Question

已知函数f（x）=(sinx+cosx)²+2cos²x–2

Answer 1

f（x）=(sinx+cosx)²+2cos²信桥碧顷x–2
=(sin^2x+cos^2x+sin2x+cos2x-1
=sin2x+cos2x
=√2sin(2x+π/滑慧猛4)

Answer 2

先将括号首旦橘内展开得者团1+sin2x，后面变形（2cos²x–1）-1
降次得
f（迟漏x）=1+sin2x+cos2x-1
=√2sin(2x+π/4)