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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](http://mathworld.wolfram.com/TrigonometryAngles.html) (see formulae 3-7 there), as well as symmetries and reflection formulae (e.g., why is $\cos\,150^\circ=-\cos\,30^\circ$?).2011-09-04
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    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?2011-09-04
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    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 $60^o + 90^o = 150^o$.2011-09-04
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    @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 $\cos(\pi+\pi/4)=-1/\sqrt2$.2011-09-04
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    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) = -1/22011-09-04
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    @John, both are valid solutions, unless you have extrinsic restrictions of the range of $x$. (You can also add or subtract any multiple of $2\pi$, of course).2011-09-05

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