# Some trig problems from Shy Guy

• April 15th 2008, 06:35 PM
topsquark
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 $(cos(\theta), sin(\theta))$. As the angle gets smaller and smaller the point on the unit circle gets closer and closer to (1, 0). Which means that $cos(0) = 1$ and $sin(0) = 0$.

Think of the tangent problems backward.
$tan \left ( \frac{\pi}{6} \right ) = \frac{\sqrt{3}}{3}$
so
$tan^{-1} \left ( \frac{\sqrt{3}}{3} \right ) = \frac{\pi}{6}$.

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
• April 15th 2008, 06:54 PM
>_<SHY_GUY>_<
i dont get how sqrt 3/3 in a 30-60-90 triangle in reference to the unit circle, is part of it...i dont know if that make sense...
because i know that 1/2 and sqrt 3/2 and the hypotenuse 1 is a 30-60-90.

and how do you know sqrt 3/3 is pi/6?
• April 16th 2008, 03:53 AM
topsquark
Quote:

Originally Posted by >_<SHY_GUY>_<
i dont get how sqrt 3/3 in a 30-60-90 triangle in reference to the unit circle, is part of it...i dont know if that make sense...
because i know that 1/2 and sqrt 3/2 and the hypotenuse 1 is a 30-60-90.

and how do you know sqrt 3/3 is pi/6?

For the $tan(\pi / 6)$ part, this is just the same as $tan(30)$, so let's work on that.

$tan(30) = \frac{sin(30)}{cos(30)} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$

$= \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

Does this help?

-Dan