A+B |
2 |
A−B |
2 |
1−cos(A+B) |
2 |
1+cos(A−B) |
2 |
4−3cos(A+B)+cos(A−B) |
2 |
∴4-3cos(A+B)+cos(A-B)=4,即3cos(A+B)=cos(A-B),
∴3cosAcosB-3sinAsinB=cosAcosB+sinAsinB,即2cosAcosB=4sinAsinB,
则tanAtanB=
sinAsinB |
cosAcosB |
2 |
4 |
1 |
2 |
故答案为:
1 |
2 |
A+B |
2 |
A−B |
2 |
A+B |
2 |
A−B |
2 |
1−cos(A+B) |
2 |
1+cos(A−B) |
2 |
4−3cos(A+B)+cos(A−B) |
2 |
sinAsinB |
cosAcosB |
2 |
4 |
1 |
2 |
1 |
2 |