纳皮尔公式2cosA*sinB
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证明:锐角三角形ABC∵∠A+∠B>90°∴∠A>90°-∠B∴sinA>sin(90°-∠B)∴sinA>cos∠B同理,sinB>cosCsinC>cosA∴sinA+sinB+sinC>cosA
sina+cosb=1/3,平方sin^2a+2sinacosb+cos^2b=1/9sinb-cosa=1/2,平方sin^2b-2cosasinb+cos^2a=1/4相加2-2(sinacosb
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因为sinA+cosA=2sina两边平方得sin^2A+cos^2A+2*sinA*cosA=4*sin^2a则2*sinA*cosA=4*sin^2a-1因为sinA*cosA=(sinb)^2则
2(sinA)^2(sinB)^2+2(cosA)^2(cosB)^2-cos2A(cosB)^2=2(sinA)^2(sinB)^2+(cosB)^2*(2(cosA)^2-cos2A)=2(sin
左边=sin(A+B)sin(B-A)+sin²C=sin(180-C)sin(B-A)+sin²C=sinCsin(B-A)+sin²C=sinC[sin(B-A)+s
左边=sin(A+B)sin(B-A)+sin²C=sin(180-C)sin(B-A)+sin²C=sinCsin(B-A)+sin²C=sinC[sin(B-A)+s
cosa+cosb+cosc=sina+sinb+sinc=0(cosa)^2=(cosb+cosc)^2=(cosb)^2+(cosc)^2+2*cosb*cosc.(1)(sina)^2=(sin
sinA+sinB=2sin(A+B)/2cos(A-B)/2cosA+cosB=2cos(A+B)/2cos(A-B)/2sinA+cosB=sinA+sin(π/2-B)=2sin(A-B+π/2
(cosA+2cosC)/(cosA+2cosB)=sinB/sinCcosAsinC+2sinCcosC=cosAsinB+2sinBcosBcosAsinC+sin2C=cosAsinB+sin2
∵cosa=cos[(a+b)/2+(a-b)/2]=cos[(a+b)/2]cos[(a-b)/2]-sin[(a+b)/2]sin[(a-b)/2].cosb=cos[(a+b)/2-(a-b)/
sina+sinb=2sin[(a+b)/2]cos[(a-b)/2]=1/2cosa+cosb=2cos[(a+b)/2]cos[(a-b)/2]=x两式相除:tan[(a+b)/2]=1/(2x)
∵2cosa=3cosb∴2cos[(a+b)-b]=3cos[(b+a)-a]∴2[cos(a+b)cosb+sin(a+b)sinb]=3[cos(a+b)·cosa+sin(a+b)sina]∴
sinA/sinB=cosB/cosA即sinAcosA=sinBcosBsin2A=sin2B2A=2B2A=180-2B.
sinA-cosB=-2sinC、cosA-sinB=-2cosC则:(sinA-cosB)²+(cosA-sinB)²=(-2sinC)²+(-2cosC)²
亲,我写给你,你要给我好评哦~~~再问:恩再问:答案呢再答:等下再答:再问:亲,答案有正和负再答:那是平方,忘了~”再问:亲,那个怎么变形等于1我看不懂再答:用三角函数再问:sina的平方/4+9/4
1/sinB+1/sinA=(sinA+sinB)/(sinA*sinB)=2(sinA+sinB)/[(sinA+sinB)^2-1]设X=sinA+sinB∈(1,根2]则上式=2/(x-1/x)
(cosA+2cosC)/(cosA+2cosB)=sinB/sinCcosAsinC+2sinCcosC=cosAsinB+2sinBcosBcosAsinC+sin2C=cosAsinB+sin2
cosacosbsinacosb-cosacosbcosasinb-sinasinbsinacosb+sinasinbcosasinb=cosacosb(sinacosb-cosasinb)-sina
显然a=b=0时,原式子成立所以cosa=1(cosa)²=1因为tana=3tanb所以sina/cosa=3sinb/cosb因为sina=2sinb所以2sinb/cosa=3sinb