Maximum of $\sqrt{\frac{1}{4}\cdot \sin^2(t)+\sin^2(t+\frac{\pi}{3})}$ Maximum of $\sqrt{\frac{1}{4}\cdot\sin^2(t)+\sin^2(t+\frac{\pi}{3})}$
In my opinion, the maximum of the sinus is 1, so I calculated $\sqrt{\frac{1}{4}\cdot1+1}$
This is wrong, why?

Other people have pointed out the OP's error.  Now, I am supplying a solution.  
Note that
$$\sin\left(t+\frac{\pi}{3}\right)=\frac{1}{2}\sin(t)+\frac{\sqrt{3}}{2}\cos(t)\,.$$
That is,
$$\begin{align}\frac{1}{4}\sin^2(t)+\sin^2\left(t+\frac{\pi}{3}\right)&=\frac{\sin^2(t)+\big(\sin(t)+\sqrt{3}\cos(t)\big)^2}{4}\\&=\frac{5}{8}+\frac{2\sqrt{3}\sin(2t)+\cos(2t)}{8}\,.\end{align}$$
We then apply the Cauchy-Schwarz Inequality to get
$$-\frac{\sqrt{13}}{8}\leq \frac{2\sqrt{3}\sin(2t)+\cos(2t)}{8}\leq +\frac{\sqrt{13}}{8}\,.$$
Consequently,
$$\sqrt{\frac{5-\sqrt{13}}{8}}\leq \sqrt{\frac{1}{4}\sin^2(t)+\sin^2\left(t+\frac{\pi}{3}\right)}\leq \sqrt{\frac{5+\sqrt{13}}{8}}\,.$$
It is not difficult to see that both the inequality on the left-hand side and the inequality on the right-hand side are sharp.
