- #1
rocomath
- 1,755
- 1
Refresh my memory please, Inverse Trig
[tex]x=\sin^{-1}\left(-\frac{\sqrt{2}}{2}\right)[/tex]
Inverse sine is defined from [tex]-\frac{\pi}{2}[/tex] to [tex]\frac{\pi}{2}[/tex] which lies in the 1st and 4th quadrant.
So [tex]x=-\frac{\pi}{4}[/tex]
[tex]x=\sin^{-1}\left(-\frac{\sqrt{2}}{2}\right)[/tex]
Inverse sine is defined from [tex]-\frac{\pi}{2}[/tex] to [tex]\frac{\pi}{2}[/tex] which lies in the 1st and 4th quadrant.
So [tex]x=-\frac{\pi}{4}[/tex]