Question

一道三角函数题已知α为锐角,且∠α≠45°,若2sinα·cosα+(1/3)sinα-(1/3)cosα=1,求以ta

Answer 1

由2sinα·cosα+(1/3)sinα-(1/3)cosα=1

sin^2α+cos^α=1

即2sinα·cosα+(1/3)sinα-(1/3)cosα=sin^2α+cos^α

(1/3)(sinα-cosα)=sin^2α+cos^α-2sinα·cosα=(sinα-cosα)^2

因为α为锐角,且∠α≠45°,所以(sinα-cosα)不等于0,等式两边同时约掉(sinα-cosα),得sinα-cosα=1/3,代入原式解得sinα·cosα=4/9.

由韦达定理知tanα+cotα=-b/a

tanα*cotα=c/a

tanα+cotα=sinα/cosα+cosα/sinα=1/sinα·cosα=9/4=-b/a

tanα*cotα=1=c/a

解得a=(-4/9)b=c 所以以tanα和cotα为两根的一元二次方程为:

(-4/9)x^2+x+(-4/9)=0

即4x^2-9x+4=0