1. ## cos pi/6

quick question

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

DL

2. ## 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 ...

$\displaystyle \cos(\theta) = \frac{opposite \, side}{adjacent \, side}$

3. ## Re: cos pi/6

Hi Skeeter,

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

cos(tau) = adjacent side / hypotenuse

4. ## Re: cos pi/6

yes, my error ... had tangent on the brain

5. ## Re: cos pi/6

Hello, davellew69!

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

Consider an equilateral triangle with side length 2.
Each vertex has a 60o angle.
Draw an altitude $\displaystyle h$. .It bisects the angle.
The diagram looks like this.
Code:
              A
*
/|\
/ | \
/30|  \
2 /   |h  \ 2
/    |    \
/     |     \
/ 60   |      \
B *-------*-------* C
1   D   1`
Pythagorus says: .$\displaystyle h^2 + 1^2 \,=\,2^2$

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

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

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

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