# How do you find the exact value of tan^(-1) (-2 cos (pi/3))?

Jun 6, 2016

First, let's convert $\frac{\pi}{3}$ to degrees. We do this with the conversion factor of $\frac{180}{\pi}$

$\frac{\pi}{3} \times \frac{180}{\pi} = {60}^{\circ}$

Now we have to calculate $\cos {60}^{\circ}$. By the special triangle of sides $1 , \sqrt{3} \mathmr{and} 2$ we have that $\cos {60}^{\circ} = \frac{1}{2}$

We can state the new expression as being ${\tan}^{-} 1 \left(- 2 \times \frac{1}{2}\right) = {\tan}^{-} 1 \left(- 1\right)$

We must calculate the ${\tan}^{-} 1 \left(- 1\right)$. We can do this by first looking in which quadrants tan will be negative. Tan is negative in quadrants $I I \mathmr{and} I V$.

Furthermore, for the ratio to be $- \frac{1}{1}$, the 45^@, 45^@, 90^@; 1, 1, sqrt(2) special triangle will be implicated.

By reference angles, we have that ${\tan}^{-} 1 \left(- 1\right) = 180 - {45}^{\circ} \mathmr{and} 360 - {45}^{\circ}$, so ${\tan}^{-} 1 \left(- 1\right) = {135}^{\circ} \mathmr{and} {315}^{\circ}$.

Depending on how your teacher wants your answer ${\tan}^{-} 1 \left(- 2 \cos \left(\frac{\pi}{3}\right)\right) = {135}^{\circ} \mathmr{and} {315}^{\circ} \mathmr{and} {\tan}^{-} 1 \left(- 2 \cos \left(\frac{\pi}{3}\right)\right) = \frac{3 \pi}{4} \mathmr{and} \frac{7 \pi}{4}$

Hopefully this helps!