the inverse sine of sin(2pi/3) = 120degrees, correct?
hell no
trick question!
it is a trick to see if you are paying attention
LOL - whatttttt??? sin-1(sinx)=x!!!
OH ITS NOT IN THE DAMN RANGE
ooops. range of arcsine is \[[-\frac{\pi}{2},\frac{\pi}{2}]\]
BALLS I SAY
it should be called trickonometry
yes. 2pi/3*180/pi 2/3*180 2*60 120
yah you have to find another value whose sine is also \[\frac{2\pi}{3}\]
so wait - am i right or wrong with my original answer of 120?
wrongo
it is \[\frac{\pi}{3}\]
oh okay - sorry. so how do i find a ....hm. range of arccosine then.....and....
dang it! i should have guessed...
\[\sin(\frac{2\pi}{3})=\frac{\sqrt{3}}{2}\] but \[\sin^{-1}(\frac{\sqrt{3}}{2})=\frac{\pi}{3}\]
so is 60 degrees
ok think like this for arcsine look on right half of unit circle from \[-\frac{\pi}{2}\] to \[\frac{\pi}{2}\]
f@ck degrees.
you are not working in degrees you are not converting degrees to radians forget about degrees
r the equation machines not working? i can't get mine up
viagra
i can't give you another medal man, but i....wood? lol
for arccosine look at upper half of unit circle. from 2 to pi
np
right, i just don't get how i can arbitrarily change the range i'm using for sine...
wood. right. i am a little slow
there is nothing arbitrary about it, it is fixed. ok i guess it is arbitrary but it does not change
and don't get the range of sine confused with the range of arcsine.
how can i apply the range of the arccosine to an arcsine?
range of arcsine is \[[-\frac{\pi}{2},\frac{\pi}{2}]\]
i am sure you know what you mean by that question but i am afraid i do not
awwww hell. i got it. youdaman
the range is the range is the range. if you want arccosine of something make sure your answer is between 0 and pi and if you want the arcsine of something your answer must be between -pi/2 andpi/2
trickonometry. i will remember that one
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