RT,求值cosπ/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11
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解决时间 2021-02-25 19:00
- 提问者网友：缘字诀
- 2021-02-25 16:04
RT,求值cosπ/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11
最佳答案
- 五星知识达人网友：神鬼未生
- 2021-02-25 17:25
原式=(cosπ/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11*2sinπ/11)/2sinπ/11=(sin2π/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11)/2sinπ/11=(1/2*sin4π/11*cos3π/11*cos4π/11*cos5π/11)/2sinπ/11=(1/4*sin8π/11*cos3π/11*cos5π/11)/2sinπ/11=(1/4*sin3π/11*cos3π/11*cos5π/11)/2sinπ/11=(1/8*sin6π/11*cos5π/11)/2sinπ/11=(1/8*sin5π/11*cos5π/11)/2sinπ/11=(1/16*sin10π/11)/2sinπ/11=1/32
全部回答
- 1楼网友：一秋
- 2021-02-25 17:41
感谢回答
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