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how do you solve arctan[tan(-7pi/9)] ?
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-7pi/9
isn't it outside of the domain? So you have to find another angle...
arctan( tan x)=x
\[\tan^{-1}(\tan(x))=x\] arctan and tan are inverses
yea
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the answer has to be within the domain\[-\pi/2\le \theta lepi/2\]
?
thats not the domain of tangent. the domain of tangent is all reals except k*pi/2, where k is an integer
oh, you want the reference angle
i see
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yes please!
so the answer would be 2pi/9?
lol im sleepy, my brain is not working right now
gotcha lol same here. the reference angle would be in the first quadrant yes? and the angle would be 2pi/9
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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