安全风险类型有几种:sin(a+b)cos(a-b)=sinacosa+sinbcosb
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sin(a+b)cos(a-b)
=(sinacosb+cosasinb)(cosacosb+sinasinb)
=sinacosa(cosb)^2+(cosa)^2sinbcosb+(sina)^2sinbcosb+sinacosa(sinb)^2
=sinacosa(cosb)^2+sinacosa(sinb)^2+(cosa)^2sinbcosb+(sina)^2sinbcosb
=sinacosa+sinbcosb
sin(a+b)cos(a-b)
=(sinacosb+cosasinb)(cosacosb+sinasinb)
=sinacosa(cosb)^2+(cosa)^2sinbcosb+(sina)^2sinbcosb+sinacosa(sinb)^2
=sinacosa(cosb)^2+sinacosa(sinb)^2+(cosa)^2sinbcosb+(sina)^2sinbcosb
=sinacosa+sinbcosb
已知cos^a-cos^B=m,那么sin(a+B)sin(a-B)=?
cos(A-B)=? sin(A-B)=?
sin(a+b)cos(a-b)=sinacosa+sinbcosb
求证:cos(a+b)cos(a-b)=cos平方b-sin平方a
若锐角a,b满足sin a-sin b=-0.5,cos a-cos b=1/3,求sin(a+b)的值
已知3sin^A+2sin^B=2sinA 求cos^A+cos^B的最值
为什么sina+sinb==2sin(a+b)/2*cos(a-b)/2
cos(a+b)=0求证sin(a+2b)=sina
cos(A+B)+sin(A-B)=(cosA+sinA)(cosB-sinB)
cos(A-B)-sin(A+B)=(cosA-sinA)(cosB-sinB)