Which is greater $x_1$ or $x_2$? $$x_1=\arccos\left(\frac{3}{5}\right)+\arccos\left(\frac{2\sqrt{2}}{3}\right)$$
$$x_2=\arcsin\left(\frac{3}{5}\right)+\arcsin\left(\frac{2\sqrt{2}}{3}\right)$$ We have to find which is greater among $x_1$ and $x_2$
If we add both we get $$x_1+x_2=\pi$$ If we use formulas we get
$$x_1=\arccos\left(\frac{6\sqrt{2}-4}{15}\right)$$ and
$$x_2=\arcsin\left(\frac{3+8\sqrt{2}}{15}\right)$$
but how to compare now?
 A: We have
$$\frac{3}{5} > \frac{1}{2} \implies \arcsin \frac{3}{5} > \arcsin \frac{1}{2} = \frac{\pi}{6}$$
and
$$\frac{2\sqrt{2}}{3} > \frac{\sqrt{3}}{2} \implies \arcsin \frac{2\sqrt{2}}{3} > \arcsin \frac{\sqrt{3}}{2} = \frac{\pi}{3},$$
hence
$$x_2 > \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}.$$
A: (I'm going to do this in degrees rather than radians as this is very intuitive about right triangles).
Obviously if $\theta = \arccos \frac ah$ then $\theta$ represents an angle of a right triangle with adjacent side $a$ and hypotenuse $h$ (and opposite side $o = \sqrt {h^2 - a^2}$).  So $\arcsin \frac ah = 90 - \theta$, the other angle of the same triangle where the side $a$ is now opposite rather than adjacent side.
[These are the triangles with sides $3,4,5$ and with sides $1, 2\sqrt{2}, 3$.]
So if $x <:> \frac{\sqrt{2}}{2}$ then $\arccos x = 45 \pm \phi$ for some positive angle $\phi$ and $\arcsin x = 45 \mp \phi$.  (Draw a picture.  It is obvious.)
So $\arccos \frac 35 + \arccos \frac {2\sqrt{2}}3 = 45 - \phi + 45 + \theta$
And $\arcsin \frac 35 + \arcsin \frac {2\sqrt{2}}3 = 45 + \phi + 45 -\theta$
So to solve this problem is simple and matter of figuring out which is larger $\phi$ or $\theta$.  I.E. which triangle has the steeper slope the the $3$, $4$, $5$ triangle, or the $1, 2\sqrt{2}, 3$ triangle.  Then answer is obviously the $1, 2\sqrt{2}, 3$ triangle 
and so $\theta > \phi$ and $\arccos \frac 35 + \arccos \frac {2\sqrt{2}}3 = 45 - \phi + 45 + \theta > \arcsin \frac 35 + \arcsin \frac {2\sqrt{2}}3 = 45 + \phi + 45 -\theta$
===== or even more straightforward =====
$\sin x = \sqrt {1 - \cos^2 x} =y$
$\arcsin y = \arccos (\sqrt{1 - y^2})$
So $x_1 = \arccos \frac 35 + \arccos \frac{2\sqrt2}{3}= \arcsin \sqrt{ 1- \frac 35^2} + \arcsin \sqrt{1- \frac {2\sqrt{2}}3^2} = \arcsin \frac 45 + \arcsin \frac 13$
$x_2 - x_1 = (\arcsin \frac  {2\sqrt 2}3 - \arcsin \frac 13)+(\arcsin \frac {3}{5} - \arcsin \frac 45)$
Now as $0 < 1/3 < 3/5  < 4/5 < \frac {2\sqrt 2} 2 < 1$ so $0 < \arcsin 1/3 <\arcsin 3/5  < \arcsin4/5 <\arcsin \frac {2\sqrt 2} 2 < 1$
So $x_2 - x_1 = (\arcsin \frac  {2\sqrt 2}3 - \arcsin \frac 13)+(\arcsin \frac {3}{5} - \arcsin \frac 45) > 0$.
So $x_2 > x_1$.
A: I'm not sure if this is actual math, or whether the intuition just pans out this time, so someone with an algebraic approach is probably better. 
Establish that $x_1 = a_1 + b_1$ so that I don't have to type that much.
Let's look at the first two terms in each:
Comparing $a_1$ to $b_1$, it is clear that $a_1$ is greater. For this, turn to geometry, and it's clear that based on the slope of the line which one is bigger. $a_1$ slope is $\frac{4}{3}$, whereas $b_1$ slope is $\frac{3}{4}$
Comparing $a_2$ to $b_2$, it is clear that $b_2$ is greater, whereas this is by a far larger factor than $a_1$ is greater than $b_1$. 
$a_2$ slope is $\frac{\sqrt{2}}{4}$ and $b_2$ slope is $2\sqrt{2}$.
If anyone can confirm that this method is "legal" that'd be great.
A: Ok fine similarly we have $$x_1=arccos\left(\frac{3}{5}\right)+arccos\left(\frac{2\sqrt{2}}{3}\right)=arcsin\left(\frac{4}{5}\right)+arcsin\left(\frac{1}{3}\right)$$
Now $$\frac{4}{5} \lt \frac{\sqrt{3}}{2}$$
hence
$$arcsin\left(\frac{4}{5}\right) \lt \frac{\pi}{3}$$  and
$$\frac{1}{3} \lt \frac{1}{2}$$ so
$$arcsin\left(\frac{1}{3}\right) \lt \frac{\pi}{6}$$
hence
$$x_1 \lt \frac{\pi}{2}$$
Finally  $$x_2 \gt \frac{\pi}{2} \gt x_1$$
