sin(a-b)cosa-cos(b-a)sina=3/5,b是第三象限角,求sin(b+5π/4),tan(b+5π/4)的值
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sin(a-b)cosa-cos(b-a)sina=3/5,b是第三象限角,求sin(b+5π/4),tan(b+5π/4)的值
sin(a-b)cosa-cos(b-a)sina=3/5,b是第三象限角,求sin(b+5π/4),tan(b+5π/4)的值
sin(a-b)cosa-cos(b-a)sina=3/5,b是第三象限角,求sin(b+5π/4),tan(b+5π/4)的值
根据sin(A-B)=sinAcosB-sinBcosA cos(b-a)=cosAcosB+sinAsinB ;
简化得到sinB=-3/5 可以得到cosB=-4/5 tanB=3/4
所以sin(B+5π/4)=sinBcos5π/4+cosBsin5π/4=-的根号2除以10;
tan(b+5π/4)=(tan5π/4+tanB)/(1-tan5π/4tanB) =5/8
cos(b-a)=cos(a-b)
所以:sin(a-b)cosa-cos(b-a)sina=3/5
即:sin(a-b-a)=3/5
所以:sin(-b)-3/5
sinb=-3/5 cosb=4/5
所以sin(b+5π/4)=sinbcos(5π/4)+cosbsin5π/4=-7√2/10
化简sin(a-b)*cosa - cos(a-b)*sina
cosb=cos[(a+b)-a]=cos(a+b)cosa+sin(a+b)sina
证明下列恒等式:sin(a+b)*cos(a-b)=sina*cosa+sinb*cosb
若a,b为两个锐角,则( ) ,A,cos (a+b)>cosa +cosb B,cos(a+b)<cosa +cosbC,cos (a+b)>sina +sinb D,cos (a+b)<sina +sin b 【说明原因】
sin(a+b)cosa-cos(a+b)sina=5/13 则sin(π-B)=?
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