# Finding values of t for sin&tan

• Oct 25th 2008, 05:57 PM
Dust
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? (Speechless)
• Oct 25th 2008, 06:05 PM
Jhevon
Quote:

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? (Speechless)

you should know by heart that $\displaystyle \sin \frac {\pi}6 = \frac 12$ and $\displaystyle \tan \frac {\pi}3 = \sqrt{3}$

now, use your knowledge of reference angles to continue

can you?
• Oct 25th 2008, 06:53 PM
Dust
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.
• Oct 25th 2008, 06:57 PM
11rdc11
sin is negative in the 3rd and 4th quad and tan is negative in the 2nd and 4th quad
• Oct 26th 2008, 06:32 AM
Dust
So i am correct?

(Worried)
• Oct 26th 2008, 08:07 AM
masters
Quote:

Originally Posted by Dust
So i am correct?

(Worried)

Yep!!

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

The reference angle in the 2nd quadrant where tan is negative is http://www.mathhelpforum.com/math-he...3b0b1c35-1.gif.

The reference angle in the 4th quadrant where tan is negative is http://www.mathhelpforum.com/math-he...ff55cbff-1.gif.