anonymous
  • anonymous
Find the exact real value of arccos(-√2/√2)? Dont want the anwser just need an explanation on how to do it. Thank you!
Precalculus
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Tell may what teta should be so as to have for cos the value \[\sqrt{2}/2\]
anonymous
  • anonymous
Tell me ..
anonymous
  • anonymous
Thats the problem I dont get this or to aleast start it off

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anonymous
  • anonymous
\[y = \cos \Theta \rightarrow \Theta=\cos^{-1} y \]
anonymous
  • anonymous
Is it sqrt(2)/2 or sqrt(2)/sqrt(2)?
anonymous
  • anonymous
Its sqrt -2/ 2 and i have been using this website for refrence I just dont know what to with the numbers given http://www.math.com/tables/algebra/functions/trig/functions.htm
anonymous
  • anonymous
i mean -sqrt(2) / 2
anonymous
  • anonymous
Yes sir
anonymous
  • anonymous
ok so let's assume that arccos(-sqrt(2) / 2) = x, for some value 'x'. We also know that arccos(x) is the inverse function of cos(x) and has a domain of [-1,1] with a range of [0, pi]. Saying that arcos(-sqrt(2)/2) = x is then the same as saying cos(x) = -sqrt(2)/2 because they are inverse function so their x and y coordinates flip. But note that x is in the interval 0 <= x <= pi because that's the range of arccos(x). For which angle is cos(x) -sqrt(2)/2? (Hint: It's a special angle.) Once you find the angle, you must then confirm that cos(x) will be negative for that angle.
anonymous
  • anonymous
@cohenbrian12
anonymous
  • anonymous
1/sqrt(2)

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