# Thread: Need some help with arcus functions

1. ## Need some help with arcus functions

Hey, I'm not sure wether this is univerity or pre-university stuff cuz I'm in university now, but I did it in high school too, so I'm sorry if this is at the wrong place.

My problem is this:
I need to calculate arcsin( 3/sqrt(10) ) + arccos( 1/sqrt(5) ) and the answer may not contain any cyklometric functions.

2. Originally Posted by Vey
Hey, I'm not sure wether this is univerity or pre-university stuff cuz I'm in university now, but I did it in high school too, so I'm sorry if this is at the wrong place.

My problem is this:
I need to calculate arcsin( 3/sqrt(10) ) + arccos( 1/sqrt(5) ) and the answer may not contain any cyklometric functions.

let $a = \arcsin\left(\frac{3}{\sqrt{10}}\right)$

$b = \arccos\left(\frac{1}{\sqrt{5}}\right)$

$\sin(a+b) = \sin{a}\cos{b} + \cos{a}\sin{b}$

$\sin(a+b) = \frac{3}{\sqrt{10}} \cdot \frac{1}{\sqrt{5}} + \frac{1}{\sqrt{10}} \cdot \frac{2}{\sqrt{5}} = \frac{1}{\sqrt{2}}
$

$a+b = \frac{3\pi}{4}$

3. Hello Vey

Welcome to Math Help Forum!
Originally Posted by Vey
Hey, I'm not sure wether this is univerity or pre-university stuff cuz I'm in university now, but I did it in high school too, so I'm sorry if this is at the wrong place.

My problem is this:
I need to calculate arcsin( 3/sqrt(10) ) + arccos( 1/sqrt(5) ) and the answer may not contain any cyklometric functions.

The best thing to do with inverse trig functions is to get rid of them as soon as you can! So:

Let:
$x = \arcsin \left(\frac{3}{\sqrt{10}}\right)$
and
$y = \arccos \left(\frac{1}{\sqrt5}\right)$
Then get rid of the inverse functions:
$\sin x = \frac{3}{\sqrt{10}}$ and $\cos y = \frac{1}{\sqrt5}$
So, using Pythagoras' Theorem:
$\cos x = \frac{1}{\sqrt{10}}$ and $\sin y = \frac{2}{\sqrt5}$
Now we need $x + y$, so, using $\cos(x+y) = \cos x \cos y - \sin x \sin y$, we get:
$\cos(x+y) = \frac{1}{\sqrt{10}}\frac{1}{\sqrt{5}}- \frac{3}{\sqrt{10}}\frac{2}{\sqrt{5}}$
$=\frac{1}{5\sqrt2}-\frac{6}{5\sqrt2}$

$= -\frac{1}{\sqrt2}$
Of course, there's an infinite number of angles whose cosine is $= -\frac{1}{\sqrt2}$, but, noting that $0 and $0 < y < \frac{\pi}{2}$, it's clear that:
$x+y = \frac{3\pi}{4}$