证明：sin(a+b)=3/5=3*1/5=3sin(a-b)即sin(a+b)=3sin(a-b)展开得sinacosb+sinbcosa=3sinacosb-3sinbcosa4sinbcosa=2sinacosb所以sinacosb=2sinbcosa把右边除到左边即为Tana/tanb=2
