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    (Original post by Zacken)
    Use the fact that some trigonometric functions are shifts of another, common examples include \sin \theta = \cos (\pi/2 - \theta), \cot \theta = \tan (\pi/2 - \theta) and the likes. This comes in useful and saves a lot of time.

    Say you were given \csc \left(\theta + \frac{\pi}{3}\right) = \sec \left(\theta - \frac{\pi}{6}\right) - this could be an 8/9/10mark A-Level question easily; want a three liner way of doing it?

    The equation is equivalent to \displaystyle \sin \left(\theta + \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{2} - \theta + \frac{\pi}{6}\right) = \sin \left(\frac{2\pi}{3} - \theta\right)

    which now becomes elementary to solve.

    Extension to this, using a STEP question, show that \sin A = \cos B \iff A = (4n+1)\frac{\pi}{2} \pm B - should be a two or three liner using this method.

    This trick has rendered one or two STEP I questions completely inane, especially the \cot \theta = \tan (\pi/2 - \theta) that's let me do a STEP I question in 4 minutes.
    For the necessary condition, why is it that taking sine of the equation gives you -sin=cos whereas taking the cosine proves it for +B. I can see the symmetry but how do I know which one to use? Pic related will explain my question better tbh.

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    (Original post by Insight314)
    For the necessary condition, why is it that taking sine of the equation gives you -sin=cos whereas taking the cosine proves it for +B. I can see the symmetry but how do I know which one to use? Pic related will explain my question better tbh.
    I'll have a look at this in a bit, I'm off to watch a movie - but the way I'd do this (the intended way) was:

    \sin X = \sin Y \iff X = Y + 2n\pi \, \text{or} \, X = \pi - Y + 2n\pi which gives us necessity and sufficiency in one fell swoop and is applicable here since \cos B = \sin (\pi/2 - B).
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    (Original post by Insight314)
    For the necessary condition, why is it that taking sine of the equation gives you -sin=cos whereas taking the cosine proves it for +B. I can see the symmetry but how do I know which one to use? Pic related will explain my question better tbh.




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    Actually - I've spotted it right away, we have \sin \left(A - (4n+1)\frac{\pi}{2}\right) = \sin \pm B.

    But this gives us 0 - \cos A = \sin \pm B which is the same thing as -\cos A = \pm \sin B which certainly gives us \cos A = \sin B.

    The +B comes from the \cos and then -B comes from the \sin.

    So remember when we have -\cos A = \pm \sin B we have cos A is the same as (+sin B or -sin B) only one of those have to be true, not both.
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    (Original post by Zacken)
    Actually - I've spotted it right away, we have \sin \left(A - (4n+1)\frac{\pi}{2}\right) = \sin \pm B.

    But this gives us 0 - \cos A = \sin \pm B which is the same thing as -\cos A = \pm \sin B which certainly gives us \cos A = \sin B.

    The +B comes from the \cos and then -B comes from the \sin.

    So remember when we have -\cos A = \pm \sin B we have cos A is the same as (+sin B or -sin B) only one of those have to be true, not both.
    Oh! Are you saying that I couldn't prove it because I was messing up my boolean logic/truth tables? So if we have +sin B or -sin B, it would be enough to say that cos A = sin B by just emitting the -sin B ?
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    (Original post by Insight314)
    Oh! Are you saying that I couldn't prove it because I was messing up my boolean logic/truth tables? So if we have +sin B or -sin B, it would be enough to say that cos A = sin B by just emitting the -sin B ?
    Precisely.
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    (Original post by Zacken)
    Precisely.
    You know how you feel really happy when you solve a math problem and learn something new? That is what I am feeling right now. Gonna start STEP III 2006, do one question and start M5 - I've been doing so much STEP today that I have forgotten I am really behind on Further Mathematics A-level.

    Thank you again. Have fun, and I hope you enjoy your movie!
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    (Original post by Insight314)
    You know how you feel really happy when you solve a math problem and learn something new? That is what I am feeling right now. Gonna start STEP III 2006, do one question and start M5 - I've been doing so much STEP today that I have forgotten I am really behind on Further Mathematics A-level.

