Question

sinθ+cosθ=1/5,θ∈(0,π)求 ①tanθ ②sinθ-cosθ ③sin^6θ-cos^6θ

已经求得1)tanx=-4/3
2)sinx-cosx=7/5
只求第三问，要详细过程
非常感谢！

Answer 1

答：
3）sinθ+cosθ=1/5
两边平方：(sinθ)^2+2sinθcosθ+(cosθ)^2=1/25
所以；sinθcosθ=-12/25

原式=(sinθ)^6-(cosθ)^6
=[(sinθ)^3+(cosθ)^3]*[(sinθ)^3-(cosθ)^3]
=(sinθ+cosθ)[(sinθ)^2-sinθcosθ+(cosθ)^2]*(sinθ-cosθ)[(sinθ)^2+sinθcosθ+(cosθ)^2]
=(1/5)(1+12/25)(7/5)(1-12/25)
=3367/15625
=0.215488

Answer 2

(sinθ+cosθ)²=1/25,
sinθcosθ=-12/25
③sin^6θ-cos^6θ
=(sin³θ)²-(cos³θ）²
=（sin³θ-cos³θ）(sin³θ+cos³θ)
=(sinθ+cosθ)(sinθ-cosθ)(1+sinθcosθ)(1-sinθcosθ)
=(1/5)(7/5)(1-sin²θcos²θ)
=(1/7)(1-144/625)
=(1/7)(481/625)
=

追问

缺了最后结果，不过还是谢啦