高一三角函数恒等变换,已知cos(π/4-α)=12/13,且π/4-α是第一象限角 ,则sin(π/2-2α)/sin(π/4+α)
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高一三角函数恒等变换,已知cos(π/4-α)=12/13,且π/4-α是第一象限角 ,则sin(π/2-2α)/sin(π/4+α)
高一三角函数恒等变换,
已知cos(π/4-α)=12/13,且π/4-α是第一象限角 ,则sin(π/2-2α)/sin(π/4+α)
高一三角函数恒等变换,已知cos(π/4-α)=12/13,且π/4-α是第一象限角 ,则sin(π/2-2α)/sin(π/4+α)
因为:cos(π/4-α)=12/13,所以:cosπ/4cosα+sinπ/4sinα=12/13,
即:(根号2/2)(sinα+cosα)=12/13,所以:sinα+cosα=(12倍根号2)/13
两遍呢同时平方得到:1+sin2α=288/169,解得:sin2α=119/169
因为:π/4-α是第一象限角 ,所以:2kπ
由cos(π/4-a)=12/13,且π/4-α是第一象限角
得sin(π/4-a)=5/13,
从而sin(π/2-2α)/sin(π/4+α)
=sin2(π/4-α)/sin(π/4+α)
=2sin(π/4-α)cos(π/4-α)/sin[π/2-(π/4-α)] (注:π/4+a=π/2-(π/4+a)),
=2sin(π/4-α)cos(π/4-α)/cos(π/4-α)
=2sin(π/4-α)
=2*(5/13)
=10/13
已知cos(π/4-α)=12/13, 且π/4-α是第一象限角 ,则sin(π/2-2α)/sin(π/4+α)=?
sin(π/2-2α)/sin(π/4+α)=cos2α/[(√2/2)(cosα+sinα)]=(cos²α-sin²α)/[(√2/2)(cosα+sinα)]
=(√2)(cosα-sinα)=2[cos(π/4)cosα-sin(π/4...
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已知cos(π/4-α)=12/13, 且π/4-α是第一象限角 ,则sin(π/2-2α)/sin(π/4+α)=?
sin(π/2-2α)/sin(π/4+α)=cos2α/[(√2/2)(cosα+sinα)]=(cos²α-sin²α)/[(√2/2)(cosα+sinα)]
=(√2)(cosα-sinα)=2[cos(π/4)cosα-sin(π/4)sinα]=2cos(π/4+α)=2cos[π/2-(π/4-α)]=2sin(π/4-α)
=2√[1-cos²(π/4-α)]=2√[1-(12/13)²]=10/13
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