问题
解答题
求函数y=
|
答案
sin3xsin3x+cos3xcos3x
=(sin3xsinx)sin2x+(cos3xcosx)cos2x
=
[(cos2x-cos4x)sin2x+(cos2x+cos4x)cos2x]1 2
=
[(sin2x+cos2x)cos2x+(cos2x-sin2x)cos4x]1 2
=
(cos2x+cos2xcos4x)1 2
=
cos2x(1+cos4x)1 2
=cos32x
所以y=
+sin2xcos32x cos22x
=cos2x+sin2x
=
sin(2x+2
).π 4
所以当sin(2x+
)=-1时,y取最小值-π 4
.2