Sin θ = [tex] \frac{1}{2} [/tex]
First we need to think, where is θ?
Because θ is the same as a variable so it will be the same for cos θ.
They didn't give us an interval to work with so there will be multiple answers.
First we need to find when sin θ is positive.
Imagine (or create) the unit circle. It's within a x and y plane. So the two quadrants where y is positive is the 1st and 2nd quadrant.
Try to devote the angles on the unit circle to memory, it helps a lot!
sin θ = [tex] \frac{1}{2} [/tex]
Needs a reference angle of [tex] \frac{ \pi }{6} [/tex]
Or in the case of the 1st quadrant it is that angle.
So, we have one angle of the two. [tex] \frac{ \pi }{6} [/tex]
Take [tex] \pi [/tex] and subtract [tex] \frac{ \pi }{6} [/tex] from it. (Don't use a calculator!)
[tex] \frac{5 \pi }{6} [/tex]
Is the number you should get.
So we have the two possible angles. Now to find the cos of those angles. Pro tip! When sin θ = [tex] \frac{1}{2} [/tex]
cos θ = [tex] \frac{+}{-} \frac{ \sqrt{3} }{2} [/tex]
Where + / - is the symbol plus or minus. Whether it is one or the other depends on the quadrant is resides in.
We have two answers! One that will be positive (quadrant 1) and one that will be negative (quadrant 2)
[tex] \frac{ \sqrt{3} }{2}, \frac{ -\sqrt{3} }{2} [/tex]
Those are our two answers. Hope this helps!
