tanβ=-3,(5sin^3 β+cos β)/(2cos^3 β+sin^2 βcos β)

(5sin³β+cos β)/(2cos³β+sin²βcosβ)
=[sinβ(5sin²β+cosβ/sinβ)]/[cosβ(2cos²β+sin²β)]
=tanβ[5sin²β+1/tanβ]/[2(1-sin²β)+sin²β]
=tanβ[5sin²β+1/tanβ]/[2-sin²β]
而sin²β
=sin²βcos²β/cos²β
=tan²βcos²β
=9cos²β
=9(1-sin²β)
=9-9sin²β
所以sin²β=9/10
代入整理得式子,
原式=(-3)[5×9/10 + 1/(-3)]/[2- 9/10]
=(-3)[9/2 - 1/3]/[11/10]
=(-3)[25/6]/[11/10]
=(-3)×25×10/(6×11)
=-125/11