Trigonometry! #146

As $\sin^2x+\cos^2x$ is always one and there is a larger coefficient in front of the $\sin^2x$ , we want to maximize the value of $\sin x$ . This yields $x=\frac{\pi}{2}$ . In order to maximize the value of $\sin\frac{x}{2}+\cos\frac{x}{2}$ we want the two values to be as close as possible as the possible values are inside a closed range. This also yields $x=\frac{\pi}{2}$ . Thus the answer is $A=4\sin^2\frac{\pi}{2}+3\cos^2\frac{\pi}{2}+\sin\frac{\pi}{2}+\cos\frac{\pi}{2}-\sqrt{2}=4+\frac{2}{\sqrt{2}}-\sqrt{2}=4+\sqrt{2}-\sqrt{2}=4$