can someone explain to me step by step how to solve an inverse trigo function? i added an image with a question i can't seem to understand why it has that certain answer... help me please!!!
this is the question:
Arcsin[sin(-8)]=8- 3π
how come? why not 3π-8???
image: http://i53.photobucket.com/albums/g58/spicefoo/invtrigo.jpg
2006-10-10 05:44:29 · 3 answers · asked by harry_potter 2 in Science & Mathematics ➔ Mathematics
uhm... is Arcsin[sin(-8)]=8? take not that i'm asking for the Arcsin not arcsin... i know there's a difference... i just forgot what...
2006-10-10 06:03:56 · update #1
sin(-x)=-sinx
so with sine functions 3pi-8 and 8-3pi are not one and the same
arc sin[sin(-8)]=8-3pi
-8=8-3pi
are you sure you have copied the sum correctly?
2006-10-10 05:48:57 · answer #1 · answered by raj 7 · 1⤊ 1⤋
Arcsin(x) will give the angle between -pi/2 and pi/2 which has the value x. In your case, -8 is clearly not in this interval, so we look for another angle x with sin(x)=sin(-8). The possibilities are
-8+2n*pi and
(pi-(-8))+2n*pi.
The problem is to find out which of these angles in in the interval from -pi/2 to pi/2.
Now, pi/2=1.57 (approx)
2pi-8=-1.717,
4pi-8=4.57,
8-pi=4.86,
8-3pi=-1.425,
so 8-3pi is *the* angle in the right interval.
2006-10-10 13:42:37 · answer #2 · answered by mathematician 7 · 1⤊ 1⤋
Arcsin[sin(-8)]=-8 +2kpi, k = 0, 1, -1,2,-2....
now since sin (-x) = sin(x +pi)
you can also write
Arcsin[sin(-8 + pi)]=-8 +pi +2kpi, k = 0, 1,-1,2,-2 ....
By convention the domain of arcsin is [0,2pi] you rteacher is doing way too difficult.
2006-10-10 13:23:07 · answer #3 · answered by gjmb1960 7 · 0⤊ 1⤋