关于三角函数的题目求证:cos6cos42cos66cos78=1/16 (全部是角度制)
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关于三角函数的题目
求证:cos6cos42cos66cos78=1/16 (全部是角度制)
求证:cos6cos42cos66cos78=1/16 (全部是角度制)
2*cosA*cosB = cos(A+B) + cos(A-B)
cosA + cosB = 2*[cos(A+B)/2]*cos[(A-B)/2]
cos6°cos42°cos66°cos78°
= (cos6°cos66°)*(cos42°cos78°)
= (cos72°+ cos60°)*(cos120°+ cos36°)/4
= (cos72°cos120°+ cos72°cos36°+ cos60°cos120°+ cos60°cos36°)/4
= [ ( cos36°- cos72°)/2 + cos72°cos36°- 1/4 )/4
= [ (cos36°- cos72°)/2 + (cos108°+ cos36°)/2 ]/4 - 1/16
= - 1/16
cosA + cosB = 2*[cos(A+B)/2]*cos[(A-B)/2]
cos6°cos42°cos66°cos78°
= (cos6°cos66°)*(cos42°cos78°)
= (cos72°+ cos60°)*(cos120°+ cos36°)/4
= (cos72°cos120°+ cos72°cos36°+ cos60°cos120°+ cos60°cos36°)/4
= [ ( cos36°- cos72°)/2 + cos72°cos36°- 1/4 )/4
= [ (cos36°- cos72°)/2 + (cos108°+ cos36°)/2 ]/4 - 1/16
= - 1/16