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Help with inverse trig functions

  • Thread starter vysero
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  • #1
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Here is my problem:

cot(arcsin(x))

my awnser:

cot= x/(1-x)^1/2

The online program were suppose to use says im wrong but im not sure what I did wrong.
 

Answers and Replies

  • #2
SammyS
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Here is my problem:

cot(arcsin(x))

my answer:

cot= x/(1-x)^1/2

The online program were suppose to use says im wrong but im not sure what I did wrong.
What's the question that you're trying to answer?

The answer you give makes no sense whatsoever.
 
  • #3
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Draw a right triangle, with θ being the acute angle in the triangle. Let θ = arcsin(x), and label the sides and hypotenuse according to this definition.

The problem boils down to evaluating cot(θ).

Your answer SHOULD NOT start with "cot =" since cot by itself is meaningless in this context. If you are doing your work on paper, your work should start with cot(arcsin(x)) = ... and end up with what you found, which is actually not too far off.
 
  • #4
HallsofIvy
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The simplest way to do this is to imagine a right triangle with one angle [itex]\theta[/itex], "opposite side" to [itex]\theta[/itex] of length x, and hypotenuse of length 1 so that [itex]sin(\theta)= x/1[/itex] and thus [itex]\arcsin(x)= \theta[/itex].

Now, what is the length of the near side and what is [itex]tan(\theta)= tan(arcsin(x))[/itex]?

You are missing a square in your answer.
 

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