tanx
=tan[(x-π/4)+π/4]
=[tan(x-π/4)+tan(π/4)]/[1-tan(x-π/4)tan(π/4)]
=(2+1)/(1-2*1)
=-3
(sin2x+cos2x)/(2cos²x-3sin2x-1)
=(2sinxcosx+cos²x-sin²x)/(2cos²x-6sinxcosx-sin²x-cos²x)
=(2sinxcosx+cos²x-sin²x)/(cos²x-6sinxcosx-sin²x)
分子分母同時(shí)除以 cos²x
=(2tanx+1-tan²x)/(1-6tanx-tan²x)
=(-6+1-9)/(1+18-9)
=-14/10
=-7/5
已知tan(x-π/4)=2 求(sin2x+cos2x)/(2cos²x-3sin2x-1)
已知tan(x-π/4)=2 求(sin2x+cos2x)/(2cos²x-3sin2x-1)
數(shù)學(xué)人氣:759 ℃時(shí)間:2020-05-04 06:26:52
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