解由sin(π-α)-cos(-α)=1/2
得sina-cosa=1/2
平方得1-2sinacosa=1/4
即sinacosa=3/8
故sin³(π+α)+cos³(2π-α)
=[-sina]³+[cos(-a)]³
=-sin³a+cos³a
=cos³a-sin³a
=(cosa-sina)(cos²a+sinacosa+sin²a)
=-(sina-sina)(1+sinacosa)
=-(1/2)(1+3/8)
=-1/2×11/8
=-11/16
解由sin(π-α)-cos(-α)=1/2
得sina-cosa=1/2
平方得1-2sinacosa=1/4
即sinacosa=3/8
故sin³(π+α)+cos³(2π-α)
=[-sina]³+[cos(-a)]³
=-sin³a+cos³a
=cos³a-sin³a
=(cosa-sina)(cos²a+sinacosa+sin²a)
=-(sina-sina)(1+sinacosa)
=-(1/2)(1+3/8)
=-1/2×11/8
=-11/16