Inverse Trig Derivatives

  • Thread starter NIZBIT
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  • #1
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I am stuck on this problem:

h(t)=sin(arccos t)

I just don't know where to start. I'm thinking chain rule. The solution shows
sqrt(1-t^2) as the next step. I am not following how they did that.
 

Answers and Replies

  • #2
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try drawing a right triangle such that cos[theta]=t (set the hypotenuse = 1) and see where that goes.
 
  • #3
TD
Homework Helper
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Or: remember that cos²a+sin²a = 1.
 
  • #4
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I don't seem to follow. I apologize if I come off the wrong way, but I don't see what the right triangle will do.
 
  • #5
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I don't think you need the chain rule, I just tried it and got 1 for some odd reason. You need to substitute something in, then use the formula [itex]sin^2{u} + cos^2{u} = 1[/itex]. I hope that helps, I may not have explained it very well. What I mean is, if t=cos u, what is sin u?
 
  • #6
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well you are taking the sine of an angle... if you can make a triangle to fit all of the data you have you can figure out what the sine is. The angle you are using is created from taking the arccosine of t. So if you fit an angle such that the cosine is t and it is a right angle triangle you can use the pythagorean theorem to figure out all of the sides and thus the sine of the original angle.
 
  • #7
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So I should solve for sin using that identity. Then sub for t and get sqrt(1-t^2)?
 
  • #8
HallsofIvy
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Or, draw a right triangle with one side of length t and the hypotenuse of length 1. The angle, [itex]\alpha[/itex] next to that side satisfies [itex]cos(\alpha)= \frac{t}{1}= t[/itex] and so [itex]\alpha= arccos t[/itex]. By the Pythagorean theorem the opposite side has length [itex]\sqrt{1- t^2}[/itex] so [itex]sin(arccos(t))= sin(\alpha)= \sqrt{1- t^2}[/itex].
 

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