锐角三角形ABC中,sin(A B)=P,sinA sinB=Q
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第一题(1)sin(A+B)/sin(A-B)=(sinAcosB+sinBcosA)/(sinAcosB-sinBcosA)=3,(tanA+tanB)/(tanA-tanB)=3,tanA+tan
因为a>b>c所以sina>sinb>sinc由二倍角sina>sinb>sinc,sina^2>sinb^2>sinc^21-cos2a>1-cos2b因为角为钝角,所以平方后要变号cos2a^2>
sin(A+B)=sinAcosB+sinBcosA=3/5...(1)sin(A-B)=sinAcosB-sinBcosA=1/5...(2)(1)=3×(2)sinAcosB+sinBcosA=3
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证明:⑴sin(A+B)=sinAcosB+sinBcosA=3/5sin(A-B)=sinAcosB-sinBcosA=1/5两式相加,得:2sinAcosB=4/5sinAcosB=2/5①则si
答:AB边上的高=√{√[(15±3√21)/2]}sin(A+B)=3/5,sin(A-B)=1/5,AB=3sin(A+B)+sin(A-B)=2sinA*cosB=3/5+1/5=4/5sinA
(1)sin(A+B)=3/5,sin(A-B)=1/5sin(a+b)=sinAcosB+sinBcosA=3/5sin(a-b)=sinAcosB-sinBcosA=1/5两式相加相减后可得:si
(1)sin(A+B)/sin(A-B)=(sinAcosB+sinBcosA)/(sinAcosB-sinBcosA)=3,(tanA+tanB)/(tanA-tanB)=3,tanA+tanB=3
把2个拆掉sin(A+B)=sinAcosB+cosAsinB=3/5(1)sin(A-B)=sinAcosB-cosAsinB=1/5(2)(1)+(2)得sinAcosB=2/5(3)(1)-(2
(I)∵sin(A+B)=sinAcosB+cosAsinB=35①sin(A−B)=sinAcosB−cosAsinB=15②①+②得:2sinAcosB=45,∴sinAcosB=25③cosAs
sin(A+B)=sinAcosB+sinBcosA=3/5...(1)sin(A-B)=sinAcosB-sinBcosA=1/5...(2)(1)=3*(2)sinAcosB+sinBcosA=3
首先,利用sin(a+b)=sinacosb+sinbcosa求出ab边上的高与ab边的交点为ab边的三等分点;然后,利用(a+b)+(a-b)=2a,可以求出sina的数值;最后,利用高h,2,和s
sin(A+B)=3/5,sin(A-B)=1/5则:sin(A+B)=3sin(A-B)sinAcosB+cosAsinB=3sinAcosB-3cosAsinB2sinAcosB=4cosAsin
无解?.再问:答案是2+根号6再答:您是老师吗知道答案还来问?再问:这本作业是我自己买的不是老师布置的,但是只写了答案没写过程,所以问。。。您不会做题可以不回答追问的再答:哦哦懂了加油学啊!好刻苦啊.
sin(A+B)=sinAcosB+sinBcosA=3/5sin(A-B)=sinAcosB-sinBcosA=1/5所以sinAcosB=2/5sinBcosA=1/5相除tanA=2tanB令A
sin(A+B)=sinAcosB+cosAsinB=3/5sin(A-B)=sinAcosB-cosAsinB=1/5两式分别相加减,得sinAcosB=2/5cosAsinB=1/5两式相除tan
证明:sin(a+b)=sin(a)cos(b)+cos(a)sin(b)=3/5①sin(a-b)=sin(a)cos(b)-cos(a)sin(b)=1/5②①+②=2sin(a)cos(b)=4
∵△ABC为锐角三角形∴cos(A-π/3)=√21/5cosAcosπ/3+sinAsinπ/3=√21/51/2cosA+√3/2sinA=√21/5——①sin(A-π/3)=2/5sinAco
高中数学:在锐角三角形ABC中,角A、B、C的对边分别为a、b、c,满足a用余弦定理换掉(a平方+c平方-b平方)的2accosB,sin(A+B)=sin(180
sin(A+B)=sinAcosB+cosAsinB=3/5(1)sin(A-B)=sinAcosB-cosAsinB=1/5(2)(1)-(2)×3可得2sinAcosB=4osAsinB,两边同时