已知0<a<π/2,0<β<π/2且tana=1/7,tanβ=3/4求a+β
问题描述:
已知0<a<π/2,0<β<π/2且tana=1/7,tanβ=3/4求a+β
答
tan(A+B) = (tanA+tanB)/(1-tanAtanB)=(1/7+3/4)/(1-1/7x3/4)=1 所以a+b=π/4
已知0<a<π/2,0<β<π/2且tana=1/7,tanβ=3/4求a+β
tan(A+B) = (tanA+tanB)/(1-tanAtanB)=(1/7+3/4)/(1-1/7x3/4)=1 所以a+b=π/4