已知cos(x+π/4)=3/5,5π/4<x<7π/4,求 sin2x+2sin²x ÷ 1-tanx 的值(sin2x+2sin²x)÷(1-tanx)
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已知cos(x+π/4)=3/5,5π/4<x<7π/4,求 sin2x+2sin²x ÷ 1-tanx 的值(sin2x+2sin²x)÷(1-tanx)
已知cos(x+π/4)=3/5,5π/4<x<7π/4,求 sin2x+2sin²x ÷ 1-tanx 的值
(sin2x+2sin²x)÷(1-tanx)
已知cos(x+π/4)=3/5,5π/4<x<7π/4,求 sin2x+2sin²x ÷ 1-tanx 的值(sin2x+2sin²x)÷(1-tanx)
(1)把要求的式子化为-sin2x,再利用已知条件利用二倍角公式求得sin2x的值,即可求得要求式子的值
(2)把要求的式子化为sin2x•tan(π/4+x),根据x的范围求出sin(π/4+x)和cos(π/4+x)的值,即可求得tan(π/4+x)的值,从而求得(sin2x+2sin2x)/(1-tanx)的值.
∵cos2(π/4+x)=cos(π/2+2x)=-sin2x,
又cos2(π/4+x)=2cos²(π/4+x)-1=2×9/25-1=-7/25,
∴sin2x=7/25.
(sin2x+2sin²x)/(1-tanx)
=sin2x(1+sinx/cosx)/(1-tanx) →sin2x·sinx/cosx=2sinxcosx·sinx/cosx=2sin²x
=sin2x(1+tanx)/(1-tanx)
=sin2xtan(π/4+x).
∵5π/4<x<7π/4,
∴3π/2<x+π/4<2π,
∴sin(π/4+x)=√(1-cos²(π/4+x)=-4/5,
∴tan(π/4+x)=sin(π/4+x)/cos(π/4+x)=-4/3.
∴(sin2x+2sin2x)/(1-tanx)
=sin2xtan(π/4+x)
=7/25×(-4/3)
=-28/75.