f(x)=2SIN^2(x+π\4)-cos(2x+π\6) 化简
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f(x)=2SIN^2(x+π\4)-cos(2x+π\6) 化简
f(x)=2SIN^2(x+π\4)-cos(2x+π\6) 化简
f(x)=2SIN^2(x+π\4)-cos(2x+π\6) 化简
f(x)=2sin^2(x+π/4)-cos(2x+π/6)
=2*[(1-cos(2x+π/2))/2] - cos(2x+π/6)
=1-cos(2x+π/2) - [cos2xcos(π/6)-sin(2x)sin(π/6)]
=1+sin2x-(√3/2)cos2x+(1/2)sin2x
=(3/2)sin2x-(√3/2)cos2x+1
=√3sin(2x-π/6)+1
(辅助角公式)
问题补充:万分感谢 偶在线的 左边项根据 cos2x=1-2sin x 来化,右边项展开 f(x)=2sin (x+π\\4)-cos(2x+π\\6) =1- cos2(x+π/4) +
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