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The function is: $f(x)= \sin(x)-\cos(x)$ over the interval $[-\pi, \pi]$.

To evaluate the increasing / decreasing portion of the function I look at where the first derivative is $>0$. $f'(x)=\cos(x) + \sin(x)$. I want to know where this is $>0$ without a graphing calculator.

I'm stuck here. I know that the derivative would be greater than zero in Quadrant 1. I know that the derivative would be $<0$ in Q3. I can intuitively think of where in Q2 and Q3 that the derivative would be $>0$, but I don't know how to show this mathematically. I also don't think I totally understand the significance of the interval, since the interval is essentially over the entire $0-2\pi$ range but stated in this unusual manner.

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    $\begingroup$ Have you tried $f^\prime(x)=0$? Because $f^\prime$ is continuous, it must have a 0 between any positive and negative values. That means that you can the interval between zeros by picking any point in them and finding if it's positive or negative. $\endgroup$ – memerson Oct 23 '18 at 20:16
  • $\begingroup$ I think part of your second sentence is missing. Can you complete. $\endgroup$ – user163862 Oct 23 '18 at 20:25
  • $\begingroup$ Let's take an example. Say $g(x)$ is continuous, $g(x)=0$ at $x=1,5$ and the sign of $g$ on the interval $[0,10]$. Let's look at $[0,1)$. Let $x,y\in[0,1)$ with $x<y$. Suppose that one of them is positive and the other is negative. By the intermediate value theorem, there is some $z\in [x,y]$ where $g(z)=0$. But that would give us another $0$, so $x$ and $y$ must have the same sign. As such, for every $x\in[0,1)$, $g(x)$ has the same sign, so if we know that $g(.5)$ is postive, so is the rest of $[0,1)$. The same applies to $(1,5)$ and $(5,10]. Does that make sense? $\endgroup$ – memerson Oct 23 '18 at 20:33
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$$\cos x+\sin x=\sqrt{2}\sin(x+\pi/4)$$

See the trigonometric identity here: https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Linear_combinations

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  • $\begingroup$ This is extremely useful. However, when I set this identity $>0$ I end up with $\endgroup$ – user163862 Oct 23 '18 at 20:43
  • $\begingroup$ When I solve for the inverse $sin$ of $0$, if I use their boundaries I miss the answer of $-\pi/4$. If I used the typical boundaries of $0$ to $2 \pi$ I get the book answers. Thank you! $\endgroup$ – user163862 Oct 23 '18 at 20:53
  • $\begingroup$ I see this now. $\endgroup$ – user163862 Oct 23 '18 at 21:10

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