Values of Inverse Trigonometic Functions without a Calculator

Question

I am taking algebra 2 with trigonometry (on ALEKS) and was asked this question : find exact value of arctan(-sqrt(3)/3). Their explanation was terrible, so I googled how to do the question and still could not figure it out. How would you calculate this?

Answer 1

What you want is $tan(\theta) = \frac{sin(\theta)}{cos(\theta)} = \frac{-\sqrt{3}}{3} = \frac{-1}{\sqrt{3}}$. Think about the unit circle. Notice that $\frac{-1}{\sqrt{3}} = \frac{-1/2}{\sqrt{3}/2}$.

Answer 2

HINT: What is the value of $\tan(-\pi/6)$? (You should know from your earlier education, when $\tan(x)$ has the value $1/\sqrt{3}$, $1$ or $\sqrt{3}$.)

Answer 3

If you draw a right triangle with legs of lengths $1$ and $\sqrt 3$ the hypotenuse will have length $2$. Surely you should recognize this as having angles $\pi/3$ and $\pi/6$. Of course, you also have to remember that $\tan(-x)=-\tan x.$