# Thread: Very quick question about sin

1. ## Very quick question about sin

Let say you want to find solutions for t(theta) of an equation. You come to this part;
sin^2 t = 1/2

To get rid of that power two, would we have to square root the 1/2 and also put plus or minus?

How about tan^2 t = 1/2? Would we have to square root the 1/2 and also put plus or minus?

Thank you so much.

2. Originally Posted by florx
Let say you want to find solutions for t(theta) of an equation. You come to this part;
sin^2 t = 1/2

To get rid of that power two, would we have to square root the 1/2 and also put plus or minus?

How about tan^2 t = 1/2? Would we have to square root the 1/2 and also put plus or minus?

Thank you so much.
Yeah that's right.

3. Which one is it? Do we have to square root and plus or minus or just one of those?

4. Originally Posted by florx
Which one is it? Do we have to square root and plus or minus or just one of those?
Depends on the domain you are trying to solve the equation on. Do you know?

5. Oh, I'm talking about something in general like 0 ≤ t ≤ 2pi.

6. For

$\sin t = \frac{1}{\sqrt{2}}$

is in quadrants 1 and 2, solutions will be $t$ and $\pi-t$

and for

$\sin t = \frac{-1}{\sqrt{2}}$

is in quadrants 3 and 4, solutions will be $\pi+t$ and $2\pi-t$

you can find $t$ by using a right angled isosceles triangle with side lengths $1,1,\sqrt{2}$