Question

sin inverse(sine 9pi/7) Trying to find the exact value?

Answer 1

It is just 9pi/7.

Because sin^(-1) ( (sin x)) = x

Answer 2

Even though 9pi/7 is bigger than pi/2?

Answer 3

Yes.

Answer 4

Becuase you're back where you started from.

Example: sin^(-1) (sin 30) = ?

well, sin 30 = 1/2

and, sin^-1 (1/2) = 30..so we're back where we started from.

Answer 5

The sin 9pi/7 on my calculator is -.7819?

Answer 6

I'm confused on how to I know which is the answer... wouldn't it be -2pi/7 then since its the inverse of the sin9pi/7?

Answer 7

Keep in mind, that sin^(-1) (x) is restricted between -pi/2 and pi/2 inclusive. Thats the reason for the discrepancy.

Answer 8

The problem says... Find the exact value, in radians, of the expression of inverse sin(sin (9pi/7))

Answer 9

? its the sin inverse of (sin(9pi/7))

Answer 10

Since sin (x) and sin^(-1) (x) are inverse functions, then by definition,

sin (sin^(-1)(x)) = x and
(sin^(-1))(sin x) = x

Answer 11

Easyaspi so then it would be the second definition? and the answer is simply 9pi/7? it seems to easy?

Answer 12

dhord; What is critical here, is the restriction of Arcsin (x) or sin^(-1)(x).

And that is it MUST be between -pi/2 and pi/2, inclusive.

Thats why there's no problem with sin^(-1)(sin 30) = 30; and sin(sin^(-1)[1/2) = 1/2..because it falls between -pi/2 and pi/2.

Look at the graph of y = sin^(-1)(x). It clearly shows that.

But in 9pi/7, thats out of the restricted area, and thats where the problem occurs.

Answer 13

\[\sin^{-1} (\sin( 9\pi/ 7 )\]

Answer 14

that's the exact expression its asking to find the exact value in radians

Answer 15

Is there no solution then? since its outside of pi/2

Answer 16

\[\sin ^{-1}(\sin \frac{9\pi}{7})=\frac{-2\pi}{7}\]

Answer 17

How did you get that?

Answer 18

If I may interject, sin (9pi/7) = - sin (2pi/7) by definition of third quadrant angle for siine.

That is, 9pi/7, is a third-quadrant angle, so its value is computed by subtracting pi radians, you get 2pi/7, but since sine is negative in third quadrant, its value is -2pi/7.

Answer 19

oh ok that makes more since

Answer 20

So now you can find sin^(-1) becuase the argument is -2pi/7, which is within the restriction of sin^(-1)(x).

Answer 21

I think the restriction of pi/2 was confusing me but now it makes since

Answer 22

sense*

Answer 23

:0