Find the exact value of the inverse trigonometric function cos(sin^-1(root 2/3))

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Find the exact value of the inverse trigonometric function cos(sin^-1(root 2/3))

Mathematics
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Let \[\Large \theta = \sin^{-1}\left(\frac{\sqrt{2}}{3}\right)\] So that means \[\Large \sin(\theta) = \frac{\sqrt{2}}{3}\]
since \[\Large \sin(\theta) = \frac{\sqrt{2}}{3}\] this means we can draw out this triangle |dw:1433388810567:dw|

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Other answers:

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

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