已知sin(x-0.75π)*cos(x-0.25π)=-0.25,求cos4x的值

问题描述:

已知sin(x-0.75π)*cos(x-0.25π)=-0.25,求cos4x的值
据说是1991年上海高考数学题-
求答案
求详细过程
谢谢!

解析:
已知sin(x-0.75π)*cos(x-0.25π)=-0.25
则sin(x-0.25π-0.5π)*cos(x-0.25π)=-0.25
即-sin[0.5π-(x-0.25π)]*cos(x-0.25π)=-0.25
cos²(x-0.25π)=0.25
0.5[1+cos(2x-0.5π)]=0.25
1+cos(0.25π-2x)=0.5
sin2x=-0.5
所以:cos4x=1-2sin²2x=1-2×0.25=0.5