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    which formulaes do i need memorise for trigonometry (c3)?
    ive done the double angle formula,
    cot2 and cosec2,
    tan2 and sec2,
    sin2 and cos2...
    anymore?
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    (Original post by obviouslystudying)
    which formulaes do i need memorise for trigonometry (c3)?
    ive done the double angle formula,
    cot2 and cosec2,
    tan2 and sec2,
    sin2 and cos2...
    anymore?
    secx = 1/cosx

    cosecx=1/sinx

    cotx = 1/tanx
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    (Original post by gottastudy)
    secx = 1/cosx

    cosecx=1/sinx

    cotx = 1/tanx
    yeah, know them but sort of forgot since i think of them as a sort of definition for sec, cosec and cot.
    thanks
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    Q5. part d)

    Find the exact values of x for which |2/x-3| = 3

    not sure how they obtained 3 2/3 and 2 1/3 fully.
    i apprently lose marks if i multiply the 3 with x-3.
    help?

    thanks!
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    (Original post by Dyslex!c Duck)
    Q5. part d)

    Find the exact values of x for which |2/x-3| = 3

    not sure how they obtained 3 2/3 and 2 1/3 fully.
    i apprently lose marks if i multiply the 3 with x-3.
    help?

    thanks!
    Getting rid of the modulus signs gives:

    \frac{2}{x-3} = 3 and \frac{-2}{x-3} = 3

    These can be solved individually to find both values of x.
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    sorry, my pc is freezing needs a format
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    AH ermm doesn't matter since i am doing AQA + its in june :P , the topic which i was iffy about was integration by substitution but i am alright now .
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    (Original post by sulexk)
    Hello there,

    need help with any edexcel c3/c4 past paper questions?

    let me know

    thank you so much!!!
    Mate how do you solve the equation sin5x+sinx=0 for x in the interval between 0 and 180 degrees?? Thank you :borat:
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    (Original post by sulexk)
    Hello there,

    need help with any edexcel c3/c4 past paper questions?

    let me know

    thank you so much!!!

    Please help with this question.

    January 2007:

    Given that: y = arccos x, -1 < x < 1 and 0 < y < pi,


    (a) express arcsin x in terms of y. (2)

    Edit: I know the answer to part (a), but I have no idea at all how to get the answer.

    (b) Hence evaluate arccos x + arcsinx. Give your answer in terms of pi. (1)

    ________

    If you can help, quote me in the reply please.
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    (Original post by ViolinGirl)
    You do-you start off by using the product rule.

    y=uv where u= x and v = e^x^2

    when you differentiate these you should get: du/dx = 1 and dv/dx= 2xe^x^2

    Then dy/dx = vdu/dx + udv/dx

    e^x^2 + 2x^2e^x^2

    This can then be simplified.

    As far as I can see, it only really involves the product rule. I may have omitted something. What is the answer?
    yeah, thats how its shown, i just cant get dv/dx
    well, how did you get that???
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    (Original post by ViolinGirl)
    You do-you start off by using the product rule.

    y=uv where u= x and v = e^x^2

    when you differentiate these you should get: du/dx = 1 and dv/dx= 2xe^x^2

    Then dy/dx = vdu/dx + udv/dx

    e^x^2 + 2x^2e^x^2

    This can then be simplified.

    As far as I can see, it only really involves the product rule. I may have omitted something. What is the answer?

    Sorry...forgot I didn't answer that...

    ok dv/dx

    v = e^x^2

    First you differentiate the power, with the chain rule, pretty much normal differentitation.

    Power is x^2 ...this becomes 2x

    But the rule with e^x is that dy/dx is e^x

    so it becomes 2xe^x^2

    Another example to make it clearer:

    e^2x would become 2e^2x

    Diferentiate power, and its that multiplies by e^ the orginial power.

    Hope that helps.
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    (Original post by obviouslystudying)
    well, how did you get that???
    Above ^^
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    (Original post by obviouslystudying)
    How do you differentiate when the base has got powers thats also got powers;
    like: 7e^{x^y}?
    This seems harder than C3 - but I think I figured it out.

