sinAsinB=sin[(A+B)/2+(A-B)/2]*sin[(A+B)/2-(A-B)/2]=[sin(A+B)/2*cos(A-B)/2+cos(A+B)/2sin(A-B)/2]*[sin(A+B)/2*cos(A-B)/2-cos(A+B)/2sin(A-B)/2]=[sin(A+B)/2*cos(A-B)/2]^2-[cos(A+B)/2sin(A-B)/2]^2=[1-cos^2(A+B)/2]cos^2(A-...
sinAsinB=sin[(A+B)/2+(A-B)/2]*sin[(A+B)/2-(A-B)/2] =[sin(A+B)/2*cos(A-B)/2+cos(A+B)/2sin(A-B)/2]*[sin(A+B)/2*cos(A-B)/2-cos(A+B)/2sin(A-B)/2] =[sin(A+B)/2*cos(A-B)/2]^2-[cos(A+B)/2sin(A-B)/2]^2 =[1-cos^2(A+B)/2]cos^2...
sinasinb在三角函数中,sinα·sinβ的乘积可通过积化和差公式转化为和差形式,其表达式为: sinα·sinβ = [cos(α−β) − cos(α+β)] / 2 该公式可将两个正弦函数的乘积转换为余弦函数的线性组合,常用于简化积分、求和等计算场景。以下是详细说明: 1. 公式推导过程 积化和差...
sinAsinB=sin[(A+B)/2+(A-B)/2]*sin[(A+B)/2-(A-B)/2] =[sin(A+B)/2*cos(A-B)/2+cos(A+B)/2sin(A-B)/2]*[sin(A+B)/2*cos(A-B)/2-cos(A+B)/2sin(A-B)/2] =[sin(A+B)/2*cos(A-B)/2]^2-[cos(A+B)/2sin(A-B)/2]^2 =[1-cos^2(A+B)/2]cos^2...
本题主要考查了正弦定理的应用,属于基础题.结果一 题目 在△ABC中,sinAsinB等于( )A.baB.abC.acD.ca 答案 由正弦定理:asinA=bsinB=csinC=2R,可得:sinA=a2R,sinB=b2R,则sinAsinB=a2Rb2R=ab.故选:B.相关推荐 1在△ABC中,sinAsinB等于( )A.baB.abC.acD.ca 反馈 收藏 ...
sinasinb=(-1/2)[cos(a+b)-cos(a-b)]拓展：两角和公式：sin(A+B)=sinAcosB+cosAsinB sin(A-B)=sinAcosB-sinBcosA cos(A+B)=cosAcosB-sinAsinB cos(A-B)=cosAcosB+sinAsinB tan(A+B)=(tanA+tanB)/(1-tanAtanB)tan(A-B)=(tanA-tanB)/(1+tanAtanB)cot(A+B)=(cotAcotB-1)...
sinAsinB=sin[(A+B)/2+(A-B)/2]*sin[(A+B)/2-(A-B)/2]=[sin(A+B)/2*cos(A-B)/2+cos(A+B)/2sin(A-B)/2]*[sin(A+B)/2*cos(A-B)/2-cos(A+B)/2sin(A-B)/2]=[sin(A+B)/2*cos(A-B)/2]^2-[cos(A+B)/2sin(A-B)/2]^2=[1-cos^2(A+B)/2]...
sinAsinB等于多少? 对于任意角的正弦都在[-1,1],根据sinasinb=1,可得sina=sinb=1或sina=sinb=-1,所以a=b=2kpi+pi/2,或者a=b=2kpi+3pi/2,那么cos(a+b)=cospi=-1或者cos(a+b)=cos3pi=-1.选a sinA×sinB等于多少? sinA×sinB=-(1/2)[cos(A+B)-cos(A-B)]。cosA×cosB=(1/2)[cos(...
2sinasinb=cos(a-b)-cos(a+b)和差化积公式：sina+sinb=2sin[(a+b)/2]cos[(a-b)/2]sina-sinb=2cos[(a+b)/2]sin[(a-b)/2]cosa+cosb=2cos[(a+b)/2]cos[(a-b)/2]cosa-cosb=-2sin[(a+b)/2]sin[(a-b)/2]这两大公式在高中数学课本不会出现，但是有时候在考试做题中，如果知道...
sinAsinB=sin[(A+B)/2+(A-B)/2]*sin[(A+B)/2-(A-B)/2]=[sin(A+B)/2*cos(A-B)/2+cos(A+B)/2sin(A-B)/2]*[sin(A+B)/2*cos(A-B)/2-cos(A+B)/2sin(A-B)/2]=[sin(A+B)/2*cos(A-B)/2]^2-[cos(A+B)/2sin(A-B)/2]^2=[1-cos^2(A+B)/2]cos^2(A-...