# In need of direction

• November 12th 2012, 05:10 PM
M670
In need of direction
• November 12th 2012, 05:42 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, 05:52 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, 05:55 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, 06:14 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, 07:19 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)?