anonymous
  • anonymous
Find the exact value of the inverse trigonometric function cos(sin^-1(root 2/3))
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
Let \[\Large \theta = \sin^{-1}\left(\frac{\sqrt{2}}{3}\right)\] So that means \[\Large \sin(\theta) = \frac{\sqrt{2}}{3}\]
jim_thompson5910
  • jim_thompson5910
since \[\Large \sin(\theta) = \frac{\sqrt{2}}{3}\] this means we can draw out this triangle |dw:1433388810567:dw|

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jim_thompson5910
  • jim_thompson5910
what is the missing side?
anonymous
  • anonymous
wait how did you know what value corresponds with what angle/side?
jim_thompson5910
  • jim_thompson5910
because sine = opposite/hypotenuse |dw:1433388989918:dw|
anonymous
  • anonymous
oh ok so you wouldn't adjacent/hypotenuse cause there's a cos?
jim_thompson5910
  • jim_thompson5910
well that comes later into play, but when setting up sine, you only deal with opposite/hypotenuse
anonymous
  • anonymous
okay, sorry I might ask a lot of question as a forewarning
jim_thompson5910
  • jim_thompson5910
that's fine
jim_thompson5910
  • jim_thompson5910
what's the missing side?
anonymous
  • anonymous
root 7?
jim_thompson5910
  • jim_thompson5910
yep
jim_thompson5910
  • jim_thompson5910
|dw:1433389408003:dw|
jim_thompson5910
  • jim_thompson5910
so saying \[\Large \cos\left(\sin^{-1}\left(\frac{\sqrt{2}}{3}\right)\right)\] is the same exact thing as saying \[\Large \cos\left(\theta\right)\]
jim_thompson5910
  • jim_thompson5910
Because remember I said "Let \[\Large \theta = \sin^{-1}\left(\frac{\sqrt{2}}{3}\right)\]" back up at the top
anonymous
  • anonymous
yeah
jim_thompson5910
  • jim_thompson5910
so whats the final answer
anonymous
  • anonymous
so is it just root 7? or am I misinterpreting?
jim_thompson5910
  • jim_thompson5910
close
jim_thompson5910
  • jim_thompson5910
|dw:1433389714910:dw|
anonymous
  • anonymous
OHHHH ok I get it now~ so it would be root 7/3?
jim_thompson5910
  • jim_thompson5910
yep \[\Large \frac{\sqrt{7}}{3}\]
anonymous
  • anonymous
cool! Thankyou!
jim_thompson5910
  • jim_thompson5910
yw

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