Question

2^(sin²x)+2^(cos²x)​

Answer 1

You know that cos2x=1−sin2xcos2⁡x=1−sin2⁡x, so you can rewrite:

2sin2x+2cos2x=2sin2x+21−sin2x=2sin2x+22sin2x2sin2⁡x+2cos2⁡x=2sin2⁡x+21−sin2⁡x=2sin2⁡x+22sin2⁡x

Now, let 2sin2x=y2sin2⁡x=y, then we basically have to maximize y+2yy+2y. But then, note that: y+2yy+2y has derivative 1−2y21−<