please help me :(
If sin Θ = negative square root 3 over 2 and π

Question

please help me :(
If sin Θ = negative square root 3 over 2 and π < Θ < 3 pi over 2, what are the values of cos Θ and tan Θ?

anonymous · Answer

cos Θ = negative 1 over 2; tan Θ = square root 3

 cos Θ = negative 1 over 2; tan Θ = −1

 cos Θ = square root 3 over 4; tan Θ = −2

 cos Θ = 1 over 2; tan Θ = square root 3

anonymous · Answer

@amistre64

anonymous · Answer

@zepdrix

zepdrix · Answer

|dw:1401137054189:dw|Theta is between Pi and 3Pi/2, so we're in the third quadrant.

zepdrix · Answer

Cosine represents our x coordinate along the circle,
is our cosine going to be positive or negative in the third quadrant?

anonymous · Answer

negative? @zepdrix

anonymous · Answer

@nanalew

anonymous · Answer

@tkhunny

anonymous · Answer

@hero

tkhunny · Answer

You should have memorized stuff about 30-60-90 right triangles.  Did you?

anonymous · Answer

no... :| @tkhunny

tkhunny · Answer

That's not good.  You should go back and do that.

$\sin(30º) = 1/2$
$\cos(30º) = \sqrt{3}/2$

$\sin(60º) = \sqrt{3}/2$
$\cos(60º) = 1/2$

anonymous · Answer

what is that? could you please explain it? @tkhunny

tkhunny · Answer

Nothing to explain.  These are the things you should have memorized.

$30º = \pi/6$
$60º = \pi/3$

anonymous · Answer

its difficult to learn math online, i was just on here to get some help and i don't know what your answer means or how you got it @tkhunny

tkhunny · Answer

Yes, it is very difficult to learn mathematics online.  You have to hang in there and try extra hard.  Just answer my questions as well as you can.

You should know these, too.

$\sin(45º) = \cos(45º) = \sqrt{2}/2$

$45º = \pi/4$

What I am trying to tell you is that you need to put these special values in your head and leave them there.  I didn't "get them" from anywhere.  They need simply to be memorized.

You have $\sin(\theta) = -\sqrt{3}/2$

The first alarm that should go off in your head is that this is a Reference Angle of $\pi/3$ in Quadrant III or Quadrant IV.  Since we are restricted to $\pi < \theta < 3\pi/2$, this puts us in Quadrant III.

Having said all that, we now know that we are talking about $\theta = 4\pi/3$.

Now, how do we solve the rest of the problem?

$\cos(3\pi/3) = ??$ < 0
$\tan(3\pi/3) = ??$ > 0