RT,求值cosπ/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11

发布时间：2021-03-10 02:31:05

RT,求值cosπ/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11

网友回答

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
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