若cos(π/4+x)=3/5,求值
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若cos(π/4+x)=3/5,求值
若x∈(17π/12,7π/4),求(sin2x+2sin^2x)/(1-tanx)的值
若x∈(17π/12,7π/4),求(sin2x+2sin^2x)/(1-tanx)的值
cos(π/4+x)=3/5,
√2/2cosx-√2/2sinx=3/5,
cosx-sinx=3√2/5,
两边平方,
1-sin2x=18/25,
sin2x=7/25,
x∈(17π/12,7π/4),在三四象限,
2x∈(17π/6,7π/2),
2x∈(2π+5π/6,2π+3π/2),在二三象限,余弦值为负,
cos2x=-√[1-(sin2x)^2]=-24/25,
sinx=-√[(1-cos2x)/2]=-7√2/10,
cosx=±√2/10,
tanx=±7,
2sin^2x=1-cos2x=49/25,
(sin2x+2sin^2x)/(1-tanx)
=(7/25+49/25)/[1-(±7)]
=-28/75,
或7/25.
√2/2cosx-√2/2sinx=3/5,
cosx-sinx=3√2/5,
两边平方,
1-sin2x=18/25,
sin2x=7/25,
x∈(17π/12,7π/4),在三四象限,
2x∈(17π/6,7π/2),
2x∈(2π+5π/6,2π+3π/2),在二三象限,余弦值为负,
cos2x=-√[1-(sin2x)^2]=-24/25,
sinx=-√[(1-cos2x)/2]=-7√2/10,
cosx=±√2/10,
tanx=±7,
2sin^2x=1-cos2x=49/25,
(sin2x+2sin^2x)/(1-tanx)
=(7/25+49/25)/[1-(±7)]
=-28/75,
或7/25.
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