How do I find the value of $\sin 2013^\circ$?

A precise decimal is not required, but must be expressed with $\sin 30^\circ,$ $\sin 45^\circ,$ and $\sin 60^\circ$ (cosine is also fine).

Hint: Use half/double angle formula.


2 Answers 2



As $2013^\circ=360^\circ\cdot 5+213^\circ$

$$\sin(2013^\circ)=\sin(213^\circ)=\sin(180^\circ+33^\circ)=-\sin33^\circ$$ as $\sin(180^\circ+x)=-\sin x$


$$\sin15^\circ=\sin(60^\circ-45^\circ)\text{ or } \sin(45^\circ-30^\circ)$$

$$\text{Similarly for }\cos15^\circ$$

$$\text{The values of }\sin18^\circ, \cos18^\circ$$ can be found here

If you insist on using using angles among $30^\circ,45^\circ,60^\circ$


Now, use $$\cos3^\circ=\cos(18^\circ-15^\circ)=\cos18^\circ\cos15^\circ+\sin18^\circ\sin15^\circ$$ and


  • 3
    $\begingroup$ This doesn't answer the question. $\endgroup$
    – Rob Arthan
    Aug 9, 2013 at 22:23
  • $\begingroup$ @RobArthan, Does the edited version answer the question? $\endgroup$ Aug 10, 2013 at 3:22
  • $\begingroup$ No. The question says that the answer must be expressed with the sines and cosines of $30^{\circ}$, $45^{\circ}$ and $60^{\circ}$. The tag says that this is a contest question, so I don't think an answer involving $\mbox{sin}(18^{\circ})$ would have been allowed. $\endgroup$
    – Rob Arthan
    Aug 10, 2013 at 20:19
  • $\begingroup$ @CommanderShepard, do you know who & why has mandated the usage of angles among $30^\circ,45^\circ,60^\circ;$. I am asking simply because it is easier to use $33=18+15$ than $30+3$ or $45-12,60-27$ etc. $\endgroup$ Aug 11, 2013 at 4:48
  • $\begingroup$ The first half leads to $\sin( 2013^\circ)= -\frac14\sqrt{8-\sqrt{28+\sqrt{80}-\sqrt{480+\sqrt{46080}}}}$ $\endgroup$
    – Henry
    Jul 1, 2022 at 21:03

As lab bhattacharjee wrote, it is $-\sin(33^\circ)$. I don't think you can write this in terms of only $\cos$ and $\sin$ of $30^\circ, 45^\circ$ and $60^\circ$ only. If you are also allowed to use $\cos$ and $\sin$ of $18^\circ$ (exact value for this is also known), then you can find $\sin(3^\circ) = \sin(18^\circ - 15^\circ)$ and $\cos(3^\circ) = \cos(18^\circ - 15^\circ)$ where you can use $\sin(15^\circ) = \sin(45^\circ - 30^\circ)$ and $\cos(15^\circ) = \cos(45^\circ - 30^\circ)$. Finally you have to use $\sin(33^\circ) = \sin(30^\circ + 3^\circ)$.

  • $\begingroup$ Can you prove that the question cannot be answered as stated? $\endgroup$
    – Rob Arthan
    Aug 9, 2013 at 22:25

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