# cos pi/6

• Sep 30th 2012, 03:24 PM
davellew69
cos pi/6
quick question

how is cos pi/6 = sq root 3 /2 I just cant work it out

DL
• Sep 30th 2012, 03:52 PM
skeeter
Re: cos pi/6
are you familiar w/ the side ratios of a 30-60-90 triangle? if so, let the hypotenuse have length = 1. also ...

$\cos(\theta) = \frac{opposite \, side}{adjacent \, side}$
• Oct 8th 2012, 05:14 AM
StefanTM
Re: cos pi/6
Hi Skeeter,

a little correction:
you did write tan(tau),

cos(tau) = adjacent side / hypotenuse
• Oct 8th 2012, 06:36 AM
skeeter
Re: cos pi/6
yes, my error ... had tangent on the brain
• Oct 8th 2012, 07:18 AM
Soroban
Re: cos pi/6
Hello, davellew69!

Quote:

$\text{How is: }\,\cos\tfrac{\pi}{6}\,=\,\frac{\sqrt{3}}{2}\,?$

Consider an equilateral triangle with side length 2.
Each vertex has a 60o angle.
Draw an altitude $h$. .It bisects the angle.
The diagram looks like this.
Code:

              A               *             /|\             / | \           /30|  \         2 /  |h  \ 2         /    |    \         /    |    \       / 60  |      \     B *-------*-------* C           1  D  1`
Pythagorus says: . $h^2 + 1^2 \,=\,2^2$

From which we get: . $h \,=\,\sqrt{3}$

Hence, in a 30-60 right triangle, we have these sides:

. . $\begin{Bmatrix}\text{opposite }30^o & 1 \\ \text{opposite }60^o\!: & \sqrt{3} \\ \text{hypotenuse:} & 2 \end{Bmatrix}$

Now we can write all the trig values for 30o and 60o.