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Mathematics 21 Online
OpenStudy (anonymous):

how do you solve arctan[tan(-7pi/9)] ?

OpenStudy (anonymous):

-7pi/9

OpenStudy (anonymous):

isn't it outside of the domain? So you have to find another angle...

OpenStudy (anonymous):

arctan( tan x)=x

OpenStudy (anonymous):

\[\tan^{-1}(\tan(x))=x\] arctan and tan are inverses

OpenStudy (anonymous):

yea

OpenStudy (anonymous):

the answer has to be within the domain\[-\pi/2\le \theta lepi/2\]

OpenStudy (anonymous):

?

OpenStudy (anonymous):

thats not the domain of tangent. the domain of tangent is all reals except k*pi/2, where k is an integer

OpenStudy (anonymous):

oh, you want the reference angle

OpenStudy (anonymous):

i see

OpenStudy (anonymous):

yes please!

OpenStudy (anonymous):

so the answer would be 2pi/9?

OpenStudy (anonymous):

lol im sleepy, my brain is not working right now

OpenStudy (anonymous):

gotcha lol same here. the reference angle would be in the first quadrant yes? and the angle would be 2pi/9

OpenStudy (anonymous):

but wouldn't it have to be positive for an angle of -7pi/9 to exist? so the original angle would be in the third quadrant

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