# Math Help - How do the quadrants affect my answer here?

I am told that $x= -7/25$ and $y = 15/8$, where x is in the second quandrant and y is in the third quandrant.

I need to find the exact value of:

$sin(x-y)$

I know I need to use the difference formula some how and I realise that sin is negative in the 3rd and 4th quadrant; however, I didn't know how to apply the negative sign for $y$. Do I apply it within the sin function or does it make the entire sin function negative? The confusion was because there are two values in different quadrants.

Also, how would I go about converting x and y into values that I can derive exact values from? Do I need to rearrange them before I apply the sum/difference formula or after?

Thanks.

Can't you just substitute $\displaystyle x$ and $\displaystyle y$ as given?

\displaystyle \begin{align*} \sin{(x - y)} &= \sin{\left(-\frac{7}{25} - \frac{15}{8}\right)} \\ &= \sin{\left(-\frac{56}{200} - \frac{105}{200}\right)} \\ &= \sin{\left(-\frac{161}{200}\right)} \\ &= -\sin{\left(\frac{161}{200}\right)}\end{align*}

This is as exact as you're going to get...

How did knowing the quadrants play any role at all then? I just figured that the statement was made for a reason. Also, I just assumed exact meant using the values for 30, 45, 60 as it also said without using a calculator.

PS. should be $-sin(431/200)$ right?

I suspect you have written the problem wrong- that the fractions are NOT x and y but are either sine or cosine of x and y. Otherwise, it makes no sense to say that a single number is in any specific quadrant.

Originally Posted by terrorsquid
How did knowing the quadrants play any role at all then? I just figured that the statement was made for a reason. Also, I just assumed exact meant using the values for 30, 45, 60 as it also said without using a calculator.

PS. should be $-sin(431/200)$ right?
Yes it should be, I added the fractions wrong ><

Originally Posted by HallsofIvy
I suspect you have written the problem wrong- that the fractions are NOT x and y but are either sine or cosine of x and y. Otherwise, it makes no sense to say that a single number is in any specific quadrant.
Ah yes sorry I read it a little fast :S

It should be:

$cos(x) = -7/25$ and $tan(y) = 15/8$

So would I draw two triangles using cosx and tany and then use the difference formula and substitute in the values from the traingles?

And would I apply the necessary sign to cos(x) and tan(y) now? ie -17/25 becomes -- = +17/25 and 15/8 stays +?

Working on it now will see what I get...

So I drew two triangles. $\bigtriangleup x = 7, 24, 25$ and $\bigtriangleup y = 8, 15, 17$.

Using the difference formula:

$sin(x-y) = sin(x)cos(y) - cos(x)sin(y)$

$= \frac{24}{25} * \frac{8}{17} - \frac{7}{25} * \frac{15}{17}$

$=\frac{87}{425}$

That's assuming the $-\frac{7}{25}$ became positive.

Or do I apply the quadrant sign when deriving values from the $sin(x)cos(y) - cos(x)sin(y)$ equation? i.e.:

$(\frac{24}{25})(-\frac{8}{17}) - (-\frac{7}{25})(-\frac{15}{17})$

$= -\frac{297}{425}$

Originally Posted by terrorsquid
Or do I apply the quadrant sign when deriving values from the $sin(x)cos(y) - cos(x)sin(y)$equation? i.e.:

$(\frac{24}{25})(-\frac{8}{17}) - (-\frac{7}{25})(-\frac{15}{17})$

$= -\frac{297}{425}$
You need to keep the numbers the way they have been given.

Originally Posted by Prove It
You need to keep the numbers the way they have been given.
So, to be clear, at what point do I use the information about which quadrant x and y are in?

Originally Posted by terrorsquid
So, to be clear, at what point do I use the information about which quadrant x and y are in?
You don't need to worry about it at all in this case. It's not asking you to find any angles, it's asking you to find the sine of a difference of angles using the information given.

Ok,fair enough. It seems strange to include the information about the quadrants in the question if they aren't going to be used though doesn't it? Maybe trying to through me off

Because $cos(x) = -\frac{7}{25}$ and $x$ is in the 2nd quadrant, where sin(x) is positive, that would mean the 7 is negative and the 25 is positive right? So when substituting values into the sum formula equation I should be using a -7 shouldn't I?

Sorry if I am talking nonsense; can't seem to get it out of my head.

Originally Posted by terrorsquid
Because $cos(x) = -\frac{7}{25}$ and $x$ is in the 2nd quadrant, where sin(x) is positive, that would mean the 7 is negative and the 25 is positive right? So when substituting values into the sum formula equation I should be using a -7 shouldn't I?
Because $x\in II~\&~y\in III$ we have:
$\sin(x)=\frac{24}{25},~\sin(y)=\frac{-15}{17},~\&~\cos(y)=\frac{-8}{17}$

Does that help?

Originally Posted by Plato
Because $x\in II~\&~y\in III$ we have:
$\sin(x)=\frac{24}{25},~\sin(y)=\frac{-15}{17},~\&~\cos(y)=\frac{-8}{17}$

Does that help?
Yes, thanks.

So for $sin(x)cos(y) - cos(x)sin(y)$ I would have:

$\left(\frac{24}{25}\right)\left(\frac{-8}{17}\right) - \left(\frac{-7}{25}\right)\left(\frac{-15}{17}\right)$

$= -\frac{297}{425}~~?$

Just out of curiosity, in $\bigtriangleup y$ where $y\in III$ you have made the two edges (15 & 8) negative and the hypotenuse (17) positive which makes tan positive and both sin & cos negative, which is correct for $III$. Does it matter that you do it this way? for example, if you were to make the hypotenuse negative and the two edges positive, you would still have tan +, sin -, cos -.

Thanks.