I'm clueless on this question. Could someone explain how to do it?
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$\begingroup$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 $$\sin(585^\circ)=-\frac{1}{\sqrt{2}}.$$
$\endgroup$ $\begingroup$$\sin (k)= \sin (360 ^\circ+k) \implies \sin(585^\circ)= \sin(225^\circ) $
$\sin (m)= -\sin (180^\circ+m) \implies \sin(180^\circ+45^\circ) =-\sin (45^\circ)$
$\endgroup$ 5 $\begingroup$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.
$\endgroup$ $\begingroup$One liner:
$\sin(585^\circ) = \sin(585^\circ-720^\circ) = \sin(-135^\circ) = \sin(-(90^\circ+45^\circ)) = -\cos(45^\circ) = -1/\sqrt{2}$
$\endgroup$ $\begingroup$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)
$\endgroup$ 2 $\begingroup$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, $\sin(585^\circ)=\sin(90^\circ\cdot 6+45^\circ)=-\sin45^\circ=-\frac1{\sqrt2}$
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