Question

若tanθ=2，求下列各式的值 (1)(sinθ+2cosθ)/(sinθ-cosθ) (2)sin^2θ+2sinθcosθ+1 麻烦写下过程

Answer 1

(1)(sinθ+2cosθ)/(sinθ-cosθ) 分子分母同除以cosθ可得(tanθ+2)/(tanθ-1)=4
(2)有tanθ=2，sin^2θ+cos^2θ=1得sin^2θ=4/5，cos^2θ=1/5，sinθcosθ=2/5
所以sin^2θ+2sinθcosθ+1 =13/5

Answer 2

(1)(sinθ+2cosθ)/(sinθ-cosθ)
=(sinθ/cosθ+2cosθ/cosθ)/(sinθ/cosθ-cosθ/cosθ)
=(tanθ+2)/(tanθ-1)
=(2+2)/(2-1)
=4

(2)sin^2θ+2sinθcosθ+1
=(1-cos2θ)/2+sin2θ+1
=1/2-cos2θ/2+sin2θ+1
=sin2θ-cos2θ/2+3/2
=2tanθ/[1+(tanθ)^2]+1/2*[1-(tanθ)^2]/[1+(tanθ)^2]+3/2
=2*2/[1+2^2]+1/2*[1-2^2]/[1+2^2]+3/2
=4/5+1/2*(-3)/5+3/2
=8/10-3/10+15/10
=2