# Trouble finding the value of $\cos(-1710^\circ)$

The technique I was taught was as follows:

Make the angle positive $\cos(-\theta) = \cos(\theta) = \cos (1710^\circ)$

Write the angle in the form of multiple of $90^\circ$.

$= \cos(90^\circ\cdot 19 + 0)$.

If the number multiplied is odd, change $\cos(\theta)$ to $\sin(\alpha)$. This means $\cos (90^\circ\cdot 19 + 0)= \sin (90^\circ) = 1$. However, the answer given is $\cos(90^\circ) = 0$. What am I doing wrong?

$-1710^\circ=-1800^\circ+90^\circ$, hence the given answer is correct. [And in one line, not 5, as in the concurrent answer :-) ].

$$\cos(-1710^{\circ})=\cos(1710^{\circ}) \because \cos (-x)=\cos(x)$$

$$\cos(1710^{\circ})=\cos(1800^{\circ}-90^{\circ})$$

$$\cos(1710^{\circ})=\cos(360^{\circ}\cdot5-90^{\circ})$$

$$\cos(1710^{\circ})=\cos(-90^{\circ})\because \cos(360^{\circ}\cdot n\pm x)=\cos(\pm x)$$

$$\cos(1710^{\circ})=\cos(90^{\circ})=0$$

The technique you have learned seem flawed, you'll always end up with $\cos$ or $\sin$ to an angle between $0^\circ$ and $90^\circ$, but that's always in $[0,1]$ and e.g. $\cos(180^\circ) = -1$. If you have only learned about $\sin$ and $\cos$ for angles, you might have a definition that avoids negative values, then the technique might work.

BTW: I don't undersand why do end up with $90^\circ$ as argument of $\sin$? The remainder is $0$.