1sorry who?
I was going through Play with graphs.
There the graph of \sin^{-1} sinx is given .
Though the graph happens to be okay.
I got one problem.
IT SUGGESTS students to learn defination of it
like
x+2Î ; -5Î /2<= x<= -3Î /2
-Î -x ; -3Î /2<=x<= -Î /2
y=\sin^{-1} sinx = x ; - Î /2<=x<=Î /2
Î -x ; Î /2<=x<=3Î /2
and so on.
My doubt is how are we getting this particular defination.
Please help
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4 Answers
1dekh eibhabe inverse trigonometry r sums koris....(doing these after a loong time so pardon me if am wrong)....
case I:
sine-1sine(x)=x ;-∩/2≤x≤∩/2
case II
when Î /2<=x<=3Î /2
add or subtract suitably so that we bring a region within -∩/2 and∩/2....
here let us subtract -∩...
-∩/2≤x-∩≤∩/2 but sine(x-∩)=-sinx so we instead multiply by -1 andtake..
-∩/2≤∩-x≤∩/2 check sine (∩-x)=sinx
case III:
-5Î /2<= x<= -3Î /2
add 2∩
-∩/2≤x+2∩≤∩/2 sine(x+2pi)=sine x
1case IV
-3Î /2<=x<= -Î /2
add pi...
we get -.5 pi≤x+pi≤.5pi
but then sine(x+pi)=-sine x
so we multiply both sides by -1
-.5pi≤-x-pi≤.5x
