1.10714...) should be -1 to have a minimum because this is what they're used to in most exam questions. It's important that you understand why in this case it should be 0.
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    (Original post by RDKGames)
    I guess, there's nothing wrong with it... but again, you go from \sin to \sin^2 when it makes more sense to do it the other way around since H(\theta) is dependent on \sin^2, so you want THIS minimised which means \sin^2 =0 and this IMPLIES \sin =0

    Maybe I'm just being pedantic here and I doubt you would lose ANY marks for this, but to minimise H you should notice that first saying "min of \sin" then going onto "min of \sin^2" doesn't flow very nicely because the min of \sin is -1 (not 0 as you claim) and this value does NOT minimise H. Whereas if you went from saying first "min of \sin^2" (which is 0 and minimised H first of all) IMPLIES "min of \sin" (which is also 0 as a result) then it makes sense. It's more logical this way.

    All in all, this bit just lacks argument as to WHY \sin(3 \theta + 1.10714...) is picked as 0 and not its minimum which would be -1.
    I don't think I completely follow I don't even know what the minimum of  sin^2 is

    All I can see is that it has to be  sin(3 \theta - 1.10714...) = 0 since if it is -1 we would get  H(\theta) = 4+100(-1)^2 = 104 which is a bigger number than if we have it as 0, therefore  H(\theta) = 4 + 100(0)^2 = 4 which is the smallest possible value we can make for the minimum.
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    (Original post by Philip-flop)
    I don't think I completely follow I don't even know what the minimum of  sin^2 is
    You REALLY need to know that the square of anything that can be 0, IS 0...
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    (Original post by Philip-flop)
    I don't think I completely follow I don't even know what the minimum of  sin^2 is

    All I can see is that it has to be  sin(3 \theta - 1.10714...) = 0 since if it is -1 we would get  H(\theta) = 4+100(-1)^2 = 104 which is a bigger number than if we have it as 0, therefore  H(\theta) = 4 + 100(0)^2 = 4 which is the smallest possible value we can make for the minimum.
    To me this looks like you understand.

    \sin^2 x = (\sin x)^2

    The minimum of \sin x is -1 but the minimum of (\sin x)^2 is 0 for the reasons you say above.

    In this case you have  sin(3 \theta - 1.10714...)^2 and the minmum of this will be the same as the minimum of (\sin x)^2.
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    (Original post by notnek)
    Philip-flop Just adding to what RDKGames said here:

    A large percentage of students would think that \sin(3 \theta + 1.10714...) should be -1 to have a minimum because this is what they're used to in most exam questions. It's important that you understand why in this case it should be 0.
    I think that's why I would normally slip up on these!

    (Original post by RDKGames)
    You REALLY need to know that the square of anything that can be 0, IS 0...
    Yeah I know that  (0)^2 always gives zero but I've never really thought about actually relating that fact to graphs. But I understand why a parabola like  y = x^2 has a minimum point at  (0, 0)
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    (Original post by notnek)
    To me this looks like you understand.

    \sin^2 x = (\sin x)^2

    The minimum of \sin x is -1 but the minimum of (\sin x)^2 is 0 for the reasons you say above.

    In this case you have  sin(3 \theta - 1.10714...)^2 and the minmum of this will be the same as the minimum of (\sin x)^2.
    Yeah that's the way I saw it. But like RDKGames is saying I probably should have known that a graph like  y = sin^2 x has a minimum value of zero so  sin^2 x = 0 due to the outputs when squaring numbers. It's almost like comparing it to  y = x^2

    But I feel like I will slip up more say if I was to adopt this approach to a fractional type Harmonic Identities question
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    Wow, I've never come across a question like this before.
    Name:  OCR C3 June 2012 Q8.png
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Size:  32.7 KB

    Part(c) I actually had no idea what to do, thank god for exam solutions though! Hopefully I'll be prepared if a similar question appears in the Edexcel C3 paper this summer!
 
 
 
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