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sin^−1(sin(4pi/3))=??
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hint: \[\huge \sin^{-1} (\sin x) = x\] \[\huge \sin (\sin^{-1} x) = x\] does that help?
nooo it doesnt equal x it equals -pi/3
uhh you do realize what i wrote is different from what you wrote right?
i was stating the "general rule"
ok then it would be 4pi/3
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keep in mind that the range of arcsine is [-pi/2, pi/2]
so -pi/3 is actually the answer
if you simplify 4pi/3 you get -pi/3
not quite, but sin(4pi/3) = sin(-pi/3)
\[\huge \sin^{-1} (\sin (\frac{4\pi}{3}) \implies \sin^{-1} (\sin (-\frac{\pi}{3}))\] that'swhat i meant
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then following the rule i said... that becomes -pi3
-pi/3*
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