# Thread: Finding values of t for sin&tan

1. ## Finding values of t for sin&tan

I have homework where i need to do the following:

3. For what values of t on the interval [0,2π] is: a. sin(t) = -1/2
b. tan(t) = -
√3

*in part 'b' the symbol is a square root, meaning the negative square root of 3*

I think it can be solved using the unit circle (..?) but I can't figure it out.
Help?

2. Originally Posted by Dust
I have homework where i need to do the following:

3. For what values of t on the interval [0,2π] is: a. sin(t) = -1/2
b. tan(t) = -
√3

*in part 'b' the symbol is a square root, meaning the negative square root of 3*

I think it can be solved using the unit circle (..?) but I can't figure it out.
Help?
you should know by heart that $\sin \frac {\pi}6 = \frac 12$ and $\tan \frac {\pi}3 = \sqrt{3}$

now, use your knowledge of reference angles to continue

can you?

3. I know that sin pi/6 is 1/2 and tan pi/3 is the square root of 3.
I understand what you said.

My problem is the negative signs are throwing me off.
Would the sine graph be 11π/6 and 7π/6 ??
And the tan graph would be 2π/3 and 5π/3 ??

If not i am truly lost.

4. sin is negative in the 3rd and 4th quad and tan is negative in the 2nd and 4th quad

5. So i am correct?

6. Originally Posted by Dust
So i am correct?

Yep!!

But what about $\tan(t)=-\sqrt{3}$?

The reference angle in the 2nd quadrant where tan is negative is .

The reference angle in the 4th quadrant where tan is negative is .