问题标题:
一道三角化简题求3/cos80°sin²40°-1/sin10°cos²40°的值
问题描述:
一道三角化简题求3/cos80°sin²40°-1/sin10°cos²40°的值
侯刚回答:
3/(cos80°·sin²40°)-1/(sin10°·cos²40°)
=3/(cos80°·sin²40°)-1/(cos80°·cos²40°)
=1/cos80°·(3/sin²40°-1/cos²40°)
=1/cos80°·(3cos²40°-sin²40°)/(sin²40°·cos²40°)
根据倍角公式
sin²40°=(1-cos80°)/2,
cos²40°=(1+cos80°)/2
cos10°=sin80°=2sin40°cos40°
分别带入上式,得到
原式
=1/cos80°·(3(1+cos80°)/2-(1-cos80°)/2)/((1/2·cos10°)^2)
=2(4cos80°+2)/(cos80°·cos²10°)
由于cos60°=1/2
所以
原式
=2(4cos80°+4cos60°)/(cos80°·cos²10°)
=8(2cos70°·cos10°)/(cos80°·cos²10°)
=(16cos70°)/(cos80°·cos10°)
=(16cos70°)/(1/2·(cos90°+cos70°))
=(16cos70°)/(1/2·(0+cos70°))
=32.
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