已知cos(15+α)=3/5,α为锐角已知:cos(15°+α)=3/5 ,α为锐角 求【tan(435°-α)+sin(α-165°)】/【cos(195°+α)×sin(105°+α)】
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已知cos(15+α)=3/5,α为锐角
已知:cos(15°+α)=3/5 ,α为锐角
求【tan(435°-α)+sin(α-165°)】/【cos(195°+α)×sin(105°+α)】
因为tana的周期是180,cosa .sina的是360
故原式=[tan(90-(15+a))+sin((15+a)-180)]/.[cos(180+15+a)>
α为锐角,cos(15°+α)=3/5>0,
则15°+α是锐角,
sin(15°+α)=4/5,
tan(15°+α)=4/3
[tan(435°-α)+sin(α-165°)]/[cos(195°+α)×sin(105°+α)]
={tan[360°-(15°+α)]+sin[-180°+(15°+α)]}/{cos[180°+(15°+α)]×sin...
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α为锐角,cos(15°+α)=3/5>0,
则15°+α是锐角,
sin(15°+α)=4/5,
tan(15°+α)=4/3
[tan(435°-α)+sin(α-165°)]/[cos(195°+α)×sin(105°+α)]
={tan[360°-(15°+α)]+sin[-180°+(15°+α)]}/{cos[180°+(15°+α)]×sin[90°+(15°+α)]}
=[-tan(15°+α)-sin(15°+α)]/[-cos(15°+α)×cos(15°+α)]
=(-4/3-4/5)/[(-3/5)*(3/5)]
=-(32/15)/(-9/25)
=160/27
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