Evaluating inverse trig function

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  • #1
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Homework Statement



We are being asked to evaluate the inverse trig function sin^-1 (1/ sqrt(2)).

Homework Equations





The Attempt at a Solution


I have no clue where to start. I have the unit circle, which makes sense to me if it was a trig function of a trig function, but when it's a trig function of a number not listed in a common unit circle diagram, I am thrown. Any guidance would be appreciated.
 

Answers and Replies

  • #2
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Homework Statement



We are being asked to evaluate the inverse trig function sin^-1 (1/ sqrt(2)).

Homework Equations





The Attempt at a Solution


I have no clue where to start. I have the unit circle, which makes sense to me if it was a trig function of a trig function, but when it's a trig function of a number not listed in a common unit circle diagram, I am thrown. Any guidance would be appreciated.
x = sin-1(1/sqrt(2)) <==> sin(x) = 1/sqrt(2)
Can you think of any angle in the first quadrant whose sine is 1/sqrt(2)?
 
  • #3
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x = sin-1(1/sqrt(2)) <==> sin(x) = 1/sqrt(2)
Can you think of any angle in the first quadrant whose sine is 1/sqrt(2)?
The closest thing I saw was pi/4, but the sine there was root(2)/2. That's what the book shows but I don't see how they get that. I am really deficient in trig ;/
 
  • #4
eumyang
Homework Helper
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They are the same. That's not a trig issue, it's an algebra issue.
[tex]\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}[/tex]
Radical expressions aren't considered simplified if there is a radical in the denominator of a fraction. So one can rationalize the denominator (you can look it up).
 
  • #5
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Thank you.
 

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