What does -3sin(arccos(2))-cos(arc cos(3)) equal?

Dec 16, 2015

Problem insolvable

Explanation:

There are no arcs that their cosine are equal to 2 and 3.

From an analytic point of view, the $\arccos$ function is only defined on $\left[- 1 , 1\right]$ so $\arccos \left(2\right)$ & $\arccos \left(3\right)$ don't exist.

Dec 20, 2015

For Real $\cos$ and $\sin$ this has no solutions, but as functions of Complex numbers we find:

$- 3 \sin \left(\arccos \left(2\right)\right) - \cos \left(\arccos \left(3\right)\right) = - 3 \sqrt{3} i - 3$

Explanation:

As Real valued functions of Real values of $x$, the functions $\cos \left(x\right)$ and $\sin \left(x\right)$ only take values in the range $\left[- 1 , 1\right]$, so $\arccos \left(2\right)$ and $\arccos \left(3\right)$ are undefined.

However, it is possible to extend the definition of these functions to Complex functions $\cos \left(z\right)$ and $\sin \left(z\right)$ as follows:

Starting with:

${e}^{i x} = \cos x + i \sin x$

$\cos \left(- x\right) = \cos \left(x\right)$

$\sin \left(- x\right) = - \sin \left(x\right)$

we can deduce:

$\cos \left(x\right) = \frac{{e}^{i x} + {e}^{- i x}}{2}$

$\sin \left(x\right) = \frac{{e}^{i x} - {e}^{- i x}}{2 i}$

Hence we can define:

$\cos \left(z\right) = \frac{{e}^{i z} + {e}^{- i z}}{2}$

$\sin \left(z\right) = \frac{{e}^{i z} - {e}^{- i z}}{2 i}$

for any Complex number $z$.

It is possible to find multiple values of $z$ that satisfy $\cos \left(z\right) = 2$ or $\cos \left(z\right) = 3$, so there could be some choices to be made to define the principal value $\arccos \left(2\right)$ or $\arccos \left(3\right)$.

To find suitable candidates, solve $\frac{{e}^{i z} + {e}^{- i z}}{2} = 2$, etc.

However, note that the identity ${\cos}^{2} z + {\sin}^{2} z = 1$ holds for any Complex number $z$, so we can deduce:

$\sin \left(\arccos \left(2\right)\right) = \pm \sqrt{1 - {2}^{2}} = \pm \sqrt{- 3} = \pm \sqrt{3} i$

I hope that it's possible to define the principal value in such a way that $\sin \left(\arccos \left(2\right)\right) = \sqrt{3} i$ rather than $- \sqrt{3} i$.

In any case, $\cos \left(\arccos \left(3\right)\right) = 3$ by definition.

Putting this all together, we find:

$- 3 \sin \left(\arccos \left(2\right)\right) - \cos \left(\arccos \left(3\right)\right) = - 3 \sqrt{3} i - 3$