Question

已知cosa+cosb=1/2,sina+sinb=1/3. 求cos(a-b)的值

Answer 1

cosa+cosb=1/2
则
cos^2a+2cosacosb+cos^2b=1/4……①
sina+sinb=1/3.
则
sin^2+2sinasinb+sin^2b=1/9……②
①+②得
1+2（cosacosb+sinasinb）+1=1/4+1/9
即 2（cosacosb+sinasinb）=13/36-2=-59/36
cosacosb+sinasinb=-59/72
cos(a-b)
=cosacosb+sinasinb
=-59/72

Answer 2

cosα+cosβ=1/2
平方得
cos²a+2cosacosb+cos²b=1/4............(1)

sinα+sinβ=1/3
平方得
sin²a+2sinasinb+sin²b=1/9............(2)

(1)+(2)得
cos²a+2cosacosb+cos²b+sin²a+2sinasinb+sin²b=1/4+1/9
1+1+2(cosacosb+sinasinb)=13/36
cos(a-b)=-59/72

Answer 3

cosa+cosb=1/2,sina+sinb=1/3.
平方得
cos^2a+cos^2b+2cosacosb=1/4
sin^2a+sin^2b+2sinasinb=1/3
相加得
2cos(a-b)=-17/12
cos(a-b)=-17/24

Answer 4

cos(a-b)=cosacosb+sinasinb
cosa+cosb=1/2,sina+sinb=1/3
(cosa+cosb)^2+(sina+sinb)^2=2cosacosb+2sinasinb+1+1=13/36
cosacosb+sinasinb=-59/72
cos(a-b)=cosacosb+sinasinb=-59/72

Answer 5

cosa+cosb=1/2,sina+sinb=1/3.
平方得
cos^2a+cos^2b+2cosacosb=1/4
sin^2a+sin^2b+2sinasinb=1/9
相加，
带诱导公式，
cos(a-b)=59/72