$y = -2\sin(x - \pi/3):\;$minimum & maximum values? I have the following function $$y = -2\sin⁡(x-\pi/3),\quad 0\leq x\leq 2\pi$$
I know that $\sin x$ has max at $\pi/2$ and min at $3\pi/2$ but how would I use this information to find the solution to the question in the title?
 A: Hints: Basic trigonometry without calculus
$$\begin{align*}&\bullet\;\;\;\sin\left(x-\frac\pi2\right)=\sin\left(-\left(\frac\pi2-x\right)\right)\\&\bullet\;\;\;\sin\left(\frac\pi2-\alpha\right)=\cos\alpha\\
&\bullet\;\;\;\sin(-x)=-\sin x\end{align*}$$
A: Let's use the brute force way:
$$y' = -2\cos(x-\pi/3)= 0$$
$$\cos(x-\pi/3)=0$$
We need the argument of $\cos$ to be an odd multiple of $\pi/2$ for $\cos$ to be zero:
$$x-\pi/3 = (2n-1)\pi/2$$
$$x = n\pi-\pi/2+\pi/3 = n\pi -\pi/6$$
So, $y$ obtains maxima / minima at the points $[\pi-\pi/6,2\pi-\pi/6]$. Plug in $x$, to find which point is which.
A: Let's say $b=x-\frac {\pi}{2}$ . Then we have $y=-2 \sin b$ . As you said, the max and min of sine occur at $\frac {\pi}{2}$ and $\frac {3 \pi}{2}$ . So for the max, $b=\frac {3 \pi}{2}$ and for the min $b=\frac {\pi}{2}$ (it's flipped because you have a $-2$ in front). Since  $b=x-\frac {\pi}{2}$ , for the max $x-\frac {\pi}{2}=\frac {3 \pi}{2}$ $\to$ $x=2\pi$ , and you do the same for the min. And of course you can add or subtract $2n \pi$ from the answer to get infinately many other answers.
A: the function has maximum when $\sin(x - π/3)$ is minimum ie,
$\sin(x - π/3)= -1$
or, $x - π/3= 3π/2$       ie,$x= 11π/6$   when $(2π ≤ x ≤ 2π)$
and, minimum at,
$\sin(x - π/3)= 1$
or,$(x - π/3)= π/2$       ie, $x =5π/6$   when $(2π ≤ x ≤ 2π)$
similarly, we can find the general solution
