HELPPP srs think about this question (1 Viewer)

[bobzz]

IM SPINING OUT
MY online tutor is telling me that sin inverse x IS THE SAME AS 1/sinx sayin
What the> This is not true ive been brought to belive that inverse functions are different to 1/function... HELPP

anyways im tryin to solve
domain and range for sin inverse (x^2)

Guys pleae help!

 

[Iruka]

If that is the case, then you need to get a new on-line tutor.

Preferably one who actually knows some mathematics.

 

[henry08]

lolz.

 

[cutemouse]

You got to be kidding me... sin^-1 IS NOT 1/sin, not mathematically anyway. Since sin means finding the ratio of the given angle, "inverse sin" means the opposite, ie. finding the angle of the given ratio.

You really do need to get a new online tutor if what you said is true.

 

[tacogym27101990]

hope your not paying for this goose

 

[bobzz]

Tell me about it! narr not paying for it, its the UWS online tutor thing...
anyways i wasted no joke like 40 mins trying to get a answer to my question, and they kept saying sin inverse was 1/sin x and i got relly frustrated. btw i still need help with my question...

come guys..

state the domain and range of
Sin inverse (x^2)

Thanks..

 

[lyounamu]

Tell me about it! narr not paying for it, its the UWS online tutor thing...
anyways i wasted no joke like 40 mins trying to get a answer to my question, and they kept saying sin inverse was 1/sin x and i got relly frustrated. btw i still need help with my question...

come guys..

state the domain and range of
Sin inverse (x^2)

Thanks..

Domain: -1<= x^2 <=1
-1<= x <= 1

0<=y<=pi

 

[bobzz]

Domain: -1<= x^2 <=1
-1<= x <= 1

0<=y<=pi

Hey thanks heaps for ur help.
I dont mean to be dumb, but can u try explain to me how u did it please
Like with the domain, i get wat u did for the first step, but then wat, y;s it just left the same

and yer thanksss if ur bothered

 

[lyounamu]

Hey thanks heaps for ur help.
I dont mean to be dumb, but can u try explain to me how u did it please
Like with the domain, i get wat u did for the first step, but then wat, y;s it just left the same

and yer thanksss if ur bothered

Let y = arcsin (x^2)
i.e. siny = x^2

so -1 <= x^2 <= 1
Draw the graph of x^2 with y= -1 and y=1. Then you will be able to find the domain.

Once you work out the domain, range is simple. From that x^2 is always positive so 0<= y <=pi because sine is positive in the first and second quadrants.

 

[bobzz]

Let y = arcsin (x^2)
i.e. siny = x^2

so -1 <= x^2 <= 1
Draw the graph of x^2 with y= -1 and y=1. Then you will be able to find the domain.

Once you work out the domain, range is simple. From that x^2 is always positive so 0<= y <=pi because sine is positive in the first and second quadrants.

i frekinnnnnn get itttt Yay
Thanks your a champ take care

 

[Trebla]

Actually, many lecturers refer to 1/x as the inverse of x (because the x is 'inverted' lol). You'll be surprised how commonly it is used...lol.
Unfortunately, for the mathematically literate, this can cause heaps of confusion....