题目

1.SinA+SinB=a,CosA+CosB=1+a,求Sin（A+B）,Cos（A+B）.
2.基本相同：SinA+CosB=二分之根三,CosA+SinB=根二,求TanA*CotB
我意思是题型差不多。

提问时间：2021-01-03

答案

1.SinA+SinB=a,CosA+CosB=1+a,求Sin（A+B）,Cos（A+B）.
sinA+sinB=a,平方之得 sin²A+2sinAsinB+sin²B=a².(1)
cosA+cosB=1+a,平方之得 cos²A+2cosAcosB+cos²B=(1+a)².(2)
(1)+(2)得cos(A-B)=[a²+(1+a)²-2]/2=a²+a-1/2.(3)
(2)-(1)得2cos(A+B)+cos2A+cos2B=2cos(A+B)+2cos(A+B)cos(A-B)
=2cos(A+B)[1+cos(A-B)]=2a+1
故cos(A+B)=(2a+1)/2[1+cos(A-B)]=(2a+1)/2(1+a²+a-1/2)=(2a+1)/(2a²+2a+1)
(sinA+sinB)(cosA+cosB)=sinAcosA+sinBcosB+sinAcosB+cosAsinB
=(1/2)(sin2A+sin2B)+sin(A+B)=sin(A+B)cos(A-B)+sin(A+B)
=sin(A+B)[cos(A-B)+1]=sin(A+B)(a²+a+1/2)=a(1+a)
∴sin(A+B)=a(1+a)/(a²+a+1/2)=2a(1+a)/(2a²+2a+1).
2.sinA+cosB=√3/2,,cosA+sinB=√2,求tanA*cotB
sinA+cosB=√3/2,平方之得 sin²A+2sinAcosB+cos²B=3/4.(1)
cosA+sinB=√2,平方之得 cos²A+2cosAsinB+sin²B=2.(2)
(1)+(2)得 sin(A+B)=3/8.(3)
(1)-(2)得 2sin(A-B)-(cos2A-cos2B)=2sin(A-B)+2sin(A+B)sin(A-B)
=2sin(A-B)[1+sin(A+B)]=2sin(A-B)(1+3/8)=(11/4)sin(A-B)=-5/4
故sin(A-B)=-5/11.(4)
tanAcotB=tanA/tanB=[sin(A+B)-sin(A-B)]/[sin(A+B)+sin(A-B)]
=(3/8+5/11) /(3/8-5/11)=-73/7