anonymous
  • anonymous
Give the measure of the angle in radians and degrees arcsin (-square root of 2/2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@satellite73 please help
anonymous
  • anonymous
@ganeshie8 please help
anonymous
  • anonymous
@tkhunny please help

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More answers

tkhunny
  • tkhunny
Do you have a "Unit Circle" to look at? You should be quite familiar with \(-\pi/4\). Why negative and not positive?
anonymous
  • anonymous
Where did you get -pi/4
anonymous
  • anonymous
Why did you use pi/4 and not something like 3pi/4?
tkhunny
  • tkhunny
Did you find a "Unit Circle", yet?
anonymous
  • anonymous
yes
tkhunny
  • tkhunny
For one, the arcsin function typically is defined only for \(-\pi/2\le arcsin(x)\le\pi/2\). This rules out quite a few options.
tkhunny
  • tkhunny
What does your unit circle tell you about the sine of \(-\pi/4\)?
anonymous
  • anonymous
Its 45 degrees and is (square root of 2/2, square root of 2/2
tkhunny
  • tkhunny
No, that's \(\pi/4\). We need \(-\pi/4\) When you find it, it will say, \(\sin(-45ยบ) = -\sqrt{2}/2\). This should look quite tantalizing in light of the original problem statement.
anonymous
  • anonymous
Would that be 7pi/4?
tkhunny
  • tkhunny
Well, that is the same place, but it is NOT the answer to the original problem statement. This is correct: \(sin(-\pi/4) = -\sqrt{2}/2\) \(sin(7\pi/4) = -\sqrt{2}/2\) However, and this may come as a surprise: \(arcsin(-\sqrt{2}/2) = -\pi/4\) and is absolutely NOT \(7\pi/4\). It is a Range Issue. The arcsin function does not contain \(7\pi/4\) in its Range.
anonymous
  • anonymous
Ok I understand, if the range was from -infinity to infinity, 7pi/4, would be a valid answer, but since it isn't you just have to slap on the negative to pi/4
tkhunny
  • tkhunny
Not quite, but that may get you to the next problem.
anonymous
  • anonymous
thanks for helping

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