the problem is Given cos2x = 7/32, tan2x<0 and -\pi/2<x<\pi/2, find sin x

$1-2sin^2x=7/32$
$-2sin^2x=7/32-1$
$-2sin^2x= -25/32$
$sin^2x=-25/32 * -1/2$
$sin^2x=25/64$
$sin x = -5/8 but not 5/8$
because
-\pi/2<x<\pi/2 means the answer would be either in the first or forth quadrant and tan2x has to be smaller than 0 so if the cos is positive, then the sin has to be negative?
Is this why there is only -5/8 as the answer instead of + or - 5/8 because of the square root when I took out the $sin^2$?

2. Originally Posted by dorkymichelle
the problem is Given cos2x = 7/32, tan2x<0 and -\pi/2<x<\pi/2, find sin x

$1-2sin^2x=7/32$
$-2sin^2x=7/32-1$
$-2sin^2x= -25/32$
$sin^2x=-25/32 * -1/2$
$sin^2x=25/64$
$sin x = -5/8 but not 5/8$
because
-\pi/2<x<\pi/2 means the answer would be either in the first or forth quadrant and tan2x has to be smaller than 0 so if the cos is positive, then the sin has to be negative?
Is this why there is only -5/8 as the answer instead of + or - 5/8 because of the square root when I took out the $sin^2$?
Look at the last line again. Should be $\sin{x}=\frac{+5}{8}$

3. The answer the teacher gave us on our study guide says -5/8 though

4. Originally Posted by VonNemo19
Look at the last line again. Should be $\sin{x}=\frac{+5}{8}$
I got positive $\frac{5}{8}$ as well. If you use the negative than $\tan{2x}$ is not less than 0

5. Originally Posted by dorkymichelle
The answer the teacher gave us on our study guide says -5/8 though
$\sqrt\frac{25}{64}=\frac{-5}{8}$ is bad math all the way around. How can taking the square root of a number give a negative result?

6. When you square something, don't you have to consider both the negative and positive, as in \sqrt 4 = +/- 2? and then you have to plug it back into the original equation to see which root works?

7. Originally Posted by dorkymichelle
When you square something, don't you have to consider both the negative and positive, as in \sqrt 4 = +/- 2? and then you have to plug it back into the original equation to see which root works?
indeed.

You are correct. I miscommunicated what I was trying to say. I was thinking about another post while I was answering this one. I apologize.

Look at 11rdc's reply. He has spoken wisely.

8. but the original said that cos2x = 7/32
so cos is positive, so don't sin have to be negative for tangent to be less than 0?

9. Originally Posted by dorkymichelle
but the original said that cos2x = 7/32
so cos is positive, so don't sin have to be negative for tangent to be less than 0?
You are correct that $\tan{x}$ would have to be negative $\sin{x}$ if $\cos{x}$ is positive, but we are not dealing with $\tan{x}$. We are dealing with $\tan{2x}$

try out for yourself and see

$\tan{2x} = \frac{2tanx}{1-tan^2{x}}$

10. Wait, but we are only given cos2x = 7/32
so there's no cos. but -\pi/2<x<\pi/2 so that means x has to be in the first or forth quadrant of the unit circle and cosine is always positive in those qudrants. So would that mean I have a pos cos, and negative sin?
I'm not sure how to solve for just the cos of x and then plug it into the tan2x formula...

11. Originally Posted by dorkymichelle
Wait, but we are only given cos2x = 7/32
so there's no cos. but -\pi/2<x<\pi/2 so that means x has to be in the first or forth quadrant of the unit circle and cosine is always positive in those qudrants. So would that mean I have a pos cos, and negative sin?
I'm not sure how to solve for just the cos of x and then plug it into the tan2x formula...
Use pythragean theorem

$\sin^2{x}+\cos^2{x} = 1$

$\sin{x} = \frac{5}{8}$

so

$\bigg(\frac{5}{8}\bigg)^2+ \cos^2{x} =1$

12. so $cos^2=39/64$
$cos = +\sqrt39/8 or -\sqrt39/8$
but whether or not sin is negative or pos would depend on where the cosin is right? since cosin is restricted to the first and forth quad, then it has to be pos. so we have to use the +\sqrt39/8 so the sin has to neg.

13. Sorry you are correct the answer would have to be $\frac{-5}{8}$

14. Thankyou! I think i've learned more in this forum than i have the last 3 weeks in classroom >_>

15. Originally Posted by dorkymichelle
so $cos^2=39/64$
$cos = +\sqrt39/8 or -\sqrt39/8$
but whether or not sin is negative or pos would depend on where the cosin is right? since cosin is restricted to the first and forth quad, then it has to be pos. so we have to use the +\sqrt39/8 so the sin has to neg.
Exactly, I kept making the mistake thinking cos was negative but didn't see that you said it had to be between $-\frac{\pi}{2}