Question

已知sin(π-α)-cos（π+α）=√2/3 (π/2＜α＜π)

求 sin^3(5π-α)+cos(6π-α)=？

Answer 1

已知sin(π-α)-cos（π+α）=(√2)/3 (π/2＜α＜π)；求 sin³(5π-α)+cos(6π-α)=？
解：sin(π-α)-cos(π+α)=sinα+cosα=(√2)sin(α+π/4)=(√2)/3
故sin(α+π/4)=1/3，α+(π/4)=π-arcsin(1/3)，α=(3π/4)-arcsin(1/3)
于是得sinα=sin[(3π/4)-arcsin(1/3)]=sin(3π/4)cosarcsin(1/3)-cos(3π/4)sinarcsin(1/3)
=(√2/2)√(1-1/9)+(√2/2)(1/3)=2/3+(√2)/6
cosα=(√2)/3-sinα=(√2)/3-[2/3+(√2)/6]=(√2)/6-(2/3)
∴sin³(5π-α)+cos(6π-α)=sin³[4π+(π-α)]+cosα=sin³(π-α)+cosα=sin³α+cosα
=[2/3+(√2)/6]³+[(√2)/6-(2/3)]=-7/27+(43√2)/108=(-28+43√2)/108