# What does sin(arccos(3))+csc(arc cot(4)) equal?

Mar 16, 2016

$\sin \left(\arccos \left(3\right)\right) + \csc \left(a r c \cot \left(4\right)\right) = 2 \sqrt{2} i + \sqrt{17}$

#### Explanation:

If $\sin$, $\cos$, etc are being treated as Real valued functions of Real values, then $\arccos \left(3\right)$ is not defined, since $3$ lies outside the range of $\cos \left(x\right)$ which is $\left[- 1 , 1\right]$.

However, inspired by Euler's formula ${e}^{i x} = \cos x + i \sin x$, it is possible to define $\sin$, $\cos$ etc for Complex values as follows:

$\sin z = \frac{{e}^{i z} - {e}^{- i z}}{2 i}$

$\cos z = \frac{{e}^{i z} + {e}^{- i z}}{2}$

With these definitions, note that ${\sin}^{2} z + {\cos}^{2} z = 1$ for any $z \in \mathbb{C}$.

Suppose $\cos z = a$ for some $a \in \mathbb{C}$. What is $z$?

Let $t = {e}^{i z}$. Then ${e}^{- i z} = \frac{1}{t}$ and we have:

$\frac{t + \frac{1}{t}}{2} = \cos z = a$

Hence:

$t + \frac{1}{t} = 2 a$

Hence:

${t}^{2} - 2 a t + 1 = 0$

Solving using the quadratic formula, we get:

${e}^{i z} = t = \frac{2 a \pm \sqrt{4 {a}^{2} - 4}}{2} = a \pm \sqrt{{a}^{2} - 1}$

So:

$i z = \ln \left(a \pm \sqrt{{a}^{2} - 1}\right)$

Hence:

$z = - i \ln \left(a \pm \sqrt{{a}^{2} - 1}\right) = \pm i \ln \left(a + \sqrt{{a}^{2} - 1}\right)$

It is reasonable that the definition of $\arccos \left(x\right)$ for $x > 1$ should choose the positive sign for this expression.

Hence:

$\arccos \left(3\right) = i \ln \left(3 + \sqrt{8}\right) = i \ln \left(3 + 2 \sqrt{2}\right)$

Then:

$\sin \left(\arccos \left(3\right)\right)$

$= \frac{{e}^{- \ln \left(3 + 2 \sqrt{2}\right)} - {e}^{\ln \left(3 + 2 \sqrt{2}\right)}}{2 i}$

$= \frac{1}{2 i} \left(\frac{1}{3 + 2 \sqrt{2}} - \left(3 + 2 \sqrt{2}\right)\right)$

$= \frac{i}{2} \left(\left(3 + 2 \sqrt{2}\right) - \left(3 - 2 \sqrt{2}\right)\right)$

$= \textcolor{b l u e}{2 \sqrt{2} i}$

$\textcolor{w h i t e}{}$
On the other hand, $\csc \left(a r c \cot \left(4\right)\right)$ is all about Real values and somewhat easier to construct.

Consider a right angled triangle with sides $1$, $4$ and $\sqrt{17}$

Let $\theta$ be the smallest angle, i.e. the one between the hypotenuse and the longer leg.

Then $\cot \left(\theta\right) = \text{adjacent"/"opposite} = \frac{4}{1} = 4$

and $\csc \left(\theta\right) = \text{hypotenuse"/"opposite} = \frac{\sqrt{17}}{1} = \sqrt{17}$

So:

$\csc \left(a r c \cot \left(4\right)\right) = \textcolor{b l u e}{\sqrt{17}}$