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I'm clueless on this question. Could someone explain how to do it?

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    $\begingroup$ are you calculating in degrees or radians ? $\endgroup$ – Dominic Michaelis Apr 7 '13 at 11:04
  • $\begingroup$ I'm calculating in degrees. $\endgroup$ – missiledragon Apr 7 '13 at 11:05
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    $\begingroup$ 585=360+120+60+45. $\endgroup$ – user67878 Apr 7 '13 at 11:06
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    $\begingroup$ If you measure angles in degrees you should write $\sin(585^\circ)$ instead of "sin585". $\endgroup$ – Christian Blatter Apr 7 '13 at 11:19
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    $\begingroup$ In the FAQ section there are directions how to use LaTeX in this site to properly write mathematics. $\endgroup$ – DonAntonio Apr 7 '13 at 11:32
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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 $$\sin(585^\circ)=-\frac{1}{\sqrt{2}}.$$

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$\sin (k)= \sin (360 ^\circ+k) \implies \sin(585^\circ)= \sin(225^\circ) $

$\sin (m)= -\sin (180^\circ+m) \implies \sin(180^\circ+45^\circ) =-\sin (45^\circ)$

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  • $\begingroup$ Thats a funny implication ;) $\endgroup$ – Dominic Michaelis Apr 7 '13 at 11:10
  • $\begingroup$ That was a typo.:P $\endgroup$ – Inceptio Apr 7 '13 at 11:12
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    $\begingroup$ you should mention you once calculate in degrees and once in radians $\endgroup$ – Dominic Michaelis Apr 7 '13 at 11:12
  • $\begingroup$ Or even better, @Inceptio: don't mix both units. $\endgroup$ – DonAntonio Apr 7 '13 at 11:33
  • $\begingroup$ Now, I chose degrees over radians.;) $\endgroup$ – Inceptio Apr 7 '13 at 11:49
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I find the circle to be a great way to understand this -

enter image description here

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.

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One liner:

$\sin(585^\circ) = \sin(585^\circ-720^\circ) = \sin(-135^\circ) = \sin(-(90^\circ+45^\circ)) = -\cos(45^\circ) = -1/\sqrt{2}$

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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)

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    $\begingroup$ In the FAQ section there are directions how to use LaTeX in this site to properly write mathematics. $\endgroup$ – DonAntonio Apr 7 '13 at 11:33
  • $\begingroup$ @DonAntonio...Thank you :) $\endgroup$ – Sreekanth Karumanaghat Apr 7 '13 at 11:35
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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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