Question

已知sin(π-a)-cos(π+a)=√2/3(π/2<a<π)求sin^3(π/2-α)+cos^3(π/2+α)的值

Answer 1

已知sin(π-a)-cos(π+a)=√2/3(π/2<a<π)求sin^3(π/2-α)+cos^3(π/2+α)的值
解：sin(π-a)-cos(π+a)=sina+cosa=(√2)/3
两边平方之得 1+2sinαcosα=2/9，故sinαcosα=(1/2)(2/9-1)=-7/18
(cosα-sinα)²=1-2sinαcosα=1+7/9=16/9，π/2<a<π，故cosα-sinα=-4/3
∴sin³(π/2-α)+cos³(π/2+α)=cos³α-sin³α=(cosα-sinα)(cos²α+sinαcosα+sin²α)
=(cosα-sinα)(1+sinαcosα)=-[4/3][1-7/18)=-(4/3)(11/18)=-22/27..

追问

能先将sin^3(π/2-α)+cos^3(π/2+α)化简做一遍吗，谢谢

追答

sin^3(π/2-α)+cos^3(π/2+α)=[sin(π/2-α)]³+[cos(π/2+α)]³=cos³α-sin³α
=(coaα-sinα)(cos²α+cosαsinα+sin²α)=(cosα-sinα)(1+sinαcosα)
=-(4/3)(1-7/18)=-22/27

Answer 2

先对sin(π-a)-cos(π+a)=√2/3进行化简，得sina+cosa=√2/3
因为sin^2(a)+cos^2(a)=1
所以sina*cosa=-7/18
所以（cosa-sina）^2=16/9
又因为π/2<a<π
所以cosa-sina=-4/3
则sin^3(π/2-α)+cos^3(π/2+α)=cos^3a-sin^3a=(cosa-sina)(cos^2(a)+cosa*sina+sin^2(a))
=-4/3*(1-7/18)=-22/27
)

追问

能先将sin^3(π/2-α)+cos^3(π/2+α)化简做一遍吗，谢谢