    Derivative of e^{f(x)} = f'(x)e^{f(x)}
    therefore if we can find the derivative of x^y, we can do the rest pretty easily.
    So, x^y is a bit tricky, but as far as I can see, x^y = e^{ylnx}
    therefore we can differentiate e^{ylnx} using the rule above and the product rule:
    (lnx(\frac{dy}{dx}) + (\frac{y}{x}))e^{ylnx}

    Therefore;
    (7(lnx(\frac{dy}{dx}) + (\frac{y}{x}))e^{ylnx})e^{x^y} = \frac{d}{dx}(7e^{x^y})

    I think

    :P
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    (Original post by Mathematician!)
    Getting rid of the modulus signs gives:

    \frac{2}{x-3} = 3 and \frac{-2}{x-3} = 3

    These can be solved individually to find both values of x.
    silly me.
    i solved it wrong before.
    i basically moved the x-3 over to the 3
    then divided the 2 and -2 with 3 whic is wrong.
    haaa
    thanks!
    (:
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    (Original post by 4WhenImBanned)
    Please help with this question.

    January 2007:

    Given that: y = arccos x, -1 < x < 1 and 0 < y < pi,


    (a) express arcsin x in terms of y. (2)

    Edit: I know the answer to part (a), but I have no idea at all how to get the answer.

    (b) Hence evaluate arccos x + arcsinx. Give your answer in terms of pi. (1)

    ________

    If you can help, quote me in the reply please.
    For this question you basically need to know cosx = sin (pi/2 -y) therefore arcsin = (pi/2 -y)

    Then knowing this arc sinx + arccos x would be (pi/2 -y) + y = pi/2

    ----

    Can anyone answer my previous question?

    1. Using this as my dy/dx = -e^2 x^-2 +e^x (so in words: minus e to the power of two multiplied by x to the power of minus two plus e to the power of x)

    How do I get the gradient of 3/4e^2 like it says in the MS if I insert in x=2?

    2. For a cosec graph with a minimum at (pi/2, −1), a maximum at ( 3pi/2 , −5) and an asymptote with equation x = pi.

    (a) Showing the coordinates of any stationary points, sketch the graph of y =
    Given that
    f : x → a + b cosec x, x ∈ , 0 < x < 2π, x ≠ π,
    (b) find the values of the constants a and b

    Thanks in advance!
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    For 1) probably get a better response if you post the question. I don't have the paper.

    2) Again think you need to type this out abit more clearly, y = ... ? is that (a+b) or a + bcosecx ? I'm guessing this is a case of a transformation of some kind.
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    Ahh I'm sorry I didn't realise it didn't paste the rest of the equation!

    1) http://www.mathsandscience.org/resources/EC3sh_H.pdf

    Question 8c.

    2)This is my second question (Q4b) :

    http://www.mathsandscience.org/resources/EC3sh_H.pdf

    Thanks in advance and sorry about any confusion!
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    (Original post by Apple24)
    For this question you basically need to know cosx = sin (pi/2 -y) therefore arcsin = (pi/2 -y)

    Then knowing this arc sinx + arccos x would be (pi/2 -y) + y = pi/2
    Hi, thanks but how do you know cosx = sin (pi/2 - y)?
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    (Original post by Toneh)
    This seems harder than C3 - but I think I figured it out.

    Derivative of e^{f(x)} = f'(x)e^{f(x)}
    therefore if we can find the derivative of x^y, we can do the rest pretty easily.
    So, x^y is a bit tricky, but as far as I can see, x^y = e^{ylnx}
    therefore we can differentiate e^{ylnx} using the rule above and the product rule:
    (lnx(\frac{dy}{dx}) + (\frac{y}{x}))e^{ylnx}

    Therefore;
    7(lnx(\frac{dy}{dx}) + (\frac{y}{x}))e^{x^y} = \frac{d}{dx}(7e^{x^y})

    I think

    :P
    power of powers is in the c3 heinemann book

    e^{f(x)} = f'(x)e^{f(x)}
    thats the rule i was searching for. thanks alot

    the violin girl too, thanks
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    (Original post by obviouslystudying)
    power of powers is in the c3 heinemann book

    e^{f(x)} = f'(x)e^{f(x)}
    thats the rule i was searching for. thanks alot
    o yer, powers of powers is alright, but x^y where y is not a constant is harder :P
 
 
 
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