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C3: Trigonometry

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    Hi,

    I have been asked to find cot(\dfrac{3\pi}{4}). However, I'm not sure how to work this out.
    I have been given the formulae cotx = (\dfrac{cosx}{sinx}) and cotx = (\dfrac{1}{tanx}). My teacher said that the latter is not always true, if that's any help. So what I did was: \dfrac{cos(3\pi)}{sin(4)}, and put that into my calculator (in radians mode), but it gave me 1.32, which is wrong as the answer is -1.

    Where am I going wrong here? Any help is appreciated!
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    (Original post by Prestoria)
    Hi,

    I have been asked to find cot(\dfrac{3\pi}{4}). However, I'm not sure how to work this out.
    I have been given the formulae cotx = (\dfrac{cosx}{sinx}) and cotx = (\dfrac{1}{tanx}). My teacher said that the latter is not always true, if that's any help. So what I did was: \dfrac{cos(3\pi)}{sin(4)}, and put that into my calculator (in radians mode), but it gave me 1.32, which is wrong as the answer is -1.

    Where am I going wrong here? Any help is appreciated!
    \dfrac{cos(\dfrac{3\pi}{4})}{sin  (\dfrac{3\pi}{4})}=-1

    And cotx \neq \dfrac{1}{tanx} when tanx = 0
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    Your teacher is wrong for a start, the definition of cot(x) is 1/tan(x).
    Basically if you re-write your question as 1/tan(3PI/4), which if you do on a calculator would give 1/(-1) which will give you the answer of -1.
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    I assume both way works - at least if tan doesn't equal 0. Thanks to the both of you, I really appreciate your speedy responses.
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    (Original post by adamjamjar)
    Your teacher is wrong for a start, the definition of cot(x) is 1/tan(x).
    No. The teacher is not wrong. Your definition is incomplete

    \cot x = \dfrac{1}{\tan x}\quad (\tan x \neq 0)

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Updated: July 7, 2012
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