# Math Help - Trigonometry again derivative

1. ## Trigonometry again derivative

I have this:
$y=3cosx-cos3x$

I need to solve it.

I got this:
$y'=3sin(3x)-3sin(x)$

I simplified it to get:
$y'=3(sin(3x)-sin(x))$

I equate it to 0 and I get:
$sin(3x)-sin(x)=0$

How do I solve this?

2. Hello,

You don't have to derivate, since it'll only (the things that could interest us) give you variations.

Transform cos(3x) into a form with cos(x) and sin(x).

cos(3x)=cos(2x+x)

So use the following formulas :

$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$

$\cos(2x)=2 \cos^2(x)-1$

$\sin(2x)=2 \sin(x)\cos(x)$

Try to get cos(3x) first...

3. Well I need to find all the maximum and minimum points so that is why I derived it.

I guess I could do what you did and then derive it. Let me try.

4. I need to solve it.
This was the only thing I knew you had to do

Well, for solving the derivative, do the same :

$\sin(3x)=\sin(2x+x)$

Formulas you need :

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

$\sin(2x)=2\sin(x)\cos(x)$

$\cos(2x)=$ hmmm, you'll see which one of the three below would be useful :

$\begin{array}{ccc} \cos(2x) & = & \cos^2(x)-\sin^2(x) \\ & = & 2 \cos^2(x)-1 \\ & = & 1-2\sin^2(x) \end{array}$

5. How do I solve:
$cos^2(x)=1/4$

Nevermind I am dumb I got it...

6. $\sin(3x)=\sin(2x)\cos(x)+\cos(2x)\sin(x)$

$\sin(2x)=2\sin(x) \cos(x)$

---> $\sin(3x)=2\sin(x) \cos^2(x)+\cos(2x)\sin(x)$

Replace $\cos^2(x)=1-sin^2(x)$ and $\cos(2x)=1-2\sin^2(x)$

7. Using all that I "simplified" it to:
$3sin(x)-4sin^3(x)$

Now I am stuck again...

8. This is correct

Now you want to solve for y'=0

So $3\sin(x)-4\sin^3(x)=0 \Longleftrightarrow \sin(x)(3-4\sin^2(x))=0$

Can you solve it ?

9. I get:
$sin(x)=0$
$sin^2(x)=3/4$

I solved the first one but how do I solve the second one.
I know that I need to do square root so that I get
$sin(x)=sqrt(3/4)$
But thats not a number that I can express... Or can I? How?

10. $\sin^{2} x = \frac{3}{4}$
$\sin x = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{\sqrt{4}} = \pm \frac{\sqrt{3}}{2}$