    Thank you again. Have fun, and I hope you enjoy your movie!
    Thanks! I'll be doing STEP III 2006 as a mock on Monday or so.

    ...I need to start Physics. :lol:

    Thanks.
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    (Original post by Zacken)
    Say you wanted to differentiate a nasty function that might require multiple uses of the product rule or such, take logs!

    Why take logs when you can take natural logs? Or is this what you meant?

    (Original post by Zacken)
    Extension to this, using the same idea, roughly - attempt \displaystyle \int_0^{\pi/2} \log \sin \theta \, \mathrm{d}\theta
    Similarly, you meant to say natural log (\ln) right?
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    (Original post by Insight314)
    Why take logs when you can take natural logs? Or is this what you meant?



    Similarly, you meant to say natural log (\ln) right?
    At any level higher than A-level \log and \ln are interchangeable, in mathematics at least.
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    (Original post by joostan)
    At any level higher than A-level \log and \ln are interchangeable, in mathematics at least.
    All right, I see. Thank you.
    I was a bit confused because I thought he meant log in base 10.
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    (Original post by Insight314)
    Why take logs when you can take natural logs? Or is this what you meant?



    Similarly, you meant to say natural log (\ln) right?
    yeah he means natural log

    edit:
    well looks like I'm late to the party
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    (Original post by Zacken)
    Using implicit differentiation can make life easier.

    Say you wanted to differentiate a nasty function that might require multiple uses of the product rule or such, take logs!

    Let's say you have f(x) = x^p(x+1)^q(x+2)^r, do you really want to differentiate this normally? I'd hope you said no. What you'd want to do is:

    \displaystyle 

\begin{equation*}\log f(x) = p \log x + q \log (x+1) + r \log (x+2) \Rightarrow f'(x) = f(x)\left(\frac{p}{x} + \frac{q}{x+1} + \frac{r}{x+2}\right)\end{equatio  n*}

    and leaving your answer in this form, i.e: as f(x)(\cdots) is quite useful in many a STEP I question that I've come across.

    As an aside, here's another place where it's useful: \frac{\mathrm{d}}{\mathrm{d}x} |f(x)|, now the standard thing to do would be to consider intervals of x that makes f positive or negative and differentiate piecewise.

    What'd I'd do is y = |f(x)| \Rightarrow y^2 = f(x)^2 \Rightarrow 2yy' = 2f'(x)f(x) \Rightarrow y' = \frac{f'(x)f(x)}{|f(x)|} - this also lets you know where the function is not differentiable.

    I'll edit it a nice problem for you lot to try in a bit.
    Do you have one right now?
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    (Original post by Insight314)
    Do you have one right now?
    Although it's brief you might like to prove the product and quotient rules for suitably defined functions using this sort of technique.
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    (Original post by Insight314)
    ...
    As above - at any level above A-Level we have that \log and \ln are interchangeable.

    Define f(x) = x^m(x-1)^n with both m, n being positive integers greater than 1. Show that the stationary point of this function in the interval (0,1) is a maximum if n is even and a minimum when n is odd.
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    (Original post by joostan)
    Although it's brief you might like to prove the product and quotient rules for suitably defined functions using this sort of technique.
    Can I just say that is a really nice avatar
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    (Original post by Zacken)
    As above - at any level above A-Level we have that \log and \ln are interchangeable.
    Except in engineering, where they often use \log = \log_{10} ...
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    (Original post by atsruser)
    Except in engineering, where they often use \log = \log_{10} ...
    And computer science, where sometimes \log = \log_{2}
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    (Original post by sweeneyrod)
    And computer science, where sometimes \log = \log_{2}
    LOL


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    (Original post by sweeneyrod)
    And computer science, where sometimes \log = \log_{2}
    And Chemistry, where \log = \log_{10} when calculating pH.
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    A STEP teacher at my college mentioned a 'loop' technique, he didn't go into any detail because he said it's not needed for A level. Does anybody know what it is?
 
 
 
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