已知5sinB=sin(2a B)

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已知5sinB=sin(2a B)
已知tan(a+b)=2tan a 证明 3sinb=sin(2a+b)

要证3sinB=sin(2A+B)即证3sin(A+B-A)=sin(A+B+A)即证3sin(A+B)cosA-3cos(A+B)sinA=sin(A+B)cosA+cos(A+B)sinA即证2s

已知5sinb=sin(2a+b) 求证 tan(a+b)/tana=3/2

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

已知tanC=(sinA+sinB)/(cosA+cosB),sin(B-A)=cosC.(1)求A,C;(2)S△AB

tanc=sinc/cosc=(sinA+sinB)/(cosA+cosB)左分子乘右分母等于右分子乘左分母,并移项得sinAcosC-sinCcosA=sinCcosB-sinBcosC即sin(A

已知sina+sinb=1/3,sinb-cosa=1/2,求sin(a-b)的值

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

化简:sin(a-B)cosa-1/2[sin(2a+B)-sinB]=?已知cos(a-π/6)+sina=4√3/5

sin(a-B)cosa-1/2[sin(2a+B)-sinB]=[sin(a-B+a)+sin(a-B-a)]/2-sin(2a+B)/2+sinB/2=sin(2a-B)/2+sin(-B)/2-

已知tana=1,3sinB=sin(2a+B),求tanB

sin2a=2tana/[1+(tana)^2]=2*1/[1+1]=1(cos2a)^2=1-(sin2a)^2(cos2a)^2=1-1(cos2a)^2=0cos2a=03sinB=sin(2a

已知sin(a+b)=13分之5 tanb=2分之1 求Sinb cosb

tanb=2分之1sinb/sinb=2分之1cosb=2sinb又因为sinb平方+cosb平方=1联立方程组sinb平方=五分之一因为b属于(0,π)所以sinb=五分之根号五cosb=五分之二倍

(1)已知:sina+cosb=3/4,cosa+sinb=-5/4,求sin(a+b).

1\Sina+cosb=3/4cosa+sinb=5/4(SinA)2+(cosB)2+2sinAcosB=9/16,(cosA)2+(sinB)2+2cosAsinB=25/16两式相加2+2(si

已知3sinB=sin(2A+B),求证tan(A+B)=2tanA

3sinb=sin(2a+b)可得sin(2a+b)-sinb=2sinb由两角正弦差的公式:sin(2a+b)-sinb=2cos[(2a+b+b)/2]sin[(2a+b-b)/2]=2cos(a

sin(a-B)cosa-1/2[sin(2a+B)-sinB]=?已知cos(a-π/6)+sina=4√3/5则si

sin(a-B)cosa-1/2[sin(2a+B)-sinB]=sin(a-B)cosa-1/2[2cos(a+b)sina]=sin(a-b)cosa-cos(a+b)sina=sinacosbc

已知ab都是锐角sin=4/5,cos=5/13,求sinb的值

题目没写清楚啊.sin和cos哪个角啊?如果是sina=4/5,cosb=5/13的话,sinb=根号下1减去5/13的平方,答案是12/13

高一数学:已知5sinB=sin(2A+B),求tan(A+B)cotB的值

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

已知sina+cosb=3/4,cosa+sinb=-5/4,求sin(a+b)

由sina+cosb=3/4,∴(sina+cosb)²=9/16,sin²a+2sinacosb+cos²b=9/16(1)由cosa+sinb=-5/4,∴(cosa

已知sina+cosb=1/5,cosa+sinb=1/3,则sin(a+b)=?

首先sin(a+b)=sina*cosb+cosa*sinbsina^2+cosa^2=1sinb^2+cosb^2=1将已知的两个等式平方得1,sina^2+2sina*cosb+cosb^2=1/

已知5sinb=sin(2a+b),tan(a+b)=9/4,则tana=

5sinB=sin(2A+B)5sin[(A+B)-A]=sin[(A+B)+A]5sin(A+B)cosA-5cos(A+B)sinA=sin(A+B)cosA+cos(A+B)sinA4sin(A

已知5sinB=sin(2A+B),求证2sin(A+B)=3tanA

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

已知tana=1,3sinB=sin(2a+B),求tan(a+b/2)

tana=1a=kπ+π/43sinB=sin(2a+B)=sin(2kπ+π/2+B)=cosB3sinB=cosBsin^2B+cos^2B=1cos^2B=9/10sin^2B=1/10sinB

已知三角行ABC的周长为2倍根号二加4且sinA+sinB=根号2倍sin求AB长

sin=sinAcosB+cosAsinB=根号2/2(sinA+sinB)cosB=cosA=根号2/2A=B=45°边长b=c=根号2/2ab=c=2a=2根号2

三角形ABC中,已知(sin^2 A-sin^2 B-sin^2 C)/(sinB sinC)=1 求A?

sin²A-sin²B-sin²C=sinBsinCa/sinA=b/sinB=c/sinC则由sin²A-sin²B-sin²C=sinB