# What is wrong with this one?

• Aug 28th 2006, 07:10 AM
a4swe
What is wrong with this one?
$\cos(4v)=-\frac{1}{\sqrt{2}} \Leftrightarrow$
$-\cos(4v)=\frac{1}{\sqrt{2}} \Leftrightarrow$
$cos(4v+\pi)=\frac{1}{\sqrt{2}} \Leftrightarrow$
$4v+\pi=+-\frac{\pi}{4}+n2\pi$
I:
$4v+\pi=\frac{\pi}{4}+n2\pi \Leftrightarrow$
$v=-\frac{3\pi}{16}+\frac{n\pi}{2}$
II:
$4v+\pi=-\frac{\pi}{4}+n2\pi \Leftrightarrow$
$v=-\frac{5\pi}{16}+\frac{n\pi}{2}$

Where I and II are my two solutions.
The answer given in my book are:
$+-\frac{3\pi}{16}+\frac{n\pi}{2}$

In this post I use a plus and a minus following eachother (like this: +-) to indicate that the following term can be either negative or positive.
• Aug 28th 2006, 07:19 AM
CaptainBlack
Quote:

Originally Posted by a4swe
$\cos(4v)=-\frac{1}{\sqrt{2}} \Leftrightarrow$
$-\cos(4v)=\frac{1}{\sqrt{2}} \Leftrightarrow$
$cos(4v+\pi)=\frac{1}{\sqrt{2}} \Leftrightarrow$
$4v+\pi=+-\frac{\pi}{4}+n2\pi$
I:
$4v+\pi=\frac{\pi}{4}+n2\pi \Leftrightarrow$
$v=-\frac{3\pi}{16}+\frac{n\pi}{2}$
II:
$4v+\pi=-\frac{\pi}{4}+n2\pi \Leftrightarrow$
$v=-\frac{5\pi}{16}+\frac{n\pi}{2}$

Where I and II are my two solutions.
The answer given in my book are:
$+-\frac{3\pi}{16}+\frac{n\pi}{2}$

In this post I use a plus and a minus following eachother (like this: +-) to indicate that the following term can be either negative or positive.

There is nothing wrong with your solution, its just another way of expressing
the solution in the book.

Note:

$
-\frac{5 \pi}{16}=\frac{3 \pi}{16} -\frac{\pi}{2}
$

RonL
• Aug 28th 2006, 07:34 AM
a4swe
Yes of course.
I thought I had missed something fundamental, that was not the case.
Thank you.
• Aug 28th 2006, 02:00 PM
Soroban
Hello, a4swe!

Just a question: why are you moving that minus-sign around?

We have: . $\cos(4v)\:=\:-\frac{1}{\sqrt{2}}\quad\Rightarrow\quad 4v \:=$ $\:\pm\frac{3\pi}{4} + 2\pi n \quad\Rightarrow\quad v \:=\:\pm\frac{3\pi}{16} + \frac{\pi}{2}n$

• Aug 28th 2006, 08:47 PM
a4swe
To show (for my self maninly) that what I do is indeed right.
I included it here just because I thought it was possible that it was somewhere there I did something wrong.
If someone would say, hey there is your fault!
Then I colud more easily trace it if I did it like I did it rather than how you did it, don't you think?

Conclusion: I do it because I am a little rusty in trigonometry.
• Aug 29th 2006, 04:48 AM
Quick
Quote:

Originally Posted by Soroban

We have: . $\cos(4v)\:=\:-\frac{1}{\sqrt{2}}\quad\Rightarrow\quad 4v \:=$ $\:\pm\frac{3\pi}{4} + 2\pi n \quad\Rightarrow\quad v \:=\:\pm\frac{3\pi}{16} + \frac{\pi}{2}n$

How do you get rid of the cos sign? is there something you multiply by?
• Aug 29th 2006, 05:48 AM
ThePerfectHacker
Quote:

Originally Posted by Quick
How do you get rid of the cos sign? is there something you multiply by?

Yes, the inverse cosine function.
Like here,
$x^2=2$ how do you remove the square.
You use the inverse function $\sqrt{x}$ which removes it.
---
However, over here that value is one of the trigonometric values that you need to have memorized.
Since,
$\cos \pi/4=\sqrt{2}/2=1/\sqrt{2}$
So when it is negative, it is either in the 2nd or 3rd quadrant so,
$\pi-\pi/4$
$\pi+\pi/4$