Question

已知sin(x-3π/4)cos(x-π/4)=-1/4,求cos4x。

已知sin(x-3π/4)cos(x-π/4)=-1/4,求cos4x。

Answer 1

解：∵sin(x-3π/4)cos(x-π/4)=-1/4
==>[sin((x-3π/4)+(x-π/4))+sin((x-3π/4)-(x-π/4))]/2=-1/4 (应用正弦积化和差公式)
==>sin(2x-π)+sin(-π/2)=-1/2
==>-sin(2x)-1=-1/2
==>sin(2x)=-1/2
∴cos(4x)=1-2sin²(2x) （应用余弦倍角公式）
=1-2(-1/2)²
=1/2

Answer 2

解：∵sin(x-3π/4)cos(x-π/4)=-1/4
==>[sin((x-3π/4)+(x-π/4))+sin((x-3π/4)-(x-π/4))]/2=-1/4
(应用正弦积化和差公式)
==>sin(2x-π)+sin(-π/2)=-1/2
==>-sin(2x)-1=-1/2
==>sin(2x)=-1/2
∴cos(4x)=1-2sin²(2x)
（应用余弦倍角公式）
=1-2(-1/2)²
=1/2

Answer 3

解：sin(x-3π/4)cos(x-π/4)=-1/4
-1/4=[sin(x-3π/4+x-π/4)+sin(x-3π/4-x+π/4)]/2
sin(2x-π)+sin(-π/2)=-1/2
-sin2x-1=-1/2
sin2x=-1/2
cos4x=1-2(sin2x)²
=1-2(-1/2)²
=1/2
应用积化和差公式与倍角公式。