Originally Posted by

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

$\displaystyle 1-2sin^2x=7/32$

$\displaystyle -2sin^2x=7/32-1$

$\displaystyle -2sin^2x= -25/32$

$\displaystyle sin^2x=-25/32 * -1/2$

$\displaystyle sin^2x=25/64$

$\displaystyle sin x = -5/8 but not 5/8$

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 $\displaystyle sin^2$?