# How do you calculate cos (2 Arcsin (sqrt3/2))?

May 13, 2015

$\arcsin \left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$ radians $= {60}^{o}$

So $\cos \left(2 \arcsin \left(\frac{\sqrt{3}}{2}\right)\right) = \cos \left(\frac{2 \pi}{3}\right) = - \frac{1}{2}$

Picture an equilateral triangle with sides of length 1. Cut the triangle in two to get two right-angled triangles.

Each of these right-angled triangles has internal angles of ${30}^{o}$, ${60}^{o}$ and ${90}^{o}$ (or $\frac{\pi}{6}$, $\frac{\pi}{3}$ and $\frac{\pi}{2}$ radians if you prefer).

The shortest side (which is opposite the ${30}^{o}$ (pi/6) angle) of one of these triangles has length $\frac{1}{2}$ and the hypotenuse is of length $1$.

So $\sin {30}^{o} = \sin \left(\frac{\pi}{6}\right) = \cos {60}^{o} = \cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$.

The remaining side of the right angled triangle can be found using Pythagoras theorem to be

$\sqrt{{1}^{2} - {\left(\frac{1}{2}\right)}^{2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

Hence $\sin {60}^{o} = \sin \left(\frac{\pi}{3}\right) = \cos {30}^{o} = \cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$.

So when you see values of $\sin \theta$ or $\cos \theta$ of the form $\pm \frac{1}{2}$ or $\pm \frac{\sqrt{3}}{2}$ then you know that $\theta$ is a multiple of ${30}^{o}$ but not a multiple of ${90}^{o}$. Then look at a rough graph of $\sin \theta$ or $\cos \theta$ to spot which multiples you need.