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$$x_1=\arccos\left(\frac{3}{5}\right)+\arccos\left(\frac{2\sqrt{2}}{3}\right)$$

$$x_2=\arcsin\left(\frac{3}{5}\right)+\arcsin\left(\frac{2\sqrt{2}}{3}\right)$$ 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?

ryan1
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Umesh shankar
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  • The formulas you have for $x_1$ and $x_2$ are incorrect. – mrnovice Apr 30 '17 at 16:03
  • I would first try to show that $x_1$ and $x_2$ are positive, and combined with the fact that $x_1 + x_2 = \pi$ we have that $x_1, x_2 \in (0, \pi).$ We know $\cos x$ is a decreasing function in this interval, so showing that $\cos x_2 > \cos x_1$ will prove that $x_2 < x_1$. – Ovi Apr 30 '17 at 16:05
  • @mrnovice why my formulas are incorrect ? – Umesh shankar Apr 30 '17 at 16:19
  • @Umeshshankar I haven't calculated the correct ones, but I made an answer assuming your formulas were correct, and it led to $x_1=x_2$ which is wrong. Plugging your formulas directly into a calculator yields $x_1=x_2$ as well, so my reasoning was correct. – mrnovice Apr 30 '17 at 16:21
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    @Umeshshankar We have $\sin x_2 = \frac{3+8\sqrt{2}}{15}$, but $x = \arcsin (\sin x)$ only holds for $x$ in the range of $\arcsin$ (which, absent other specifications, defaults to $[-\pi/2, \pi/2]$ for the principal branch). – Daniel Fischer Apr 30 '17 at 16:25
  • See https://math.stackexchange.com/questions/672575/proof-for-the-formula-of-sum-of-arcsine-functions-arcsin-x-arcsin-y – lab bhattacharjee Apr 30 '17 at 17:38

4 Answers4

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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

$$x_2 > \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}.$$

Daniel Fischer
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1

(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 $x_2 > x_1$.

fleablood
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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.

John Lou
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  • It's legal if you formalize and prove it. Given an angle theta you have a basic right triangle with sides a, b, h and $a^2 + b^2 = h^2$ and $a/h = \cos \theta = \sin (90 - \theta)$ and $b/h = \sin \theta = \cos (90 - \theta)$ and $\tan \theta = b/a =$ slope of hypotenuse. And from that, everything you say follows. – fleablood Apr 30 '17 at 18:51
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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 $$x_2 \gt \frac{\pi}{2} \gt x_1$$

Umesh shankar
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