Question

How do you find the exact values `cos (pi/6)` using the special triangles?

Answer 1

As below.

Explanation:

`(pi/6)^c = (pi/6) * (180/pi) = 30^@`

`cos (pi/6)^c = cos (30)^@`

In the above special right triangles,

`"Angles are " 30^@, 60^@, 90^@ " and the sides are "a, a sqrt3, 2a`

`cos theta = " Adj. side / Hypotenuse" = (AB) / (AC)`

`cos 30 = (cancela sqrt3) / (2 cancela) = sqrt3 / 2`