keelyjm
help with inverse trig functions?
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keelyjm
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Find the exact value of sin(arctan(2)). For full credit, explain your reasoning.
AravindG
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well first try to make than tan inverse inside the bracket to sin inverse so that we get sin (sin-1)
AravindG
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drawing a triangle can help you do that
keelyjm
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Drawing a triangle will help change tan-1 to sin-1?
hartnn
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Let arctan 2 = x
so, tan x = 2 /1
since you know tan x = opposite side/adjacent side
make a right angled triangle with one leg =2 and other leg = 1
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now can you find the hypotenuse ??
AravindG
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yep :) just draw ..you will see
hartnn
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once you find the hypotenuse, find sin x.
keelyjm
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sqrt 5
hartnn
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because sin(arctan(2)). = sin x
hartnn
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yes, so sin x = sin (arctan 2) =...?
keelyjm
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sqrt5?
hartnn
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thats the hypotenuse .
sin ratio = opposite side / hypotenuse =... ?
keelyjm
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oh so it's 2/sqrt5
hartnn
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yup .
hartnn
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any doubts ?
keelyjm
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alright and what about finding the exact real value of arccos(sqrt2/2)?
keelyjm
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no I understand that now that you have to use a tiangle to solve it
hartnn
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arccos(sqrt2/2) =x
cos x = sqrt2/2 = 1/sqrt 2
for what value of angle 'x' is cos x = 1/sqrt 2
its one of the standard angle....
hartnn
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also, next time please ask new question in new post...
keelyjm
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okay sorry. I am not sure how you got cos x = sqrt2/2 = 1/sqrt 2 ?
hartnn
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you know unit circle ?
which defines standard values of sin/cos for standard angles...
hartnn
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i suggest you learn and remember all standard angles and values..
keelyjm
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yes i know the unit circle. is that pi/4? 45 degrees?
hartnn
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yes, exactly :)
keelyjm
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so pi/4 would be the exact real value?
anonymous
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\[\arccos(\frac{\sqrt{2}}{2})\] see if you recall an angle whose cosine is \(\frac{\sqrt{2}}{2}\)
hartnn
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yes.
pi/4 in radians
45 in degrees.
keelyjm
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Thank you so much!
hartnn
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welcome ^_^