Find the value of $\sin 2013^\circ$

Question

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.

Answer 1

HINT:

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

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

Now,$$\sin33^\circ=\sin(18^\circ+15^\circ)=\sin18^\circ\cos15^\circ+\sin15^\circ\cos18^\circ$$

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

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

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

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

$$\sin33^\circ=\sin(30^\circ+3^\circ)=\sin30^\circ\cos3^\circ+\sin3^\circ\cos30^\circ=\frac{\cos3^\circ+\sqrt3\sin3^\circ}2$$

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

$$\sin3^\circ=\sin(18^\circ-15^\circ)=\sin18^\circ\cos15^\circ-\cos18^\circ\sin15^\circ$$

Answer 2

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 $\sin(33^\circ) = \sin(30^\circ + 3^\circ)$.