Some trig problems from Shy Guy

Quote:

Originally Posted by **>_<SHY_GUY>_<**

one thing to just clear out...im kinda blank today...why is cosine of 0 = 1 and sine of 0 = 0?

how is arctan square root of 3 divided by 3 = pi/6?

how is (arctan x = pi/4) equal to 1?

how is (arctan x = -pi/6) equal to the negative square root of 3 divided by 3?

and how is square root 3 /3 part of the 30-60-90 triangle? i though it was sqaureroot 3/2?

Imagine the unit circle. A point (x, y) has coordinates $\displaystyle (cos(\theta), sin(\theta))$. As the angle gets smaller and smaller the point on the unit circle gets closer and closer to (1, 0). Which means that $\displaystyle cos(0) = 1$ and $\displaystyle sin(0) = 0$.

Think of the tangent problems backward.

$\displaystyle tan \left ( \frac{\pi}{6} \right ) = \frac{\sqrt{3}}{3}$

so

$\displaystyle tan^{-1} \left ( \frac{\sqrt{3}}{3} \right ) = \frac{\pi}{6}$.

The other inverse tangent problems follow a similar logic.

The 30 - 60 - 90 triangle contains a 30 degree angle. tan(30) = sqrt(3) / 3. You are thinking of either sin(60) or cos(30), which are both sqrt(3) / 2.

-Dan