sin(a-b)cosa-cos(b-a)sina=12/13,求cos(5π/4-b)和cos(π/3+2b)
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sin(a-b)cosa-cos(b-a)sina=12/13,求cos(5π/4-b)和cos(π/3+2b)sin(a-b)cosa-cos(b-a)sina=12/13,求cos(5π/4-b)
sin(a-b)cosa-cos(b-a)sina=12/13,求cos(5π/4-b)和cos(π/3+2b)
sin(a-b)cosa-cos(b-a)sina=12/13,求cos(5π/4-b)
和cos(π/3+2b)
sin(a-b)cosa-cos(b-a)sina=12/13,求cos(5π/4-b)和cos(π/3+2b)
sin(a-b)cosa-cos(a-b)sina=12/13
sin(a-b-a)=-sinb=12/13
sinb=-12/13
cos2b=1-2sin^2b=(169-288)/169=-119/169
cosb=±5/13
sin2b=2sinbcosb=120/169
cos(5π/4-b)
=cos5π/4cosb+sin5π/4sinb
=-√2/2cosb-√2/2sinb
.
cos(π/3+2b)=cosπ/3cos2b-sin(π/3)sin2b
,.
化简sin(a-b)*cosa - cos(a-b)*sina
sin(a-b)cosa-cos(b-a)sina=12/13,求cos(5π/4-b)和cos(π/3+2b)
求证cos(a+b)cos(a-b)=cos^2b-sin^2a
在△ABC中,下列各表达式中为常数的是( )A.sin(A+B)+sinC B.cos(B+C)-cosA C.tan(A+B)/2×tanC/2 D.cos(B+C)×sinA/2
cosb=cos[(a+b)-a]=cos(a+b)cosa+sin(a+b)sina
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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】
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