1.cos^2 73°+cos^2 47°+cos73°cos47°=多少?2.如果A+B+C=π ,求证:cosA+cosB+cosC=1+4sin A/2 sin B/2 sin C/2
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1.cos^2 73°+cos^2 47°+cos73°cos47°=多少?2.如果A+B+C=π ,求证:cosA+cosB+cosC=1+4sin A/2 sin B/2 sin C/2
1.cos^2 73°+cos^2 47°+cos73°cos47°=多少?
2.如果A+B+C=π ,求证:
cosA+cosB+cosC=1+4sin A/2 sin B/2 sin C/2
1.cos^2 73°+cos^2 47°+cos73°cos47°=多少?2.如果A+B+C=π ,求证:cosA+cosB+cosC=1+4sin A/2 sin B/2 sin C/2
(1)先证明如下公式:
cos²a+cos²b
=[1+cos(2a)+1+cos(2b)]/2
=[cos(2a)+cos(2b)]/2+1
=[2cos(a+b)cos(a-b)]/2+1(和差化积)
=cos(a+b)cos(a-b)+1
cosacosb=[cos(a+b)+cos(a-b)]/2
故cos² 73°+cos² 47°+cos73°cos47°
=cos(73°+47°)cos(73°-47°)+1+[cos(73°+47°)+cos(73°-47°)]/2
=cos120°cos26°+cos120°/2+cos26°/2+1
=cos120°/2+1
=-1/4+1
=3/4
(2)利用和差化积公式和二倍角公式
因A+B+C=π
故C=π-(A+B)
故cosA+cosB+cosC
=cosA+cosB-cos(A+B)
=2cos[(A+B)/2]cos[(A-B)/2]-2cos²[(A+B)/2]+1
=2cos[(A+B)/2]{cos[(A-B)/2]-cos[(A+B)/2]}+1
=2sinC/2{[sinB/2sinC/2]+1
=1+4sin A/2 sin B/2 sin C/2