Question

已知sin(5π-θ)-sin(5π/2-θ)=1/5

（1）求sinθcosθ的值 （2）求sin³θ-cos³θ的值（3）当-π＜θ＜0时求tanθ的值

Answer 1

已知sin(5π-θ)-sin(5π/2-θ)=1/5
那么sinθ-cosθ=1/5
(1)
两边平方得(sinθ-cosθ)²=1/25
即sin²θ-2sinθcosθ+cos²θ=1/25
又sin²θ+cos²θ=1
所以1-2sinθcosθ=1/25
故sinθcosθ=12/25

(2)
sin³θ-cos³θ
=(sinθ-cosθ)(sin²θ+sinθcosθ+cos²θ)
=(1/5)*(1+12/25)
=37/125

(3)
sinθ-cosθ=1/5①
-π＜θ＜0
那么sinθ＜0
故cosθ＜0
因为sinθcosθ=12/25
所以1+2sinθcosθ=49/25
即(sinθ+cosθ)²=49/25
所以sinθ+cosθ=-7/5②

由①②解得sinθ=-3/5,cosθ=-4/5
所以tanθ=sinθ/cosθ=(-3/5)/(-4/5)=3/4

Answer 2

sin(5π-θ)-sin(5π/2-θ)=1/5
sin(4π+π-θ)-sin(2π-π/2-θ)=1/5
sin(π-θ)+sin(π/2+θ)=1/5
sinθ+cosθ=1/5

sinθ+cosθ=1/5平方
(sinθ+cosθ)²=1/25
sin²θ+cos²θ+2sinθcosθ=1/25
1+2sinθcosθ=1/25
2sinθcosθ=-24/25
sinθcosθ=-12/25

(sinθ-cosθ)²=(sinθ+cosθ)²-4sinθcosθ
(sinθ-cosθ)²=1/25-4*(-12/25)
(sinθ-cosθ)²=49/25
sinθ-cosθ=±7/5

sin³θ-cos³θ
=(sinθ-cosθ)(sin²θ+sinθcosθ+cos²θ)
=±7/5*(1+sinθcosθ)
=±7/5*(1-12/25)
=±7/5*13/25
=±91/125

当-π＜θ＜0时求tanθ的值
sinθcosθ=-12/25
sinθ,cosθ异号
所以-3π/2＜θ＜0
tanθ<0
sinθcosθ/1=-12/25
sinθcosθ/(sin²θ+cos²θ)=-12/25分子分母同时除以cos²θ
(sinθcosθ/cos²θ)/(sin²θ/cos²θ+cos²θ/cos²θ)=-12/25
tanθ/(tan²θ+1)=-12/25
12(tan²θ+1)=-25tanθ
12tan²θ+25tanθ+12=0
(3tanθ+4)(4tanθ+3)=0
tanθ=-4/3或tanθ=-3/4