# Trigonometric Function

• Nov 6th 2007, 10:24 PM
ktcrew
Trigonometric Function
Ok I'm asked to find the other Trig. Functions given one value.
Here's the question:

sec = -2 tan > 0

So sec = 1/x, so I get 1/(-1/2) = -2.
If I'm understanding it correctly, x = cos, therefore x = -1 and r = 2?
Also since tan > 0, that means y is also negative since tan = y/x and x is already -1?
I just want to know if I'm understanding it correctly.
Thanks.
• Nov 7th 2007, 03:47 AM
topsquark
Quote:

Originally Posted by ktcrew
Ok I'm asked to find the other Trig. Functions given one value.
Here's the question:

sec = -2 tan > 0

So sec = 1/x, so I get 1/(-1/2) = -2.
If I'm understanding it correctly, x = cos, therefore x = -1 and r = 2?
Also since tan > 0, that means y is also negative since tan = y/x and x is already -1?
I just want to know if I'm understanding it correctly.
Thanks.

You really need to work on your notation. (Worried)

I would translate what you wrote into something more meaningful, but there are too many conceptual errors in there to make it worth the effort.

We know that
$sec(q) = -2$ and $tan(q) > 0$.

Thus
$cos(q) = \frac{1}{sec(q)} = \frac{1}{-2} = -\frac{1}{2}$.

This means that q is in either quadrant II or quadrant III. Given that $tan(q) > 0$, we can reduce this to q being in quadrant II.

So, to find $sin(q)$, just use $sin^2(q) + cos^2(q) = 1$. The value of $sin(q)$ will be positive since q is in QII.

Now the rest is easy:
$tan(q) = \frac{sin(q)}{cos(q)}$

$csc(q) = \frac{1}{sin(q)}$

$cot(q) = \frac{1}{tan(q)}$

-Dan
• Nov 7th 2007, 04:58 AM
ktcrew
Sorry about the notation, I haven't really learned all the tags yet.
But I don't quite understand $tan(q)$ being in quadrant 2.
Shouldn't $sin(q)$ be negative since $cos(q)$ is a negative value? And a negative numerator divided by a negative denominator would result in a positive number because the signs cancel each other, no?
• Nov 7th 2007, 05:08 AM
topsquark
Quote:

Originally Posted by ktcrew
Sorry about the notation, I haven't really learned all the tags yet.

My comment wasn't about the lack of the fancy looking tags. It was about statements like
"sec = -2 and tan > 0"

You need to place an angle in that statement. I'll admit it's a little fussy, but it's just good practice at being clear in your writing.

Quote:

Originally Posted by ktcrew
But I don't quite understand $tan(q)$ being in quadrant 2.
Shouldn't $sin(q)$ be negative since $cos(q)$ is a negative value? And a negative numerator divided by a negative denominator would result in a positive number because the signs cancel each other, no?

Yes, you are right. I should have said that angle q is in QIII. The tangent function is positive in quadrants I and III, it is negative in QII. My apologies.

-Dan
• Nov 7th 2007, 05:18 AM
ktcrew
Ok, thanks.