Solve $\arccos(11/14)+\arcsin(-1/7)$ with complex numbers using:
$$\arg (z_1\cdot z_2)= \arg(z_1) + \arg(z_2)$$
$$\arg (z_1\cdot z_2)= \arccos(11/14) + \arcsin(1/7)$$
but I can't find the arg for both terms. I've created reference angles but get the wrong answer.
I made $\arccos(11/14)$ into $\arctan(-1/\sqrt{48})$
and $\arcsin(1/7)$ into $\arctan(11/\sqrt{75})$, which would give $$(\sqrt{48}-i)(\sqrt{75}+11i) = 71+39\sqrt{3}i,$$ but it's an angle of $46.43$. The answer is: $\pi/6$.
Best Answer
Let $\phi = \arccos\frac {11}{14}$. Then $\cos^2 \phi + \sin^2 \phi =1$ and $\sin^2 \phi = 1-\frac {121}{ 196} = \frac {75}{196}$ and $\sin \phi = \frac {5\sqrt 3}{14}$ (don't forget $\arccos$ is always between $0, \pi$ so $\sin (\arccos x)$ is always non-negative).
So $\tan \phi = \frac {5\sqrt 3}{11}$. And $\phi = \arctan \frac {5\sqrt 3}{11}$
Let $\psi = \arcsin \frac 17$ and do the same. $\cos^2 \psi = 1-\frac 1{49} = \frac {48}{49}$ and $\cos \phi = \frac {4\sqrt 3}{7}$ (again the range of $\arcsin$ is between $-\frac \pi 2, \frac \pi 2$ so $\cos(\arcsin x))$ is non-negative)
So $\psi = \arctan \frac 1{4\sqrt 3}$.
Trick is that for any non-zero $a+bi$ we can convert to polar coordinates by $a+bi= r(\cos \phi + i\sin \phi)=re^{i\phi}$ where $r = \sqrt{a^2 + b^2}$ and $\phi = \arg(a+bi)=\arctan \frac ba$.
So $\phi = \arg (11+ 5\sqrt 3i)$ and $\psi = \arg(4\sqrt 3+ i)$.
And so $\phi -\psi = \arg (\frac {11+5\sqrt 3i}{4\sqrt 3+i})$
And $\frac {11+5\sqrt 3i}{4\sqrt 3+i}=\frac {(11+5\sqrt 3i)(4\sqrt 3 -i)}{(4\sqrt 3+i)(4\sqrt 3-i)}=$
$\frac {(44\sqrt 3+5\sqrt 3)+ (20\cdot 3-11)i}{16\cdot 3 + 1}=\frac {49\sqrt 3 + 49i}{49} = \sqrt 3 + i$
ANd $\arg (\sqrt 3+i) = \arctan \frac 1{\sqrt 3}= \frac \pi 6$.