sin²B=2sinA×sinC

sin^2a+sin^b=sina-sin(2a)/2=sina-sinacosa假设f(a)=sina-sinacosa求导得f'(a)=cosa-((cosa)^2-(sina)^2)=-2(co

[sin(2A+B)]/sinA-2cos(A+B)=[sin(A+B+A)]/sinA-2cos(A+B)=[sin(A+B)cosA+cos(A+B)sinA]/sinA-2cos(A+B)=[s

明如下由COS(A+B)=0,有：COS(A+B)=cosAcosB-sinAsinB=0所以cosAcosB=sinAsinBSIN(A+2B)=sinAcos2B+sin2BcosA=sinA[(

sin²A=sin²B+sin²C,a/sinA=b/sinB=c/sinC=2R(a/2R)^2=(b/2R)^2+(c/2R)^2a^2=b^2+c^2,ABC是直角

应当是sin^2A+sin^2B【+】sin^2C=sinB*sinC+sinC*sinA+sinA*sinB吧括号中是要改的.两边同乘以22sin²A+2sin²B+2sin&s

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

因为cos(a+B)+1=0所以a+B=180度所以sin(2a+B)+sinB=sin(360度-B)+sinB=-sinB+sinB=0

证明:要证sin(a+b)sin(a-b)=(sina+sinb)(sina-sinb)只须证(sina*cosb+cosa*sinb)(sina*cosb-cosa*sinb)=(sina+sinb

a+b=a+b/2+b/2,a=a+b/2-b/2sin(a+b)-sina=sin[（a+b/2）+(b/2)]-sin[（a+b/2）-(b/2)]展开=sin（a+b/2）*cosb/2+cos

sin(2a+b)=sin2acosb+cos2asinb=2sinacosacosb+(1-2sin²a)sinb2cos(a+b)=2cosacosb-2sinasinb代入原式得：2c

cos(a+B)*cosa+sin(a+B)*sina=-4/5得到cos[(a+b)-a]=-4/5...cosb=-4/5...sin(π/2-b)=cosb=-4/5

ina+sinb==2sin(a+b)\

由正弦定理,原式可化为a^2+c^2-ac=b^2即[(a^2+c^2-b^2)/2ac]=0.5即cosB=0.5∴B=π/3

sin(2a+b)/sina-2cos(a+b)=sinb/sina==>左边=[sin(2a+b)-2cos(a+b)sina]/sina=[sina*cos(a+b)+sin(a+b)*cosa-

sina+sinb=sin[(a+b)/2+(a-b)/2]+sin[(a+b)/2-(a-b)/2]=2sin(a+b)/2*cos(a-b)/2

左边用积化和差公式=(cos2B-cos2A)/2=(1-2sinB^2-1+2sinA^2)/2=sinA^2-sinB^2

应该是sinA+sinB=2sin[(A+B)/2]cos[(A-B)/2]A=(A+B)/2+(A-B)/2.B=(A+B)/2-(A-B)/2所以sin(A+B)/2cos(A-B)/2+cos(

sin(A+B)-sinA=2cos(A+B/2)sin(B/2)和差化积sin(A+B)-sinA=1/2[cos(A+B)/2]*[sin(A+B-A)/2]=2cos(A+B/2)sin(B/2

这里要用到三角函数中和差化积的公式：sinA-sinB=2cos[(A+B)/2]sin[(A-B)/2]=2sin(C/2)sin[(A-B)/2]sinA+sinB=2sin[(A+B)/2]co

和差化积公式sinα+sinβ=2sin[(α+β)/2]cos[(α-β)/2]sinα-sinβ=2cos[(α+β)/2]sin[(α-β)/2]cosα+cosβ=2cos[(α+β)/2]c