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    Guys, I need help with an issue on solving trig equations using identities (for C2 maths). Before study leave, my teacher was going on about something to do with 'not losing solutions' when simplifying a trig equation? I have an IDEA of what she meant but I'm not 100% sure. So could someone please clarify? Thanks
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    (Original post by GCSE-help)
    Guys, I need help with an issue on solving trig equations using identities (for C2 maths). Before study leave, my teacher was going on about something to do with 'not losing a trig function' when simplifying a trig equation? I have an IDEA of what she meant but I'm not 100% sure. So could someone please clarify? Thanks
    I'm pretty sure they were talking about dividing through by a factor which could produce values..

    EG;

    2cos^2x-4cosx=0



2cosx(cosx-2)=0



cosx=0 or cosx=2

    In other words, don't forget about the cosx=0 part!
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    (Original post by krishkmistry)
    Help me ? Attachment 215628
    Just use the trapezium rule formula for the first part.

    For the second part, it's an underestimate because the shape of the graph is concave down.

    In general,

    If the shape of the graph is concave up (i.e. f''(x) > 0), the trapezium rule will overestimate the area.

    If the shape of the graph is concave down (i.e. f''(x) < 0), the trapezium rule will underestimate the area.

    Part (i), just sketch the graph... look it up in your textbook.

    Part (ii),

    4\sin x = 3\cos x

    Divide both sides by cos, then divide both sides by 4:

    \dfrac{\sin x}{\cos x} = \dfrac{3}{4}

    Can you do this now?

    And this ? Attachment 215630
    I don't think this sort of question will come out in the real exam, but I'll try and help you out anyway.

    Treat it like a normal straight line, so:

    \log_{10} y = mx + c
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    (Original post by justinawe)
    Just use the trapezium rule formula for the first part.

    For the second part, it's an underestimate because the shape of the graph is concave down.

    In general,

    If the shape of the graph is concave up (i.e. f''(x) > 0), the trapezium rule will overestimate the area.

    If the shape of the graph is concave down (i.e. f''(x) < 0), the trapezium rule will underestimate the area.



    Part (i), just sketch the graph... look it up in your textbook.

    Part (ii),

    4\sin x = 3\cos x

    Divide both sides by cos, then divide both sides by 4:

    \dfrac{\sin x}{\cos x} = \dfrac{3}{4}

    Can you do this now?



    I don't think this sort of question will come out in the real exam, but I'll try and help you out anyway.

    Treat it like a normal straight line, so:

    \log_{10} y = mx + c
    Thank you so much


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    (Original post by krishkmistry)
    Thank you so much


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    Are you sure about the trapezium rule because if the second derivative is less than 0 it's a max point so it's a concave up graph ?


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    Afternoon guys. How did M1/C1 go yesterday?
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    (Original post by DJMayes)
    Afternoon guys. How did M1/C1 go yesterday?
    Hurriedly I would say

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    (Original post by MAyman12)
    I actually can't believe that I did the stupidest mistakes in that exam. First in the moments question, I only think it's hard because I've never did anything like it before, but I'm quite happy that I got it all right except the final answer because of a lousy mistake. And the question about the woman in the lift I didn't substitute (-2) in the equation, instead I substituted (2) I hope I could get a few marks at least.
    We shall just have to wait until results day now
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    (Original post by DJMayes)
    Afternoon guys. How did M1/C1 go yesterday?
    I'm ashamed to say I'm going to study Maths next year as I wrote down the area of a triangle wrong, but other than that it was all good some different questions though apparently.
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    (Original post by justinawe)
    In general, Just use the trapezium rule formula for the first part.

    For the second part, it's an underestimate because the shape of the graph is concave down.


    If the shape of the graph is concave up (i.e. f''(x) > 0), the trapezium rule will overestimate the area.

    If the shape of the graph is concave down (i.e. f''(x) < 0), the trapezium rule will underestimate the area.

    Are you sure about this?
    I think that was an overestimate

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    (Original post by StUdEnTIGCSE)
    Are you sure about this?
    I think that was an overestimate

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    Yes.

    http://en.wikipedia.org/wiki/Trapezoidal_rule

    It follows that if the integrand is concave up (and thus has a positive second derivative), then the error is negative and the trapezoidal rule overestimates the true value. This can also be seen from the geometric picture: the trapezoids include all of the area under the curve and extend over it. Similarly, a concave-down function yields an underestimate because area is unaccounted for under the curve, but none is counted above.
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    (Original post by MathsNerd1)
    I'm ashamed to say I'm going to study Maths next year as I wrote down the area of a triangle wrong, but other than that it was all good some different questions though apparently.
    :lol:

    I've been receiving mixed reviews of the paper, where mixed here means that one person is telling me it was a very nice paper and everyone else thinks that it was somewhat awkward.
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    (Original post by justinawe)
    Yes.
    Hey that was concave up!
    http://tutorial.math.lamar.edu/Class...GraphPtII.aspx


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    (Original post by DJMayes)
    :lol:

    I've been receiving mixed reviews of the paper, where mixed here means that one person is telling me it was a very nice paper and everyone else thinks that it was somewhat awkward.
    Well everyone bar 2-3 people including myself at my school found it quite hard and couldn't answer half the questions when they were averaging at a high B beforehand, but I'm not sure whether it'll lower the grade boundaries significatnly enough to change what 100 UMS is.
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    Ah, yes, you have me there

    My apologies, I should have looked at the graph better to make sure.
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    Can you draw both graphs and explain it ?


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    (Original post by MathsNerd1)
    Well everyone bar 2-3 people including myself at my school found it quite hard and couldn't answer half the questions when they were averaging at a high B beforehand, but I'm not sure whether it'll lower the grade boundaries significatnly enough to change what 100 UMS is.
    From what I gather there's one person looking at full UMS in my year. I haven't properly looked at the paper though so I can't really comment on it.
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    (Original post by krishkmistry)
    Can you draw both graphs and explain it ?


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    I made a mistake, it was actually concave up, so it was an overestimate.

    Check out the link StudentIGCSE posted: http://tutorial.math.lamar.edu/Class...GraphPtII.aspx
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    (Original post by DJMayes)
    From what I gather there's one person looking at full UMS in my year. I haven't properly looked at the paper though so I can't really comment on it.
    The reactions in the M1 thread is brutal!, 9/10 of the people I've asked said it was really hard!
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    (Original post by Boy_wonder_95)
    The reactions in the M1 thread is brutal!, 9/10 of the people I've asked said it was really hard!
    That's typical, to be honest. People always think the real thing was harder than the past papers, but generally the only reason for this is because they weren't under the same kind of pressure when doing the past papers.
 
 
 
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