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\cos 40 \cos 80 \cos 160 = \frac { - 1 } { 8 } |
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Answer» cos 40.cos 80.cos 160 = (1/2) (2 cos 40.cos 80).cos 160= (1/2) ( cos(40+80) + cos(40 - 80) ).cos 160 { 2cosAcosB = cos(A + B) + cos(A - B) }= (1/2) ( cos(120) + cos(-40) ).cos 160= (1/2) ( -1/2 + cos(40) ).cos 160 { cos(-A) = cos(A) }= (1/2) ( -cos(160)/2 + cos(160)cos(40) )= (1/2) ( -cos(160)/2 + (1/2)2*cos(160)cos(40) )= (1/2) ( -cos(160)/2 + (1/2) [ cos(200) + cos(120) ] )= (1/2) ( -cos(160)/2 + cos(200)/2 + cos(120)/2 )= (1/2) ( -cos(160)/2 + cos(200)/2 - 1/4 )= (1/2) ( 1/2 [cos(200) - cos(160)] - 1/4 )= (1/2) ( 1/2 [-2sin(180)*sin(20))] - 1/4 ) { cos(A)-cos(B)= -2(sin(A+B)/2)*(sin(A-B)/2) }= (1/2) ( 0 - 1/4 ) { sin(180) = 0 }= -1/8 Tq |
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