# Thread: What is wrong with this one?

1. ## 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.

2. 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

3. Yes of course.
I thought I had missed something fundamental, that was not the case.
Thank you.

4. Hello, a4swe!

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

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

5. 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.

6. 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?

7. 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,
$\pi-\pi/4$
$\pi+\pi/4$