Isn't like when the second derivative is positive then it is a minimum, not maximum?

Yeah. Think he just made a typo. The second derivative should be

*negative* when \(x=1\), and hence we have a local maximum.

This formula is not taught in the 2U course. It's a valid cheat from a 3U context, but we cannot assume 2U students know it.

Hi,

why does sin (x+) = -sin(x)?

Your LaTeX isn't showing up, but if like DrDusk said you meant \(\sin(x+\pi)\) on the left, you basically need to think back to ASTC.

Throughout the course, you've done things like \( \sin \left( \frac{5\pi}{4} \right) = -\sin \frac\pi4\), and \( \sin \frac{3\pi}{4} = \sin \frac\pi4\). These were done identifying things like \( \frac{5\pi}4\) is in the 3rd quadrant and \( \frac{3\pi}{4}\) is in the 2nd quadrant. But you may have learnt it as nothing more but a trick, without thinking about why they work.

In reality, the identity \( \boxed{\sin(\pi + x) = -\sin x} \) for the 3rd quadrant, along with other identities for the 2nd and 4th quadrants, are the formulas that let you do this in the first place. In general, depending on which quadrant you're in, you can use the `ASTC' trick to figure out whether the trig ratio is positive or negative. But the actual formulae being used are:

\begin{align*}

\sin(\pi - x) &= \sin x\tag{+'ve in 2nd quad}\\

\sin(\pi + x) &= -\sin x\\

\sin (2\pi - x) &= -\sin x

\end{align*}

\begin{align*}

\cos(\pi - x) &= -\cos x\\

\cos(\pi + x) &= -\cos x\\

\cos(2\pi - x) &= \cos x\tag{+'ve in 4th quad}

\end{align*}

\begin{align*}

\tan(\pi - x) &= -\tan x\\

\tan (\pi + x) &= \tan x\tag{+'ve in 3rd quad}\\

\tan(2\pi - x) &= -\tan x

\end{align*}

If you're genuinely interested in derivations of such results, you have to go back to the unit circle definition of the trigonometric ratios.

This video can help get you started.