1. ## Sin Graph Help

Hey I'm not sure how you solve this. Calculate to the nearest degree the values of x when sin x = 0.2, for -180degrees < x < 180degrees.

2. Originally Posted by digideens
Hey I'm not sure how you solve this. Calculate to the nearest degree the values of x when sin x = 0.2, for -180degrees < x < 180degrees.
It depends on what you mean by calculate. I would normaly use the inverse
sin function on a calculator to do this. It can be done using Newton Raphson
or a number of other techniques.

There is also the problem that there are two solutions in the specified range.

RonL

3. Originally Posted by digideens
Hey I'm not sure how you solve this. Calculate to the nearest degree the values of x when sin x = 0.2, for -180degrees < x < 180degrees.
You need to use a function on your calculator called the arc-sin or sometimes written as $\sin^{-1}$ it tell you the inverse operation.
Thus, for
$\sin x=.2$
You find, (have calculator in degree mode).
$x=\sin^{-1}(.2)\approx 11.54^o$
But that is still not it. Because the sine is also positive in the second Quadrant. In the second quadrant the value is,
$180^o-11.54^o\approx 168.46^o$

4. Thanks for the quick replies guys, I got it now.