Question

若tanθ=2,求下列各式的值 (1)(sinθ+2cosθ)/(sinθ-cosθ) (2)sin^2θ+2sinθc

Answer 1

(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