Question

高一数学三角函数问题

sin(a+b)=1/2 sin(a-b)=1/10 求tana/tanb 请用正反思路详细求解 谢谢

Answer 1

tana/tanb
=(sina/cosa)/(sinb/cosb)
=(sina*cosb)/(cosa*sinb)
=[sin(a+b)+sin(a-b)]/[sin(a+b)-sin(a-b)]
=(1/2+1/10)/(1/2-1/10)
=2/3
用到公式积化和差
sina*cosb=[sin(a+b)+sin(a-b)]
cosa*sinb=[sin(a+b)-sin(a-b)]

Answer 2

sinacosb+cosasinb=1/2, sinacosb-cosasinb=1/10,
2sinacosb=3/5,
2cosasinb=2/5

tana/tanb= 3/2

Answer 3

• sin(a+b)=1/2        原式sinacosb+cosasinb=1/2             sin(a-b）=1/10原式sinacosb-cosasinb=1/10        tana/tanb=sina/cosa*cosb/sinb          sinacosb=1/2-cosasinb,sinacosb=1/10+cossinb          1/2-cosasinb=1/10+cosasinb    cosasinb=1/5                      sinacosb=3/10       tana/tanb=sinacosb/cosasinb=3/10*1/5=3/50