已知tan（π/4+θ）+tan（π/4-θ)=4,且-π＜θ＜-π/2,求sin²θ-2sinθcosθ-cos²θ的值

數(shù)學(xué)人氣：390 ℃時間：2020-05-23 22:11:22

優(yōu)質(zhì)解答

tan( π/4 + θ ) + tan( π/4 - θ ) = 4
[ sin( π/4 + θ )cos( π/4 - θ ) + cos( π/4 + θ )sin( π/4 - θ )] / [cos( π/4 + θ )cos( π/4 - θ )] = 4
sin( π/2 ) / [cos( π/4 + θ )cos( π/4 - θ )] = 4
cos( π/4 + θ )cos( π/4 - θ ) = 1 / 4
( cos π/2 + cos 2θ ) / 2= 1 / 4
cos 2θ = 1 / 2
sin 2θ = √( 1 - cos² 2θ ) = √3 / 2
sin² θ - 2sin θ cos θ - cos² θ = - ( sin 2θ + cos 2θ ) = - (√3 + 1) / 2