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    Simplify cot (pi +x)
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    1/tan(pi+x) and use compound angle
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    Explain please, step by step

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    (Original post by swellow5)
    Explain please, step by step

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    Reciprocate (flip it upside down) the double angle formula for tan (this comes from 1/tan(pi+x))using the values in your OP it will simplify when you do this. It is better for you to be given hints and do it yourself otherwise you won't learn. HINT: tan(pi)=0 in radians.
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    Oh I see, but what if it was cot(pi - x)

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    (Original post by L33t)
    Reciprocate (flip it upside down) the double angle formula for tan (this comes from 1/tan(pi+x))using the values in your OP it will simplify when you do this. It is better for you to be given hints and do it yourself otherwise you won't learn. HINT: tan(pi)=0 in radians.
    Probably better to just understand the fact that \tan(\pi+x)=\tan x instead of mechanically resorting to the double angle formula. It's one of the most important properties of the function (\pi is its period) and it's very easy to see with the unit circle. It's always best to use the unit circle as much as possible since this enforces understanding of the trig functions as opposed to memorisation of formulae.
    But yeah your method is still valid (although strictly speaking \cot x \not\equiv \dfrac{1}{\tan x}).
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    (Original post by IrrationalRoot)
    Probably better to just understand the fact that \tan(\pi+x)=\tan x instead of mechanically resorting to the double angle formula. It's one of the most important properties of the function (\pi is its period) and it's very easy to see with the unit circle. It's always best to use the unit circle as much as possible since this enforces understanding of the trig functions as opposed to memorisation of formulae.
    But yeah your method is still valid (although strictly speaking \cot x \not\equiv \dfrac{1}{\tan x}).
    Either method gets you the right answer though, doesn't it? Twas the way I was taught to do it anyhow.

    I don't mind either way, I've finished my maths A-Level now thank god!
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    (Original post by L33t)
    Either method gets you the right answer though, doesn't it?

    I don't mind either way, I've finished my maths A-Level now thank god!
    Right answer for this question, but with that mentality one could struggle with a lot of other questions due to poor understanding of how the functions work.
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    (Original post by IrrationalRoot)
    Right answer for this question, but with that mentality one could struggle with a lot of other questions due to poor understanding of how the functions work.
    Why is everything on TSR a competition all of a sudden? I was never taught the method you used (edexcel board maths), not that that's an excuse for my apparent sloppy mathematical methods, but sorry my mathematical standards don't live up to your expectations. In future I guess I must try harder.

    It is my understanding that strictly speaking \cot x \not\equiv \dfrac{1}{\tan x} is beyond the scope of C3 understanding?


    Have a nice day
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    (Original post by swellow5)
    Oh I see, but what if it was cot(pi - x)

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    I'd use the same method with the tan(A-B) identity although apparently my methods are inferior!
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    (Original post by L33t)
    Why is everything on TSR a competition all of a sudden? I was never taught the method you used (edexcel board maths), not that that's an excuse for my apparent sloppy mathematical methods, but sorry my mathematical standards don't live up to your expectations. In future I guess I must try harder.

    It is my understanding that strictly speaking \cot x \not\equiv \dfrac{1}{\tan x} is beyond the scope of C3 understanding?


    Have a nice day
    Not sure why you're getting so wound up; this was mainly for OP's benefit than yours.
    Neither am I sure why helping people to improve their mathematical understanding would be considered a 'competition'.
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    (Original post by L33t)
    I'd use the same method with the tan(A-B) identity although apparently my methods are inferior!
    I've just realised you probably misunderstood what I meant when I said this:

    "Right answer for this question, but with that mentality one could struggle with a lot of other questions due to poor understanding of how the functions work."

    When I said 'with that mentality' I didn't mean the method which you used (which is perfectly valid), I meant the mentality of 'if it gets the right answer, it's fine and you should move on.' Sometimes getting the right answer alone isn't ideal, for the reasons I mentioned before.
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    tan (and, by extension, cot) has period of pi. i.e. tanx = tan(x - pi) = tan(x + pi) = tan(x + k*pi) for any integer k. This is obvious from the graph.
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    (Original post by IrrationalRoot)
    I've just realised you probably misunderstood what I meant when I said this:

    "Right answer for this question, but with that mentality one could struggle with a lot of other questions due to poor understanding of how the functions work."

    When I said 'with that mentality' I didn't mean the method which you used (which is perfectly valid), I meant the mentality of 'if it gets the right answer, it's fine and you should move on.' Sometimes getting the right answer alone isn't ideal, for the reasons I mentioned before.
    Fair dos, i'm not getting wound up, I genuinely misunderstood what you said then. I thought you were saying I was stupid for using that method and didn't understand what I was saying. I recognise that understanding the trig functions and the domains in which they are repeated is a better and quicker way to deal this question.
 
 
 
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