题目

已知tanθ=2,那么sin²θ+sinθcosθ+cos²θ的值为?

提问时间：2020-11-03

答案

sin²θ+sinθcosθ+cos²θ
=(sin²θ+sinθcosθ+cos²θ)/1
=(sin²θ+sinθcosθ+cos²θ)/(sin²θ+cos²θ)分子分母同时除以cos²θ
=(sin²θ/cos²θ+sinθcosθ/cos²θ+cos²θ/cos²θ)/(sin²θ/cos²θ+cos²θ/cos²θ)
=[(sinθ/cosθ)²+sinθ/cosθ+1]/[(sinθ/cosθ)²+1]
=[tan²θ+tanθ+1]/[tan²θ+1]
=(2²+2+1)/(2²+1)
=(4+3)/(4+1)
=7/5