已知向量a=(1-cosθ,√2),b=(sinθ,1),且a//b,则tanθ/2的值为

问题描述:

已知向量a=(1-cosθ,√2),b=(sinθ,1),且a//b,则tanθ/2的值为

a∥b,即:(1-cosθ)/sinθ=√2
令:t=tan(θ/2)
sinθ=2t/(1+t^2)
cosθ=(1-t^2)/(1+t^2)
即:1-cosθ=2t^2/(1+t^2)
即:2t^2/(1+t^2)/(2t/(1+t^2))=t=√2
即:tan(θ/2)=√2嗯明白了谢谢