# Thread: Demonstration involving sine and cosine

1. ## Demonstration involving sine and cosine

Hi there, I want to demonstrate that

$1\leq{|\cos x|+|\sin x|\leq{\sqrt[]{2}}}$

How should I proceed?

Bye and thanks.

2. Originally Posted by Ulysses
Hi there, I want to demonstrate that

$1\leq{|\cos x|+|\sin x|\leq{\sqrt[]{2}}}$

How should I proceed?

Bye and thanks.
You can use a unit circle and/or the identity $\cos^2x+\sin^2x=1$.

3. How do I fit that trig identity intro my inequality?

Thanks.

4. Originally Posted by Ulysses
How do I fit that trig identity intro my inequality?

Thanks.
Hmm I wonder if I'm overcomplicating things.

$|\cos x| + |\sin x| = |\cos x| + \sqrt{1-\cos^2x} = |\cos x| + \sqrt{1-|\cos x|^2}$

Recall also that $0\le |\cos x|\le 1$

You can substitute $y = |\cos x|$ and show that the maximum on this interval is approximately $\sqrt{2}$ using a graphing calculator (I'm assuming you haven't learned calculus yet, which you could use to find the maximum exactly). Since it says "demonstrate" and not "prove" this should be enough.

5. I've learn some calculus, I'm actually trying to use this for a limit involving two variables :P I thought of using max and min, but I didn't cause I thought it would appear the sine and the cosine again driving me nowhere.

6. For the minimum, it's easy to find it. Put x = 0, then cos(x) = 1, and sin(x) = 0. 1 + 0 = 1.

For the maximum, you might not be familiar with what value of x to make cos(x) + sin(x) maximum.

I would find the derivative of the function first.

$\dfrac{d(cos(x)+sin(x))}{dx} = -sin(x) + cos(x)= cos(x) - sin(x)$

Then, solve for zero to find the maximum and minimum values.

However, we'll have to convert this into a simpler form.

$cos(x) - sin(x) = Rcos(x + \alpha)$

Where R and alpha are constants.

Expanding on the right side gives:

$R[cos(x)cos(\alpha) - sin(x)sin(\alpha)] = Rcos(x)cos(\alpha) - Rsin(x)sin(\alpha)$

Equate:

$cos(x) = Rcos(x)cos(\alpha)$
$-sin(x) = - Rsin(x)sin(\alpha)$

Giving:

$1= Rcos(\alpha)$ ---1
$1 = Rsin(\alpha)$ ---2

Take 2/1:
$1 = tan(\alpha)$

$\alpha = 45^o$

Take 1^2 + 2^2;

$1^2 + 1^2 = R^2cos^2(\alpha) + R^2sin^2(\alpha)$

$2 = R^2$

$R = \sqrt{2}$

So, $cos(x) - sin(x) = \sqrt{2}cos(x + 45^o)$

Set to zero now:

$\sqrt{2}cos(x + 45^o) = 0$

$cos(x + 45^o) = 0$

The maximum value of this solution is when $x + 45^o = 90^o$

$x = 45^o$