Sketching the graph of a trig function How would I sketch the following function.
$f(x)=\cos(x)+\sin(x)$ on $[0,2\pi]$
for my first derivative I got
$f'(x)=-\sin(x)+\cos(x)=0$ 
but How would I find the critical points I mean I know $\cos^{-1}=0$ can be 90 or 270 degrees but I am not sure if this is correct.
The second derivative is 
$f''(x)=-\cos(x)-\sin(x)$
But how would I find the inflection point If I set the function zero.
 A: There's a much easier way of sketching such a curve. 
My claim is that we can write $\cos x + \sin x$ as $\sqrt{2}\cos\left(x-\frac{\pi}{4}\right)$. 
The graph is a shifted cosine curve, moving between $-\sqrt{2}$ and $\sqrt{2}$. The shift is to the right by $\frac{\pi}{4}$.
The aim is to write it in the form $R\cos(x - \alpha)$ for suitable $R$ and $\alpha$. First apply the formula:
$$R\cos(x-\alpha) \equiv R\cos x \cos \alpha + R \sin x \sin \alpha$$
We want $\cos x + \sin x \equiv R\cos(x-\alpha)$ and so we need to find an $R$ and an $\alpha$ for which $R\cos\alpha = 1$ and $R\sin\alpha  = 1$. We can find $R$ and $\alpha$ by using Pythagoras' Theorem and some trigonometry. First, 
divide both of the terms:
$$\frac{R\sin\alpha}{R\cos\alpha} \equiv \tan \alpha = \frac{1}{1} = 1$$
It follows tha $\alpha = \frac{\pi}{4}$. To find $R$, we can square both terms. First notice that
$$(R\cos\alpha)^2 + (R\sin\alpha)^2 = 1^2 + 1^2$$
Expanding and then using the identity $\cos^2\alpha + \sin^2\alpha \equiv 1$ gives $R^2 = 2$, hence $R = \sqrt{2}$. We have:
$$\cos x + \sin x \equiv \sqrt{2}\cos\left(x-\frac{\pi}{4}\right)$$
A: For any $x\in [0,2\pi]$ you get $f'(x)=\cos (x) -\sin (x)$ and $f''(x)=-\sin (x)-\cos (x)$. Therefore 
$$\begin{align} 
f'(x)=0&\iff \cos (x) -\sin(x)=0\\
 &\iff \cos (x) =\sin (x)\\
 &\iff x\in \{\frac{\pi}{4}, \frac{5\pi}{4}\}
\end{align}$$
Which gives you the critical points $\displaystyle \frac{\pi}{4}$ and $\displaystyle \frac{5\pi}{4}$.
The last equivalence is easy to see geometrically if you look at the unit circle.

I don't understand why you're messing with $\cos ^{-1}$. Care to explain so we can help you?
Similarly, for $f''$ you'll get the zeros $\displaystyle \frac{3\pi}{4}$ and $\displaystyle \frac{7\pi}{4}$.
A: $-\sin(x)+\cos(x)=0 \iff \sin(x)=\cos(x)$ which should be relatively easy to see when that happens, right?  
Consider $x=\frac{\pi}{4},-\frac{\pi}{4}$ and look at the values of sine and cosine in these cases.
For the inflection points, notice that the zeroes of the second derivative are the same as the zeroes of the initial function which are likely to happen at the points where $\sin(x)=-\cos(x)$ which could be thought of as the mirror images of the previous case.  You could divide by $\cos(x)$ and get the -1 which is also relatively solvable by considering the geometric idea I mentioned earlier here as the first derivative's zeroes are where $y=x$ and the second case is the intersection of $y=-x$.
If it helps, remember the formulas for $\sin(\theta)=y/r$ and $\cos(\theta)=x/r$.
A: The first derivative shows you your min and max points. You know you will see at min or a max whenever $\cos(x) = \sin(x)$. This happens at intervals of $\frac{\pi}{4}$.
I would recommend you try putting in all the multiples of $\frac{\pi}{4}$ from zero to $2\pi$, and then you will get a set of points that corresponds to all the minima and maxima.

