﹙1/cos²80-3/cos²10﹚·1/cos20
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﹙1/cos²80-3/cos²10﹚·1/cos20
(1/cos^280-3/cos^210)*(1/cos20)
=(1/cos^2(90-10)-3/cos^210)(1/cos20)
=(1/sin^210-3/cos^210)*(1/cos20)
=(cos^210-3sin^210)/(sin^210*cos^210)*(1/cos20)
=(4cos^210-3)/(sin^210*cos^210)*(1/cos20)
=(2cos20-1)/(sin^210*cos^210)*(1/cos20)
=4(2cos20-1)/(sin^210*cos^210)*(1/cos20)
=8 (cos20 - cos60 ) / sin^220 *(1/cos20)
=16sin40sin20 / sin^220 *(1/cos20)
=32 cos20 *(1/cos20)
=32
=(1/cos^2(90-10)-3/cos^210)(1/cos20)
=(1/sin^210-3/cos^210)*(1/cos20)
=(cos^210-3sin^210)/(sin^210*cos^210)*(1/cos20)
=(4cos^210-3)/(sin^210*cos^210)*(1/cos20)
=(2cos20-1)/(sin^210*cos^210)*(1/cos20)
=4(2cos20-1)/(sin^210*cos^210)*(1/cos20)
=8 (cos20 - cos60 ) / sin^220 *(1/cos20)
=16sin40sin20 / sin^220 *(1/cos20)
=32 cos20 *(1/cos20)
=32
求证 1/cos^100 - 3/sin^100=32cos20
2sin20°cos20°-2cos²25°
求证:(1/sin方10°)—(3/cos方10°)=32cos20°
求证:1/(sin^210度)-3/(cos^210度)=32cos20度
已知sinα,sinβ是二次方程x²-(根号2cos20°)x+(cos²20°-1/2)=0的两根
化简根号下:1-2sin20°cos20°.除以2cos^10°-根号下1-cos^160°.-1
化简根号下:1-2sin20°cos20°.除以2cos^10°-根号下1-cos^160°.-1得
求值根号1-2sin20cos20/cos20-根号1-cos平方160
若cos(α+β)cos(α-β)=1/3 则cos²α-sin²β等于
三角函数难题 已知α+β=3/4 证明:cos²α+cos²β+cosα×cosβ=1/2
(sin50(1+根号3*tan10)-cos20)/cos80根号(1-cos20)
根号[(3/sin20')-(1/cos20')],