已知cos（75゜+α）=3/5，180゜<α<270゜，求sin(30゜-2α)-cos(α-45゜)的值

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∵180゜0∴ ∠(75゜+α)的終邊在第四象限∴270゜<75゜+α<360゜∴-360゜<-75゜-α<-270゜∴-360゜+90゜<-75゜-α+90゜<-270゜+90゜，-270゜<15゜-α<-180゜∴∠(15゜-α)的終邊∈Ⅱ又∵sin(15゜-α)=cos[90゜-(15゜-α)]=cos(75゜+α)=3/5∴cos(15゜-α)=-√[1-(sin^2)(15゜-α)]=-√[1-(3/5)^2]=-4/5∴sin(30゜-2α)=2sin(15゜-α)cos(15゜-α)=2*(3/5)*(-4/5)=-24/25 cos(α-45゜)=cos(45゜-α)=cos[(15゜-α)+30゜] =cos(15゜-α)cos30゜-sin(15゜-α)sin30゜ =(-4/5) ×(√3)/2-(3/5) ×(1/2) =-(2√3)/5-3/10∴ sin(30゜-2α)-cos(α-45゜) =-24/25+2(√3)/5+3/10 =(2√3)/5-33/50 =[(20√3)-33]/50。

熱心網(wǎng)友

答案：(2√3)/5-33/50

熱心網(wǎng)友

=(2√3)/5-33/50

熱心網(wǎng)友

1802250---270sin(a+75)=-4/5sin(30-2a)-cos(a-45)=sin[2(15-a)]-cos[(a+75)-120]=2sin(15-a)cos(15-a)-[cos(a+75)cos120+sin(a+75)sin120]=2cos(75+a)sin(75+a)-[3/5*(-1/2)+(-4/5)(3^.5)/2]=2*3/5*(-4/5)-[-(3+4*3^.5)/2]=-24/25+3/2+2*3^.5=27/50+2*3^.5

熱心網(wǎng)友

解：因?yàn)?80゜0,所以， 角(75゜+α)的終邊在第四象限即270゜<75゜+α<360゜。所以，-360゜<-75゜-α<-270゜。所以，-360゜+90゜<-75゜-α+90゜<-270゜+90゜，即-270゜<15゜-α<-180゜,所以角(15゜-α)的終邊在第二象限。又因?yàn)閟in(15゜-α)=cos[90゜-(15゜-α)]=cos(75゜+α)=3/5，所以，cos(15゜-α)=-√[1-(sin^2)(15゜-α)]=-√[1-(3/5)^2]=-4/5。所以，sin(30゜-2α)=2sin(15゜-α)cos(15゜-α)=2*(3/5)*(-4/5)=-24/25。cos(α-45゜)=cos(45゜-α)=cos[(15゜-α)+30゜]=cos(15゜-α)cos30゜-sin(15゜-α)sin30゜=(-4/5) ×(√3)/2-(3/5) ×(1/2)=-(2√3)/5-3/10。所以， sin(30゜-2α)-cos(α-45゜)=-24/25+2(√3)/5+3/10=(2√3)/5-33/50。。

熱心網(wǎng)友

180<α<270，所以，255<75+α<360，so -360<-75-α<-255,so -270<15-α<-145，cos（75-α）=sin(90-75-α)=3/5=sin(15-α)so cos(15-α)=-4/5<0sin(30-2α)=2sin(15-α)cos(15-α)=-24/25cos(α-45)=cos(α-15-30)=cos(α-15)cos30+sin(α-15)sin30=(-4/5) ×√3/2+(-3/5)/2=-[(4√3)+3]/10so sin(30-2α)-cos(α-45)=-24/25+[(4√3)+3]/10=[(20√3)-33]/50