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Edexcel A2 C4 Mathematics June 2016 - Official Thread Watch

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    (Original post by Euclidean)
    What part of trigonometry are you struggling with?

    If it's spotting the identities then you're going to want to go over them again and do a lot of problems involving them (if you get bored try some trig. integrals)

    If it's just general solving, perhaps review what the sine, cosine and tangent formulae do so that you have a better understanding of exactly what you're doing when evaluating
    Yeah, identities and not making stupid mistakes when solving trig equations (i.e. remembering when values are negative/positive, forgetting about domains etc). I'll just have to work a bit harder at trig questions I guess.

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    (Original post by Euclidean)

    \tan(x-53)=\cot(x+35) where 0\leq x \leq360
    Isn't that elementary?

    \tan \left(\frac{\pi}{2} - x\right) = \cot x so \cot (x + 35^{\circ}) = \tan (55^{\circ} - x), hence your equation is nothing but:

    \tan (x - 53^{\circ}) = \tan (55^{\circ} - x) whereupon a standard comparing of arguments and adding \pi because periodicity gets you your solutions? Seems too easy for C3 trig, no?
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    (Original post by Euclidean)
    What part of trigonometry are you struggling with?

    If it's spotting the identities then you're going to want to go over them again and do a lot of problems involving them (if you get bored try some trig. integrals)

    If it's just general solving, perhaps review what the sine, cosine and tangent formulae do so that you have a better understanding of exactly what you're doing when evaluating

    edit: There was a good trig problem posted by 13 1 20 8 42 somewhere I believe, it ran something like this:

    \tan(x-53)=\cot(x+35) where 0\leq x \leq360
    I think I remember solving that, but not posting it..oh well, I have a terrible memory lol
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    (Original post by Zacken)
    Isn't that elementary?

    \tan \left(\frac{\pi}{2} - x\right) = \cot x so \cot (x + 35^{\circ}) = \tan (55^{\circ} - x), hence your equation is nothing but:

    \tan (x - 53^{\circ}) = \tan (55^{\circ} - x) whereupon a standard comparing of arguments and adding \pi because periodicity gets you your solutions? Seems too easy for C3 trig, no?
    Seems about the level of C3 trig tbh, not like C3 trig is particularly difficult
    Isn't cot introduced at C3...I'd say otherwise it is more C2 style I guess.
    The first identity used there is one I doubt everyone will automatically know doing C3 also.
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    (Original post by 13 1 20 8 42)
    Seems about the level of C3 trig tbh, not like C3 trig is particularly difficult
    Isn't cot introduced at C3...I'd say otherwise it is more C2 style I guess.
    The first identity used there is one I doubt everyone will automatically know doing C3 also.
    Guess maybe I've been doing too much STEP! :lol: Cool proof for anybody (else, obviously not you ) reading this:

    \displaystyle \tan x = \frac{\sin x}{\cos x} = \frac{\cos \left(\frac{\pi}{2} - x\right)}{\sin \left(\frac{\pi}{2} - x\right)} = \cot \left(\frac{\pi}{2} -x\right)
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    (Original post by Zacken)
    Isn't that elementary?
    Depends entirely on your target audience

    (Original post by Zacken)
    \tan (x - 53^{\circ}) = \tan (55^{\circ} - x) whereupon a standard comparing of arguments and adding \pi because periodicity gets you your solutions? Seems too easy for C3 trig, no?
    I just realised that the initial problem was posed along the lines of 'by using \tan(x)=\frac{\sin(x)}{\cos(x)} only...' which I had forgotten about, this leads to a more interesting exercise I suppose
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    (Original post by Euclidean)
    ...
    Fair enough. I love finding shortcuts that turn 8 mark A-Level questions into a 2 lined answer that gets the examiner ripping out their hair. :lol:
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    (Original post by Zacken)
    Fair enough. I love finding shortcuts that turn 8 mark A-Level questions into a 2 lined answer that gets the examiner ripping out their hair. :lol:
    I always find myself doing that and then trying to write more steps into my working afterwards to make it look like I'm deserving of all the marks :lol:
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    (Original post by Princepieman)
    Yeah, identities and not making stupid mistakes when solving trig equations (i.e. remembering when values are negative/positive, forgetting about domains etc). I'll just have to work a bit harder at trig questions I guess.

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    Just keep doing all the past papers you can possibly find
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    If you have seen the associated STEP question (I think that is where I recalled this idea from anyway..), this will be too easy..but otherwise a nice test of your "feel" for trig derivatives/integrals. Think the hint was given in the STEP question, but it makes things a bit trivial so it can be optional..

