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C3: Trigonometry

Hi,

I have been asked to find cot(3π4)cot(\dfrac{3\pi}{4}). However, I'm not sure how to work this out.
I have been given the formulae cotx=(cosxsinx)cotx = (\dfrac{cosx}{sinx}) and cotx=(1tanx)cotx = (\dfrac{1}{tanx}). My teacher said that the latter is not always true, if that's any help. So what I did was: cos(3π)sin(4)\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!
(edited 11 years ago)
Reply 1
Original post by Prestoria
Hi,

I have been asked to find cot(3π4)cot(\dfrac{3\pi}{4}). However, I'm not sure how to work this out.
I have been given the formulae cotx=(cosxsinx)cotx = (\dfrac{cosx}{sinx}) and cotx=(1tanx)cotx = (\dfrac{1}{tanx}). My teacher said that the latter is not always true, if that's any help. So what I did was: cos(3π)sin(4)\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!


cos(3π4)sin(3π4)=1\dfrac{cos(\dfrac{3\pi}{4})}{sin(\dfrac{3\pi}{4})}=-1

And cotx1tanxcotx \neq \dfrac{1}{tanx} when tanx = 0
(edited 11 years ago)
Reply 2
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. :smile:
Reply 3
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.
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

cotx=1tanx(tanx0)\cot x = \dfrac{1}{\tan x}\quad (\tan x \neq 0)

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