# Find the exact value of s in the given interval that has the given function value

• Feb 25th 2013, 09:02 PM
boltage619
Find the exact value of s in the given interval that has the given function value
a) [pi, 3pi/2 ] ; sin(s) = -(sqrt of 2)/2

b) [pi/2, pi] ; csc(s) = 2

c) [3pi/2, 2pi] ; cos(s) = (sqrt of 3)/2
• Feb 25th 2013, 09:06 PM
Prove It
Re: Find the exact value of s in the given interval that has the given function value
Quote:

Originally Posted by boltage619
a) [pi, 3pi/2 ] ; sin(s) = -(sqrt of 2)/2

b) [pi/2, pi] ; csc(s) = 2

c) [3pi/2, 2pi] ; cos(s) = (sqrt of 3)/2

The focus angle will be the angle in the first quadrant, so

\displaystyle \displaystyle \begin{align*} \sin{(\theta)} &= \frac{\sqrt{2}}{2} \\ \theta &= \frac{\pi}{4} \end{align*}.

We know that the sine value is negative in the third and fourth quadrants, and from the domain given, we know the angle needs to be in the third quadrant.

So the angle is \displaystyle \displaystyle \begin{align*} \pi + \frac{\pi}{4} = \frac{5\pi}{4} \end{align*}.

Can you do something similar for the other two questions?
• Feb 25th 2013, 09:34 PM
boltage619
Re: Find the exact value of s in the given interval that has the given function value
Well the square root of 2/2 is actually negative, not positive.
• Feb 25th 2013, 09:36 PM
Prove It
Re: Find the exact value of s in the given interval that has the given function value
You obviously didn't read what I posted. You need to work out the FOCUS angle first, which is in the first quadrant, where the sine values (and all other values) are positive. Then once you have the focus angle you add/subtract it to whatever you need to so that you end up in the correct quadrant. In this case, since you need to sweep out a full semicircle to go from the first quadrant to the third, you need to add an angle of \displaystyle \displaystyle \begin{align*} \pi \end{align*} to your focus angle.
• Feb 25th 2013, 09:41 PM
boltage619
Re: Find the exact value of s in the given interval that has the given function value
Alright, thanks for explaining that to me!