In need of direction

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November 12th 2012, 04:10 PM
M670

In need of direction

Suppose
http://webwork.mathstat.concordia.ca...dc804d39a1.png
and http://webwork.mathstat.concordia.ca...c4ab36f8c1.png is negative. Here are some small variations on the previous problems:
http://webwork.mathstat.concordia.ca...b69be1bdf1.png
http://webwork.mathstat.concordia.ca...96990ee631.png
http://webwork.mathstat.concordia.ca...d042b0e5f1.png
http://webwork.mathstat.concordia.ca...846bc07dc1.png
http://webwork.mathstat.concordia.ca...85572f1991.png

Can someone please point me in a direction towards solving this problem, I am not sure what they are asking?
November 12th 2012, 04:42 PM
skeeter

Re: In need of direction

$\cos{u} > 0$ and $\sin{u} < 0$ indicates $u$ is a quad IV angle.

$\cos{u} = \frac{5}{13} \, , \, \sin{u} = -\frac{12}{13}$

to calculate the rest, use the difference identity for sine and cosine ...

$\sin(a - b) = \sin{a}\cos{b} - \cos{a}\sin{b}$

$\cos(a-b) = \cos{a}\cos{b} + \sin{a}\sin{b}$
November 12th 2012, 04:52 PM
HallsofIvy

Re: In need of direction

You can also note that a right triangle with hypotenuse of length 13 and one leg of length 5 has cosine of the angle between those two side 5/13. What is the length of the other leg? And from that, what is the sine of the angle? This is treating the angle as between 0 and $\pi/2$ . Use the quadrant information to find the sign.
November 12th 2012, 04:55 PM
M670

Re: In need of direction

Quote:

Originally Posted by HallsofIvy

You can also note that a right triangle with hypotenuse of length 13 and one leg of length 5. What is the length of the other leg?

12 ahh I see how you are looking at it...
November 12th 2012, 05:14 PM
M670

Re: In need of direction

so then for sin(u-pi) it should look like this =sin(u)cos(pi)-cos(u)sin(pi)
November 12th 2012, 06:19 PM
M670

Re: In need of direction

or would it be sin(-(12/13)-pi)= sin-(12/13)cos(pi)-cos-(12/13)sin(pi)?

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