Range of a function $|\sin x|+|\cos x|$ What is the range of function $Y=[|\sin x|+|\cos x|]$ where $[ ]$ denotes the greatest integer function.
And what is range of function $Y=|\sin x|+|\cos x|$
For the second one, 
I have tried squaring :
I got  $Y^{2}=1+|\sin 2x|$
and range will be between $1$ and $\sqrt{2}$
Please help me with this...!!! 
 A: I don't understand your description of the second solution of the second question, but your first solution of that question is correct, the range is $[1,\sqrt{2}]$.
The method for solving the first question is to follow definitions and think logically.
You know that $|\sin(x)|  \in [0,1]$ and that $|\cos(x)| \in [0,1]$. Therefore $$|\sin(x)| + |\cos(x)| \in [0,2]
$$ 
It follows that 
$$[|\sin(x)| + |\cos(x)|] \in \{0,1,2\}
$$
In other words, the range of your function is a subset of $\{0,1,2\}$. 
So now you have three further questions to pursue:


*

*Does the range contain $0$?

*Does the range contain $1$?

*Does the range contain $2$?


Question 3 is the easiest: $[|\sin(x)| + |\cos(x)|] = 2$ if and only if $|\sin(x)| = 1$ and $|\cos(x)| = 1$, and I'm sure that you can convince yourself that this is impossible, no matter value of $x$ you pick. Therefore $2$ is not in the range of the function.
Question 2 is also easy: I'm sure that you can find a value of $x$ such that one of $|\sin(x)|$, $|\cos(x)|$ equals $0$ and the other equals $1$, so their sum equals $1$. Therefore $1$ is in the range of the function.
Question 1 is the trickiest. $[|\sin(x)| + |\cos(x)|] = 0$ if and only if $[|\sin(x)| + |\cos(x)|] \in [0,1)$. Can you find a value of $x$ for which this is true? Or perhaps can you prove that this is false for all values of $x$? 
Perhaps that's enough for you to proceed.
A: By C-S $$|\sin{x}|+|\cos{x}|\leq\sqrt{(1^2+1^2)(\sin^2x+\cos^2x)}=\sqrt2.$$
The equality occurs for $x=y=45^{\circ},$ which says that we got a maximal value.
In another hand,
$$|\sin{x}|+|\cos{x}|=\sqrt{\left(|\sin{x}|+|\cos{x}|\right)^2}=\sqrt{1+|\sin2x|}\geq1.$$
The equality occurs for $x=90^{\circ},$ which says that we got a minimal value and since $Y$ is a continuous function, we got the answer:
$$[1,\sqrt2].$$
A: The most intuitive way to proceed might be to sketch the shape of the level sets
$$ |x|+|y| = 0 \\
|x|+|y| = 1  \\
|x|+|y| = 1.5\\
|x|+|y| = 2 $$
and so forth. Which of them have points in common with the unit circle?
A: For any real $x$ there exists $x'\in [0,\pi/2]$ such that $|\sin x|=|\sin x'|$ and $|\cos x|=|\cos x'|.$ Since $\sin x'\ge 0$ and $\cos x'\ge 0$ We have  $$(|\sin x|+|\cos x|)^2=(|\sin x'|+|\cos x'|)^2=$$ $$=(\sin x' +\cos x')^2=$$ $$=(\sin^2 x'+\cos^2 x')+2\sin x' \cos x'=$$ $$=1+\sin 2x'.$$ As $x$ ranges over $\Bbb R,\;$ $\, x'$  takes all, and only, values in $[0,\pi/2],$ so $2x'$ takes all, and only, values in $[0,\pi],$ so $\sqrt {1+\sin 2x'}\,$ takes all, and only, values in $[1,\sqrt 2\,]$.
Now $1<\sqrt 2\,<2,$ so for all real $x$, the greatest integer not exceeding $|\sin x|+|\cos x|$ is $1.$ 
