Need some help with arcus functions

**Vey** · January 2nd 2010, 07:16 AM

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.

Thanks in advance!

**skeeter** · January 2nd 2010, 07:24 AM

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.

Thanks in advance!

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

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

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

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

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

**Grandad** · January 2nd 2010, 07:41 AM

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.

Thanks in advance!

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 $0<x<\frac{\pi}{2}$ and $0 < y < \frac{\pi}{2}$ , it's clear that:

$x+y = \frac{3\pi}{4}$

Grandad

**Vey** · January 2nd 2010, 08:03 AM

Thanks guys, the second method proved useful!

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