    Find \displaystyle \int\frac{cosx}{cosx+sinx}dx
    Hint:
    Spoiler:
    Show
    Consider also \displaystyle  \int\frac{sinx}{cosx+sinx}dx
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    (Original post by 13 1 20 8 42)
    If you have seen the associated STEP question (I think that is where I recalled this idea from anyway..), this will be too easy..but otherwise a nice test of your "feel" for trig derivatives/integrals. Think the hint was given in the STEP question, but it makes things a bit trivial so it can be optional..

    Find \displaystyle \int\frac{cosx}{cosx+sinx}dx
    Hint:
    Spoiler:
    Show
    Consider also \displaystyle  \int\frac{sinx}{cosx+sinx}dx
    Spoiler:
    Show
    Adding both integrals and subtracting them again, yeah? It's 3 am here so I won't try it
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    (Original post by 13 1 20 8 42)
    If you have seen the associated STEP question (I think that is where I recalled this idea from anyway..), this will be too easy..but otherwise a nice test of your "feel" for trig derivatives/integrals. Think the hint was given in the STEP question, but it makes things a bit trivial so it can be optional..

    Find \displaystyle \int\frac{cosx}{cosx+sinx}dx
    Hint:
    Spoiler:
    Show
    Consider also \displaystyle  \int\frac{sinx}{cosx+sinx}dx
    I posted this a few days ago, if you want - another way of doing it is:
    Spoiler:
    Show
    \displaystyle \int \frac{\cos x}{\cos x + \sin x} \, \mathrm{d}x = \frac{1}{2} \int \frac{2\cos x}{\cos x + \sin x} \, \mathrm{d}x

    \displaystyle = \frac{1}{2} \int \frac{\cos x + \sin x - (\sin x - \cos x)}{\sin x + \cos x} \, \mathrm{d}x= \frac{1}{2} \int 1 + \frac{\cos x - \sin x}{\cos x + \sin x} \, \mathrm{d}x
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    This is a favourite of mine:

    \displaystyle \int_{-1}^{1} \frac{\mathrm{d}x}{x^2}
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    (Original post by Zacken)
    This is a favourite of mine:

    \displaystyle \int_{-1}^{1} \frac{\mathrm{d}x}{x^2}
    Oooh ln x^2!!!!
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    (Original post by aymanzayedmannan)
    Oooh ln x^2!!!!
    omg yas!!! genius
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    (Original post by Zacken)
    omg yas!!! genius
    No but seriously your clever addition of zeroes is genius. I'm just wondering how the heck you spot it
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    (Original post by aymanzayedmannan)
    No but seriously your clever addition of zeroes is genius. I'm just wondering how the heck you spot it
    So do I, most of the time! Now seriously, give my integral I posted a go!
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    (Original post by Zacken)
    I posted this a few days ago, if you want - another way of doing it is:
    Spoiler:
    Show
    \displaystyle \int \frac{\cos x}{\cos x + \sin x} \, \mathrm{d}x = \frac{1}{2} \int \frac{2\cos x}{\cos x + \sin x} \, \mathrm{d}x

    \displaystyle = \frac{1}{2} \int \frac{\cos x + \sin x - (\sin x - \cos x)}{\sin x + \cos x} \, \mathrm{d}x= \frac{1}{2} \int 1 + \frac{\cos x - \sin x}{\cos x + \sin x} \, \mathrm{d}x
    Oops. Good stuff
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    (Original post by Zacken)
    This is a favourite of mine:

    \displaystyle \int_{-1}^{1} \frac{\mathrm{d}x}{x^2}
    I feel like a fool because I tried this, had to ignore some \ln(-1) terms (bad maths I know) and got \frac{4}{3} only to check on WolframAlpha that the integral doesn't converge...

    I got trolled by Zacken

    In hindsight I should have known as 0 is an asymptote of the curve but it's late.
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    (Original post by Euclidean)
    I feel like a fool because I tried this, had to ignore some \ln(-1) terms (bad maths I know) and got \frac{4}{3} only to check on WolframAlpha that the integral doesn't converge...
    Fam... \int x^{-2} \, \mathrm{d}x = -x^{-1}, where'd you get \ln from?

    Edit to add: Remember that when you integrate, you need to put in modulus terms for logarithms, so even if you managed a logarithmic term, you'd have \ln |-1| = \ln 1 = 0 (which isn't bad maths at all, there's a justification for us placing moduli on logarithm arguments, btw, if you're interested)
 
 
 
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