# How do you solve tan^ (-1) (1)+ cos ^(-1)(-1/2) + sin ^-1(-1/2)?

Sep 12, 2017

${\tan}^{-} 1 \left(1\right) + {\cos}^{-} 1 \left(- \frac{1}{2}\right) + {\sin}^{-} 1 \left(- \frac{1}{2}\right) = \frac{3 \pi}{4}$

#### Explanation:

Note that typical trigonometric functions are in the form $\sin \left(\theta\right) = x$, where $\theta$ is an angle and $x$ is a ratio.

Inverse trigonometric functions are then in the form $\theta = {\sin}^{-} 1 \left(x\right)$, where the argument of the inverse trig function is instead a ratio and the function yields an angle measure.

Then, for the three inverse trig functions we have here, note that:

• $\alpha = {\tan}^{-} 1 \left(1\right) \text{ "=>" } \tan \left(\alpha\right) = - 1$
• $\beta = {\cos}^{-} 1 \left(- \frac{1}{2}\right) \text{ "=>" } \cos \left(\beta\right) = - \frac{1}{2}$
• $\phi = {\sin}^{-} 1 \left(- \frac{1}{2}\right) \text{ "=>" } \sin \left(\phi\right) = - \frac{1}{2}$

where $\alpha , \beta , \phi$ are my made up names for some yet-unknown angles.

Our knowledge of trigonometric functions tells us that:

• $\tan \left(\frac{\pi}{4}\right) = 1$
• $\cos \left(\frac{2 \pi}{3}\right) = - \frac{1}{2}$
• $\sin \left(- \frac{\pi}{6}\right) = - \frac{1}{2}$

Note that, for example, $\tan \left(\frac{5 \pi}{4}\right) = 1$ as well. However, we do say that $\alpha = \frac{\pi}{4}$ and NOT that $\alpha = \frac{5 \pi}{4}$ due to the range restrictions on the inverse trigonometric functions. If one input had multiple outputs, they would cease to be functions. The ranges are:

• range of $\alpha = {\tan}^{-} 1 \left(x\right)$: $\text{ } - \frac{\pi}{2} \le \alpha \le \frac{\pi}{2}$
• range of $\beta = {\cos}^{-} 1 \left(x\right)$: $\text{ } \textcolor{w h i t e}{-} 0 \le \beta \le \pi$
• range of $\phi = {\sin}^{-} 1 \left(x\right)$: $\text{ } - \frac{\pi}{2} \le \phi \le \frac{\pi}{2}$

Then:

• $\alpha = \frac{\pi}{4}$
• $\beta = \frac{2 \pi}{3}$
• $\phi = - \frac{\pi}{6}$

Thus:

${\tan}^{-} 1 \left(1\right) + {\cos}^{-} 1 \left(- \frac{1}{2}\right) + {\sin}^{-} 1 \left(- \frac{1}{2}\right)$

$= \alpha + \beta + \phi$

$= \frac{\pi}{4} + \frac{2 \pi}{3} - \frac{\pi}{6}$

$= \frac{3 \pi}{4}$

Sep 12, 2017

Undefined question.

#### Explanation:

We can't solve this question. The question just ask us to add up
3 specific quantities. To me, this problem is trigonometrical undefined. Reasons:
arctan (1) --> $\frac{\pi}{4} \mathmr{and} \frac{5 \pi}{4}$
$\arccos \left(- \frac{1}{2}\right)$ --> $\pm \frac{3 \pi}{4}$
$\arcsin \left(- \frac{1}{2}\right)$ --> $- \frac{\pi}{6} \mathmr{and} \frac{- 5 \pi}{6}$
With the same variable arc x, there are no indications about different
domains for periodic functions.
However, we can use calculator to add up the results by ignoring the trig aspect of the question.
45^@ + 120^@ - 30^@ = 135^@