How to determine the exact value of $\sin(585^\circ)$? I'm clueless on this question. Could someone explain how to do it?
 A: $\sin (k)= \sin (360 ^\circ+k) \implies \sin(585^\circ)= \sin(225^\circ) $
$\sin (m)= -\sin (180^\circ+m) \implies \sin(180^\circ+45^\circ) =-\sin (45^\circ)$
A: I find the circle to be a great way to understand this -

Since the sine function is repetitive, in a 360 degree cycle, it's the same as 225 degrees. 
I am 50, and don't recall using this circle in trig class. It's a great way to visualize the function for both Sine and Cosine and can easily be memorized if need be. 
A: One liner:
$\sin(585^\circ) = \sin(585^\circ-720^\circ) = \sin(-135^\circ) = \sin(-(90^\circ+45^\circ)) = -\cos(45^\circ) = -1/\sqrt{2}$
A: When calculating in degrees, $\sin$ is periodic with a period of 360 degrees. Hence
$$\sin(585^\circ)=\sin(225^\circ).$$
In particular, $\sin(x+180^\circ)=-\sin(x)$.
Hence $$\sin(225^\circ)=\sin(45^\circ+180^\circ)=-\sin(45^\circ).$$
On the other hand, we know that $\sin(45^\circ)=\cos(45^\circ)=\frac{1}{\sqrt{2}}$.
Hence 
$$\sin(585^\circ)=-\frac{1}{\sqrt{2}}.$$
A: NB - All angle references are in degrees...
Since sin function has a period of 360,
sin 585 = sin 225
        = sin(180+45)
        = sin180*cos45+cos180sin45
        = 0+ -1*1/sqrt(2)
        = -1/sqrt(2)
A: Observe that $585^\circ\equiv 225^\circ\pmod{360^\circ}=180^\circ+45^\circ$ 
$180^\circ<180^\circ+45^\circ<270^\circ$
So, $585^\circ$ lies in the $3$rd Quadrant.
Using All-Sin-Tan-Cos formula or here, $\sin(585^\circ)<0$
Now, $585^\circ=90^\circ\cdot 6+45^\circ $
as the multiplicand of $90^\circ$ is even, sine will remain sine 
So, $\sin(585^\circ)=\sin(90^\circ\cdot 6+45^\circ)=-\sin45^\circ=-\frac1{\sqrt2}$
