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

Řídící
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missiledragon
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    are you calculating in degrees or radians ? – Dominic Michaelis Apr 07 '13 at 11:04
  • I'm calculating in degrees. – missiledragon Apr 07 '13 at 11:05
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    585=360+120+60+45. –  Apr 07 '13 at 11:06
  • Now it is easy with that decomposition. –  Apr 07 '13 at 11:07
  • I can do it even with subtraction, look! 585=720-180+45 ;-D –  Apr 07 '13 at 11:13
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    If you measure angles in degrees you should write $\sin(585^\circ)$ instead of "sin585". – Christian Blatter Apr 07 '13 at 11:19
  • Not every one knows how to put the mathematical symbols like that it would be helpfull if some one provided a guidance – Sreekanth Karumanaghat Apr 07 '13 at 11:31
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    In the FAQ section there are directions how to use LaTeX in this site to properly write mathematics. – DonAntonio Apr 07 '13 at 11:32
  • @missledragon : if you write an angle without a degree symbol then it is in radians, regardless of what you are thinking. I penalize my students when they omit the degree symbol (except for the angle $0$, which is the same in radians and degrees. For some nonzero numbers $x$, $\sin(x^\circ) = \sin(x)$ and $\cos(x^\circ) = \cos(x)$, but these are very special numbers and I haven't figured out yet what they are. – Stefan Smith Apr 07 '13 at 17:31
  • @Stefan Smith Where did you get information about those numbers that have equal value of sine and/or cosine irrespectively of the representation in degrees or in radians? –  Apr 07 '13 at 18:13
  • @Thus See here. – Řídící Apr 07 '13 at 18:31
  • @Gugg Oh, I forgot to ask Mr. Wolfram, I see now, and they are not especially special numbers (at least not to me). Just some irrationals in the world of irrationals. –  Apr 07 '13 at 18:38
  • @Gugg : Thanks, but I don't think this is enough if you want the sine and cosine to be the same. You need $\frac{2\pi}{360}x-x$ to be an integer multiple of $2\pi$, so $x$ has the form $\frac{360n\pi}{\pi-180}$ for some integer $n$ (Wolfram's first answer). Wolfram's second answer doesn't work. – Stefan Smith Apr 07 '13 at 20:33
  • @Thus : see my comment to Gugg. – Stefan Smith Apr 07 '13 at 20:34
  • @StefanSmith I ignored/overlooked the cosine bit, perhaps because I didn't (and still don't) see the relevance of cos to (the stating of) the original question. Wolfram with cosine now. – Řídící Apr 07 '13 at 20:44

6 Answers6

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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 $$\sin(585^\circ)=-\frac{1}{\sqrt{2}}.$$

Ilmari Karonen
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Dominic Michaelis
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$\sin (k)= \sin (360 ^\circ+k) \implies \sin(585^\circ)= \sin(225^\circ) $

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

Inceptio
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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.

JTP - Apologise to Monica
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One liner:

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

Christian Mann
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igumnov
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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)

Sreekanth Karumanaghat
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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, $\sin(585^\circ)=\sin(90^\circ\cdot 6+45^\circ)=-\sin45^\circ=-\frac1{\sqrt2}$

lab bhattacharjee
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