How to find minimum of this sine wave Can someone help me with this question. I've found the maximum which is really easy. However, the minimum gives me some trouble. I want to solve it algebraically, rather than solving it through trial and error (the problem is done without calculator so I can just select values of "x" that yield sin(special angle, ex. 0deg, 30deg, 45deg, etc.). According to the answer sheet the minimum is sqrt(3)/2 which is the y-intercept as seen in the Desmos graph. 



 A: To show that $\frac {\sqrt3} 2$ is the minimum value on $[0,\frac {2\pi} 3]$ you can observe that the only point where the derivative is $0$ is $x =\frac {\pi} 3$. This implies that the minimum of the function on $[0,\frac {2\pi} 3]$ is nothing but the minimum of the numbers $f(0),f(\frac {\pi} 3), f(\frac {2\pi} 3)$ and this minimum is indeed $\frac {\sqrt3} 2$.
A: As $f(x)=f(\pi-x)$, rewrite the wave as follows:- $$sin(\pi-x) +sin(\frac 23\pi -(\pi-x))$$
$$=\sin(x)+\sin(x+\frac \pi3)$$
$$=\sin(x)+\frac {\sin(x)}2+\frac{\sqrt3}2\cos(x)$$
$$=\frac32\sin(x)+\frac{\sqrt3}2\cos(x)$$
Notice this is the form $asinx+bcosx$, so multiply and divide by $\sqrt{a^2+b^2}$
$$=\sqrt3(\sin(t)\sin(x)+\cos(t)\cos(x))$$ where $$\sin(t)=\frac{\frac32}{\sqrt{(\frac32)^2+(\frac{\sqrt3}2)^2}},\cos(t)=\frac{\frac{\sqrt3}2}{\sqrt{(\frac32)^2+(\frac{\sqrt3}2)^2}}$$
$$=\sqrt3(\cos(x-t))$$
$$=\sqrt3(\cos(x-\arctan(\frac{\frac32}{\frac{\sqrt3}2}))$$
$$=\sqrt3(\cos(x-\frac{\pi}3)$$
For maxima, you have done absolutely correct. For minima, note that in $[0,\frac{\pi}2]$ as $x$ rises, the function decreases. So we input minimum value of $x$ as we can which is $0$ to get
$$\sqrt3(\cos(\frac{\pi}3))=\frac{\sqrt3}2$$
A: Other than the critical points,  you have the endpoints to check for extrema.  Since the critical point is a max, in this case the min occurrs at the endpoints. 
