已知x=x+y2+x−y2 , y=x+y2−x−y2 , sinx−siny=sin(x+y2+x−y2)−sin(x+y2−x−y2)=sinx+y2cosx−y2+cosx+y2sinx−y2−sinx+y2cosx−y2+cosx+y2sinx−y2=2cosx+y2sinx−y2 cosx−cosy=cos(x+y2+x−y2)−cos(x+y2−x−y2)=cosx+y2co...
【解析】-|||-B-|||-解析:两式平方相加有:cos(x-y)=,又x,y为锐-|||-9-|||-角,-|||-sinxsiny,故x-yE(-,0),sin(x-y)=-2i4,tan-|||-2-|||-9-|||-(x-y)=-214.-|||-5 结果一 题目 【题目】已知sinx-siny=5/9 cosx-cosy=π/(2) ,,且x,y为锐角,则tan(x-y)的值是...
sinx-siny=2cos[(x+y)/2]sin[(x-y)/2]；cosx+cosy=2cos[(x+y)/2]cos[(x-y)/2]；cosx-cosy=-2sin[(x+y)/2]sin[(x-y)/2]；2.三角函数的积化和差公式（4个）sinxcosy=(1/2)[sin(x+y)+sin(x-y)]；cosxsiny=(1/2)[sin(x+y)-sin(x-y)]；cosxcosy=(1/2)[cos(x+y)+c...
【答案】 分析: 已知等式左边利用两角和与差的余弦函数公式化简,求出cos(x-y)的值,所求式子利用二倍角的余弦函数公式化简后,将cos(x-y)的值代入计算即可求出值. 解答: 解:∵cosxcosy+sinxsiny=cos(x-y)= , ∴cos(2x-2y)=cos2(x-y)=2cos 2(x-y)-1=- . 故答案为:- 点评: 此题考查了两角...
两式分别求平方得：sinx2-2sinxsiny+siny2=1/4，cosx2-2cosxcosy+cosy2=1/4 相加得：2-2（sinxsiny+cosxcosy）=1/2，sinxsiny+cosxcosy=3/4 cos（x-y）=sinxsiny+cosxcosy=3/4，因为x，y为锐角，所以（x-y)是锐角 所以sin（x-y）=√1-3/4X3/4=√7/4，四分之根号七啊 ...
|PQ|2=|QB|2+|BP|2=(1−|OB|2)+(1−|OB|)2=(1−cos2(y−x))+(1−cos(y−x))2 因此有(siny−sinx)2+(cosy−cosx)2=(1−cos2(y−x))+(1−cos(y−x))2 即2−2sinysinx−2cosycosx=2−2cos(y−x) ...
解析 【解析】 sinx-siny=2cos[(a+b)/2]sin) cosx+cosy= 2cos[(a+b)/2]cos[(a-b)/2] cosx-cosy=-2sin[(a+b)/2]sin[(a-b)/2] sinxcosy=[sin(a+b)]+sin(a-b)] cossiny= [sin(a+b)-sin(a-b)]/2 cosxcosy=[cos(a+b)]/2 sinxsiny=-[cos(a+b)-cos(a-b)]/2 ...
解:(sinx-siny)2=sin2x-2sinxsiny+sin2y=,① (cosx-cosy)2=cos2x-2cosxcosy+cos2y=,② 由①+②得cos(x-y)=, 又∵sinx-siny=-<0, cosx-cosy=>0,x、y为锐角,∴-π<x-y<0. ∴sin(x-y)=- . ∴tan(x-y)= =- . 练习册系列答案 ...
首先推导出两角和公式:sin(x+y)=sinxcosy+cosxsiny令x=θ/2,y=θ/2sin(θ/2+θ/2)=sinθ/2cosθ/2+cosθ/2sinθ/2得到:cosθ/2=sinθ/2sinθ/2sin(x-y)=sinxcosy-cosxsiny令x=θ,y=θ/2sin(θ-θ/2)=sinθcosθ/2-cosθsinθ/2sinθ/2=sinθ(sinθ/2sinθ/2)-cosθsinθ/2...
sinx+siny=2sin[(x+y)/2]cos[(x-y)/2]sinx-siny=2cos[(x+y)/2]sin[(x-y)/2]cosx+cosy=2cos[(x+y)/2]cos[(x-y)/2]cosx-cosy=-2sin[(x+y)/2]sin[(x-y)/2]COSX