Can someone help me out with this Trig question? When sinx = -8/17 and x lies in Quadrant III and cosy = -4/5 and y lies in Quadrant II, what is cos(x-y?) So far, I did cos(-8/17 - -4/5) = cos-8/17cos-4/5 + sin-8/17sin-4/5 And now I'm stuck.
we need \[\cos(x-y)=\cos(x)\cos(y)+\sin(x)\sin(y)\] so we want \[\cos(x), \sin(x), \cos(y), \sin(y)\]
I think so.
\[\sin(x)=-\frac{8}{17}\]
sorry that was a type, but before we start let me point out something i think is confusing youi
the sine of x IS -8/17, you do not want the sine of -8/17 you just need these numbers, not the sine or cosine of these numbers
cos(x) = -15/17, you wanted that earlier, right?
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So... I'm not supposed to be using the formula I used...?
yes, but you are confused i think about the formula. you need the numbers themselves, not the cosine and sine of the numbers
in other words just work with the numbers \[\sin(x)=-\frac{8}{15}\] \[\cos(x)=-\frac{15}{17}\] \[\cos(y)=-\frac{4}{2}\] \[\sin(y)=\frac{3}{5}\]
typo there \[\cos(y)=-\frac{4}{5}\] of course now plug the numbers directly into the formula
and in quadrant II sine is positive
so your answer is \[-\frac{8}{17}\times -\frac{4}{5}+(-\frac{15}{17})\times \frac{4}{5}\]
clear right? those are the sines and cosines, you do not take the sines and cosines of those numbers
Uh... I'm a little confused, hold on a sec x_x;;;
ok i will wait. this is a common mistake, i have seen it before
point is that \[\cos(x)=-\frac{3}{5}\] you do not take the cosine of that number. that number is your cosine
If it's cos(x)cos(y), shouldn't it be -15/17 x -4/5?
yes sorry that was a typo on my part
you are right just don't do this cos-8/17cos-4/5 + sin-8/17sin-4/5 instead do this -8/17 *-4/5 + -8/17*4/5
According to my calculator, that's zero, unless there's another typo ><....
no it is not zero
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