化简:(3/sin^20)-(1/cos^20)+64sin^20
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/11/12 23:23:11
化简:(3/sin^20)-(1/cos^20)+64sin^20
(3/sin^20)-(1/cos^20)+64sin^20
=[3(cos20)^2-(sin20)^2]/(sin20)^2(cos20)^2+64(sin20)^2
=[(√3cos20+sin20)(√3cos20-sin20)]/(sin20cos20)^2+64(sin20)^2
=[2sin(60+20)*2sin(60-20)]/[(sin40)/2]^2+64(sin20)^2
=16sin80sin40/sin40*sin40+64(sin20)^2
=16sin80/sin40+64(sin20)^2
=32sin40cos40/sin40+64(sin20)^2
=32cos40+64(sin20)^2
=32[1-2(sin20)^2]+64(sin20)^2
=32
=[3(cos20)^2-(sin20)^2]/(sin20)^2(cos20)^2+64(sin20)^2
=[(√3cos20+sin20)(√3cos20-sin20)]/(sin20cos20)^2+64(sin20)^2
=[2sin(60+20)*2sin(60-20)]/[(sin40)/2]^2+64(sin20)^2
=16sin80sin40/sin40*sin40+64(sin20)^2
=16sin80/sin40+64(sin20)^2
=32sin40cos40/sin40+64(sin20)^2
=32cos40+64(sin20)^2
=32[1-2(sin20)^2]+64(sin20)^2
=32
计算[3/(sin^2)20°-[1/(cos^2)20°]+64(sin^2)20°
化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x
利用三角函数定义证明(COSα-SINα+1)/(COSα+SINα+1)=(1-SINα)/COSα悬赏20分
化简求值sin(-20/3π)cos(43/6π)+sin(35/6π)cos-(17/3π)
化简:1+sinθ+cosθ+2sinθcosθ /1+sinθ+cosθ
化简 1/2cos x-根号3/2sin x 根3sin x+cos x
化简 (1-2sin@cos@)/(cos@^2-sin@^2)*(1+2sin@cos@)/(1-2sin@^2) 两
化简:tanα*(cosα-sinα)+[sinα(sinα+tanα)/1+cosα]
化简:tanα(cosα-sinα)+sinα(sinα+tanα)/1+cosα.
sinα=-2cosα,求sin^2α-3sinαcosα+1
化简sin^2(20')+cos^2(80')+根号下3*sin20'*cos80'
(1+sinα+cosα)*[sin(α/2)-cos(α/2)]/√(2+2cosα)化简 (3/2*π