f(x)=sin(x-π/4)+2√2sin2x+sin(x-π/4)-2
=2sin(x-π/4)+2√2cos[2(x-π/4)]-2
=2√2[1-2sin(x-π/4)^2]+2sin(x-π/4)-2
=-4√2sin(x-π/4)^2+2sin(x-π/4)+2√2-2
=-4√2[sin(x-π/4)-√2/8]^2+17√2/8-2
化簡f(x)=sin(x-π/4)+4*(根號(hào)2)sinxcosx-cos(x+π/4)-2
化簡f(x)=sin(x-π/4)+4*(根號(hào)2)sinxcosx-cos(x+π/4)-2
數(shù)學(xué)人氣:705 ℃時(shí)間:2020-04-09 17:56:22
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