Solving $\sin{(\cdot)}, \cos{(\cdot)}, \tan{(\cdot)}, \cot{(\cdot)} \dots$ without a calculator.

For example solve $\cos{\left(\frac{5\pi}{4}\right)}$ without a calculator or solve $\cos{(x)} = -\frac{1}{2}$.

I remember vaguely that the method involves referring to a triangle, but im not sure. Could you be kind enough to either explain how to go about these questions or to forward me to a website which explains this in detail.

Thanks a lot!

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You would want to at least remember the most common values (see formulae 3-7 there), as well as symmetries and reflection formulae (e.g., why is $\cos\,150^\circ=-\cos\,30^\circ$?). – J. M. Sep 4 '11 at 13:19
Whats that thing with the triangle then? The most common values arnt asked for...How do i get from a common value to whatever they are asking for? – John Sep 4 '11 at 13:35
You could use the geometrical interpretation of the sine and cosine to solve such questions. For example, $(\cos(\theta), \sin(\theta))$ is a point on the unit circle with angle $\theta$ from the $x$-axis. If $\cos(x) = -1/2$ then you get an equilateral triangle formed by the points $(0,0), (-1,0)$ and $(\cos(\theta), \sin(\theta))$, so $\theta$ is $60^o + 90^o = 150^o$. – TMM Sep 4 '11 at 13:36
@John, the most common values may not be what you asked for, but they are what you need in order to solve such equations symbolically. For example, if you know that $\cos(\pi/3)=1/2$ (one of the common values in question), this will allow you to see that $\cos(\pi-\pi/3)=-1/2$. And if you know that $\cos(\pi/4)=1/\sqrt2$, this allows you to see $\cos(\pi+\pi/4)=-1/\sqrt2$. – Henning Makholm Sep 4 '11 at 15:16
Alright thanks, iv come pretty far now, but cos -1/2 exists in both quadrant 2 and 3, how do I know which one is correct? Im refering to the question, solve cos(x) = -1/2 – John Sep 4 '11 at 20:51

\begin{align*} \sin\frac{\pi}{4} &= \frac{\sqrt{2}}{2}\\ \sin\frac{\pi}{6} &= \frac{1}{2}\\ \sin\frac{\pi}{3} &= \frac{\sqrt{3}}{2} \end{align*}