已知tanθ=2,求sin²θ+2sinθ*cosθ+1的值
问题描述：
已知tanθ=2,求sin²θ+2sinθ*cosθ+1的值
答
sin²θ+2sinθ*cosθ+1=(sin²θ+2sinθ*cosθ+sin²θ+cos²θ)/(sin²θ+cos²θ)
=(2tan²θ+2tanθ+1)/(tan²θ+1) (同除以cos²θ)
=(2*4+4+1)/(4+1)=13/5
答
tanθ=sinθ/cosθ=2
sinθ=2cosθ
sin^2θ+cos^2θ=1
5cos^2θ=1
sin²θ+2sinθ*cosθ+1
=4cos^2θ+4cos^2θ+1
=4/5+4/5+1
=13/5
答
tanθ=2
sin²θ+2sinθ*cosθ+1
= (sin²θ+2sinθ*cosθ)/1 + 1
= (sin²θ+2sinθ*cosθ)/(sin²θ+cos²θ) + 1
前边分子分母分别除以cos²θ：
= (tan²θ+2tanθ)/(tan²θ+1) + 1
= (2²+2×2)/(2²+1) + 1
= 13/5