What is the solution to cot(2x)=0.5(cot(x)-tg(x))?

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SUMMARY

The discussion centers on solving the equation cot(2x) = 0.5(cot(x) - tg(x)). Participants clarify the correct notation for trigonometric functions, emphasizing that cot(arc tan(x)) should be written as cot(arctan(x)). The relationship between cotangent and tangent is established, with cot(θ) defined as 1/tan(θ). Additionally, the conversation highlights the importance of visual aids, such as right triangles, for simplifying expressions involving arctan.

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  • Familiarity with inverse trigonometric functions, particularly arctan.
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  • Learn how to simplify expressions involving sin(arctan(x)) and cos(arctan(x)).
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Sobhan
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The answer to Cot(arc tan x)
 
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Sobhan said:
The answer to Cot(arc tan x)

How else can you express \cot{\theta}? Also, arctan is a function and every function is only ever one word, so the way you should be writing this is \cot\left(\arctan(x)\right)
 
Cot(arc tan (x))?
 
Sobhan said:
Cot(arc tan (x))?

I was simply explaining that functions such as cot(x), arctan(x), sin(x), ln(x) etc. are always written without any spaces.

arc tan (x) is wrong
arctan(x) is right

cot(arc tan (x)) is wrong
cot(arctan(x)) is right

Ok, so now that we have that out of the way, let's go back to the problem at hand. You want to simplify \cot(\arctan(x)). Firstly, how else can you express \cot(x)? That function has a very obvious relationship with tan(x) because it was defined that way. What is it?
 
Strange, I suggested you should make a little drawing to understand the question, but it seems to have gone lost in transmission


ArxTangent.jpg
 
Do you not know that, by definition, cot(\theta)= \frac{1}{tan(\theta)}?
 
is it x^-1?
 
thanks for making me think.
 
  • #10
What is cot(0.5×arctan(x))
 
  • #11
Search for how else you can represent
\tan(\theta / 2)

and decide whether this can help you. Sometimes expressions can't be simplified further, but they can be transformed regardless.

To help you on this journey, you might also need to figure out how to simplify \sin(\arctan(x)) and \cos(\arctan(x)). The way you can do this is by drawing a right triangle, and then let one angle be \theta. This angle will actually be \arctan(x) though, which means that \theta=\arctan(x) hence \tan(\theta)=x. If \tan(\theta)=x then can you label every side of the triangle in terms of x?
 
  • #12
I have reached a equation to express cot(x) by cot(2x) and it is not an easy one the equation is this (2cot(2x)+(-)(4cot(2x)^2+4)^0.5))×2^-1
Is this right?
 
  • #13
Did you make a drawing again ?
 
  • #14
Sobhan said:
I have reached a equation to express cot(x) by cot(2x) and it is not an easy one the equation is this (2cot(2x)+(-)(4cot(2x)^2+4)^0.5))×2^-1
Is this right?

Did you derive that for yourself or take it from somewhere else? It also has an obvious common factor of 2 in the numerator and denominator.

BvU said:
Did you make a drawing again ?
Sorry BvU, I forgot all about your drawing further up. It could've saved me quite a bit of explaining in my last post if I just referred to your diagram.
 
  • #15
Is it clear to all that if ##\cot(\arctan x)## is a very simple expression, then perhaps ##\cot ({1\over 2} \arctan x)## mght also be a simple expression ?

[edit] LaTeX justifiably does not recognize atan as a mathematical function name. It indeed should be arctan (no space) and not atan as I first typed ... .:rolleyes:

[edit] woops, all wrong, sorry o:). Wake up first, then check alerts! Not the other way around.
 
Last edited:
  • #16
I got that from this : cot(2x)=0.5(cot(x)-tg(x))
 
  • #17
Sobhan said:
I got that from this : cot(2x)=0.5(cot(x)-tg(x))
It's definitely been a while since I've seen tg.

Ok awesome, so can you finish it off?
 

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