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is the inverse sine of sine of (-(4pi)/9)) equal pi/9?
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sin(-x) = -sinx
-sin^-1 (sin 4pie/9)
so -4pie/9
Why not just -pi/9?
But pi/9 is in that range?
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Oh. Wait It is.
What is?
lol. So yeah \[\frac{-4 \pi }{9} \in [\frac{-\pi}{2}, \frac{\pi}{2}] =>\sin^{-1}(\sin(\frac{-4 \pi}{9}))=\frac{-4 \pi}{9} \]
\[-.444 \pi \in [-.5 \pi , .5 \pi]\] Since \[-.444 \in [-.5,.5]\]
-4/9 is approximately -.444 :)
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