Hi there, I want to demonstrate that
$\displaystyle 1\leq{|\cos x|+|\sin x|\leq{\sqrt[]{2}}}$
How should I proceed?
Bye and thanks.
Hmm I wonder if I'm overcomplicating things.
$\displaystyle |\cos x| + |\sin x| = |\cos x| + \sqrt{1-\cos^2x} = |\cos x| + \sqrt{1-|\cos x|^2}$
Recall also that $\displaystyle 0\le |\cos x|\le 1$
You can substitute $\displaystyle y = |\cos x|$ and show that the maximum on this interval is approximately $\displaystyle \sqrt{2}$ 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.
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.
$\displaystyle \dfrac{d(cos(x)+sin(x))}{dx} = -sin(x) + cos(x)= cos(x) - sin(x)$
Then, solve for zero to find the maximum and minimum values.
However, we'll have to convert this into a simpler form.
$\displaystyle cos(x) - sin(x) = Rcos(x + \alpha)$
Where R and alpha are constants.
Expanding on the right side gives:
$\displaystyle R[cos(x)cos(\alpha) - sin(x)sin(\alpha)] = Rcos(x)cos(\alpha) - Rsin(x)sin(\alpha)$
Equate:
$\displaystyle cos(x) = Rcos(x)cos(\alpha)$
$\displaystyle -sin(x) = - Rsin(x)sin(\alpha)$
Giving:
$\displaystyle 1= Rcos(\alpha)$ ---1
$\displaystyle 1 = Rsin(\alpha)$ ---2
Take 2/1:
$\displaystyle 1 = tan(\alpha)$
$\displaystyle \alpha = 45^o$
Take 1^2 + 2^2;
$\displaystyle 1^2 + 1^2 = R^2cos^2(\alpha) + R^2sin^2(\alpha)$
$\displaystyle 2 = R^2$
$\displaystyle R = \sqrt{2}$
So, $\displaystyle cos(x) - sin(x) = \sqrt{2}cos(x + 45^o)$
Set to zero now:
$\displaystyle \sqrt{2}cos(x + 45^o) = 0$
$\displaystyle cos(x + 45^o) = 0$
The maximum value of this solution is when $\displaystyle x + 45^o = 90^o$
$\displaystyle x = 45^o$