cos pi/6

**davellew69** · September 30th 2012, 02:24 PM

quick question

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

DL

**skeeter** · September 30th 2012, 02:52 PM

are you familiar w/ the side ratios of a 30-60-90 triangle? if so, let the hypotenuse have length = 1. also ...

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

**StefanTM** · October 8th 2012, 04:14 AM

Hi Skeeter,

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

cos(tau) = adjacent side / hypotenuse

**skeeter** · October 8th 2012, 05:36 AM

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

**Soroban** · October 8th 2012, 06:18 AM

Hello, davellew69!

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

Consider an equilateral triangle with side length 2.
Each vertex has a 60^o 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 30^o and 60^o.

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