已知sina=2sin〔a+b〕,求证tan(a+b)=sinb\/(cosb-2)
tan(a+b)=(tana+tanb)\/(1-tanatanb)=[2sinb\/(1-2cosb)+sinb\/cosb]\/[1-2(sinb)^2\/cosb(1-2cosb)]分子分母同乘以cosb(1-2cosb)=(2sinbcosb+sinb-2sinbcosb)\/[cosb-2(cosb)^2-2(sinb)^2]=sinb\/(cosb-2)
已知sina=2sin(a+2b),求证tan(a+b)=-3tanb
解由sina=2sin(a+2b),得sin(a+b-b)=2sin(a+b+b),sin(a+b)cosb-cos(a+b)sinb=2[sin(a+b)cosb+cos(a+b)sinb]即 sin(a+b)cosb-cos(a+b)sinb=2sin(a+b)cosb+2cos(a+b)sinb 即-3cos(a+b)sinb=sin(a+b)cosb 即-3sinb\/cosb=sin(a+b)\/cos(a+b)即-3tanb=tan(a...
sinA=2sinB, sin(A+B).tan(A-B)=1-则tanAtanB等于多少?
由于sinA = 2sinB,因此可以将sin2B表示为sinA\/2,代入上式得:sin(A+B) = sinA\/2 + 1\/tan(A-B)又因为:tan(A-B) = 1\/sin(A-B)= cos(A-B)\/sin(A-B)代入上式得:sin(A+B) = sinA\/2 + sin(A-B)\/cos(A-B)将上式化简得:2sin(A+B)cos(A-B) = sinAcos(A-B) +...
已知2sinA=sin(2A+B),求tan(A+B)\/tanB的值
n-m=2sina sina\/cosa=tana 所以cosa=sina\/tana=(n-m)\/(n+m)选A f(π\/3+x)=f(-x),则f(π\/6+x)=f(π\/6-x)所以x=π\/6是对称轴 所以x=π\/6,sin取最值±1 所以f(π\/6)=3×(±1)选B
已知3sinB=sin(2A+B),求证:tan(A+B)=2tan A。
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+b)sina 因此:cos(a+b)sina=sinb,即:cos(a+b)=sinb\/sina 则:sin(2a+b)=sin[a+(a+b)]=sinacos(a+b)+cosasin(a+b)……(*)将sin(2a+b)=3sinb,cos(a+b)sina=sinb代入等式(*...
三角函数万能公式sinA+sinB=2sin(A+B)\/2cos(A-B)\/2怎么证明
sin2A + sin2B + sin2C = 4sinAsinBsinC 三角函数公式应用比索(一)不等式 情况下,称为A,B,C是一个三角形 确认COTA三个内角+ cotB + COTC> =√3 COTA + cotB + COTC = COTA + cotB-COT(A + B)> COTA + cotB-COT(B)= COTA> 0比索(COTA + cotB + COTC)^ 2> = ...
已知sin(b)=sin(2a+b),求证tan(a+b)=2tan(a)
你好 您应该是打漏了一个3 题目应该是 已知3sin(b)=sin(2a+b),求证tan(a+b)=2tan(a)解:3sinB=sin(2A+B)3sinB=sinAcos(A+B)+cosAsin(A+B)3sinB=sinA(cosAcosB-sinAsinB)+cosA(sinAcosB+cosAsinB)3sinB=sinAcosAcosB-sinAsinAsinB+cosAsinAcosB+cosAcosAsinB 3sinB=2cosAsinAcosB...
tanA=2tanB,sin(A+B)=¼,则sin(A-B)=?
sin(A+B)=sinA.cosB+cosA.sinB 1\/4=2cosA.sinB +cosA.sinB 3cosA.sinB =1\/4 cosA.sinB =1\/12 (1)Also,sin(A+B)=sinA.cosB+cosA.sinB 1\/4=sinA.cosB+(1\/2)sinA.cosB sinA.cosB = 1\/6 (2)sin(A-B)=sinA.cosB - cosA.sinB =1\/6-1\/12 =1\/12 ...
求一道高一证明题:已知3sina=sin(2b+a),求证tan(a+b)=2tanb._百度...
3sina=sin(2b+a)=>3sin[(a+b)-b]=sin[(a+b)+b]=3sin(a+b)cosb-3sinbcos(a+b)=sin(a+b)cosb+sinbcos(a+b)=>2sin(a+b)cosb=4sinbcos(a+b)=>tan(a+b)=2tanb
sinA+sinB=2sin((A+B)\/2)cos((A-B)\/2的推导过程是怎样的?
由两角和与差的正 弦公式得:sin(a+b)=sinacosb+cosasinb,sin(a-b)=sinacosb-cosasinb,两式相加得:sin(a+b)+sin(a-b)=2sinacosb,设a+b=A,a-b=B,解得:a=(A+B)\/2,b=(A-B)\/2,代入上式即得:sinA+sinB=2sin[(A+B)\/2]cos[(A-B)\/2]....