A little question about absolute value

[harieche]

Define the absolute value of sine

|sin x| = sinx , if x >=0
-sinx , if x<0


Since sinx>=0, for x ∈ [2k[FONT=&quot]π[/FONT],(2k+1)[FONT=&quot]π[/FONT]] , k∈?]
sinx<0, for x ∈ [(2k-1)[FONT=&quot]π[/FONT],2k[FONT=&quot]π[/FONT]] , k∈?]

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Now the question, if i need to define the absolute value of cosine,

Can I just say the first derivative of sine is cosine?

 

[skaa]

Yes, it is correct.

 

[Ishuda]

Define the absolute value of sine

|sin x| = sinx , if x >=0
-sinx , if x<0


Since sinx>=0, for x ∈ [2kπ,(2k+1)π] , k∈?]
sinx<0, for x ∈ [(2k-1)π,2kπ] , k∈?]

__________________________________________________________________
Now the question, if i need to define the absolute value of cosine,

Can I just say the first derivative of sine is cosine?

First, the initial statement is not true although you correct it later. That is x=\(\displaystyle 3\pi/2\, \gt\, 0\) but sin(\(\displaystyle 3\pi/2)=-1\, \lt\, 0\).

Next, you need to be careful about your intervals, i.e. open, closed, or 'half open/closed'. For example, the following is NOT true
sinx<0, for x ∈ [(2k-1)π,2kπ] , k∈?]
Had you made that sin(x)\(\displaystyle \le\)0 or an open interval
sinx<0, for x ∈ ((2k-1)π,2kπ) , k∈?]
you would have a true statement.

And finally, no. Noting that the cosine is the derivative of the sine is not sufficient to give the same type of statement for the cosine.

 

[Ishuda]

Define the absolute value of sine

|sin x| = sinx , if x >=0
-sinx , if x<0


Since sinx>=0, for x ∈ [2kπ,(2k+1)π] , k∈?]
sinx<0, for x ∈ [(2k-1)π,2kπ] , k∈?]

__________________________________________________________________
Now the question, if i need to define the absolute value of cosine,

Can I just say the first derivative of sine is cosine?

BTW: What may be wanted is

\(\displaystyle |sin(x)|\, =\,
\left\{
\begin{array}{ll}
sin(x) & \quad sin(x) \ge 0 \\
-sin(x) & \quad sin(x)\, \le\ 0
\end{array}
\right.\)