已知sin(a-b)等于3/5cos
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5sinB=sin(2A+B)=sin(A+B+A)=sin(A+B)cosA+cos(A+B)sinA,sinB=sin(A+B-A)=sin(A+B)cosA-cos(A+B)sinA5sin(A
a∥b,则4/3=cosα/sinα则tanα=sinα/cosα=3/4
首先,由a·b=0并化简可得5/4*cos(a+b)=cos(a-b);然后,展开移项sin*sin=1/9cos*cos;最后可得tgA*tgB=1/9.公式自己去背,别问我!
原式=sin[(a+b)+b]-sin[(a+b)-b]=sin(a+b)cosb+cos(a+b)sinb-[sin(a+b)cosb-cos(a+b)sinb]=2cos(a+b)sinb=0
3(sinA)^2+2(sinB)^2=5sinA(sinA)^2+(sinB)^2=5sinA/2-(sinA)^2/25sinA/2-(sinA)^2/2=-(1/2)(sinA-5/2)^2+2
这个题是不是有问题啊?
sin^2(a)+cos^2(a)=1可以求出a
sin(α+π/6)=sinα·cos(π/6)+cosα·sin(π/6)=(√3/2)sinα+(1/2)cosα;所以sin(α+π/6)+cosα=(√3/2)sinα+(3/2)cosα;即
sin(A+B)=3/5,sin(A-B)=1/5则:sin(A+B)=3sin(A-B)sinAcosB+cosAsinB=3sinAcosB-3cosAsinB2sinAcosB=4cosAsin
(sinC)^2=(sinA)^2+(sinB)^2,由正弦定理,c^2=a^2+b^2,(1)C=90°.(2)(1/2)ab=√3,a^2+b^2=16,∴(a+b)^2=16+4√3,∴a+b=
1、sina+cosb=4/5sin²a+2sinacosb+cos²b=16/25cosa+sinb=3/5cos²a+2cosasinb+sin²b=9/2
1sin(a+b)=2/3sina*cosb+cosa*sinb=2/3-----------(1)sin(a-b)=1/5sina*cosb-cosa*sinb=1/5-----------(2)联
解sin(π/2-b)*cos(a+b)-sin(π+b)*sin(a+b)=3/5即cosbcos(a+b)+sinbsin(a+b)=cos[b-(a+b)]=cos(-a)=cosa∴cosa=
证明:5sin[(A+B)-A]=sin[A+(A+B)]5sin(A+B)cosA-5cos(A+B)sinA=sin(A+B)cosA+cos(A+B)sinA4sin(A+B)cosA=6cos
sina+cosb=1/3,sinb-cosa=1/2平方得(sina+cosb)^2=(1/3)^2,(sinb-cosa)^2=(1/2)^2开展,并将两式相加得2+sinacosb-cosasi
sin(a+b)=sinacosb+cosasinb=1/2sin(a-b)=sinacosb-cosasinb=1/3所以sinacosb=(1/2+1/3)/2=5/12cosasinb=(1/2
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)cosa-cos(b-a)sina=-[sin(b-a)cosa+cos(b-a)sina]=-sin(b-a+a)=-sinb=>sinb=-3/5=>cosb=-4/5sin(b+
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