求高手帮忙把cos²B+Cos²(2π/3-B)化为1-sin(2B-π/6)答案上是一部到位的,我算半天弄不出来,求高手赐教.
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/30 00:38:11
![求高手帮忙把cos²B+Cos²(2π/3-B)化为1-sin(2B-π/6)答案上是一部到位的,我算半天弄不出来,求高手赐教.](/uploads/image/z/14016918-30-8.jpg?t=%E6%B1%82%E9%AB%98%E6%89%8B%E5%B8%AE%E5%BF%99%E6%8A%8Acos%26%23178%3BB%EF%BC%8BCos%26%23178%3B%EF%BC%882%CF%80%2F3%EF%BC%8DB%EF%BC%89%E5%8C%96%E4%B8%BA1%EF%BC%8Dsin%EF%BC%882B%EF%BC%8D%CF%80%2F6%EF%BC%89%E7%AD%94%E6%A1%88%E4%B8%8A%E6%98%AF%E4%B8%80%E9%83%A8%E5%88%B0%E4%BD%8D%E7%9A%84%2C%E6%88%91%E7%AE%97%E5%8D%8A%E5%A4%A9%E5%BC%84%E4%B8%8D%E5%87%BA%E6%9D%A5%2C%E6%B1%82%E9%AB%98%E6%89%8B%E8%B5%90%E6%95%99.)
求高手帮忙把cos²B+Cos²(2π/3-B)化为1-sin(2B-π/6)答案上是一部到位的,我算半天弄不出来,求高手赐教.
求高手帮忙把cos²B+Cos²(2π/3-B)化为1-sin(2B-π/6)答案上是一部到位的,我算半天弄不出来,求高手赐教.
求高手帮忙把cos²B+Cos²(2π/3-B)化为1-sin(2B-π/6)答案上是一部到位的,我算半天弄不出来,求高手赐教.
cos²B+cos²(2π/3-B)
=(1/2)(2cos²B-1)+(1/2)[2cos²(2π/3-B)-1]+1
=cos2B+cos(4π/3-2B)+1
=cos2B+cos4π/3cos2B+sin4π/3sin2B+1
=cos2B-(1/2)cos2B-(√3/2)sin2B+1
=1-(sin2Bcosπ/6-cos2Bsinπ/6)
=1-sin(2B-π/6)
原式=cos²B+(cos2π/3cosB+sin2π/3sinB)²
=cos²B+(-1/2cosB+根号3/2sinB)²
=5/4cos²B+3/4sin²B-根号3/2sinBcosB 然后用二倍角公式的变形
=5/4[(1+cos2B)/2]+3/4[(1-co...
全部展开
原式=cos²B+(cos2π/3cosB+sin2π/3sinB)²
=cos²B+(-1/2cosB+根号3/2sinB)²
=5/4cos²B+3/4sin²B-根号3/2sinBcosB 然后用二倍角公式的变形
=5/4[(1+cos2B)/2]+3/4[(1-cos2B)/2]-根号3/4sin2B 展开
=1-(根号3/4sin2B -1/4cos2B) 辅角公式
=1-1/2(根号3/2sin2B-1/2cos2B)
=1-1/2(sin2Bcosπ/6-cos2Bsinπ/6)
=1-1/2sin(2B-π/6) 你是不是少写了个1/2啊
收起