化简:√[1/2-1/2√(1/2-1/2cos2α)],270
来源:学生作业帮 编辑:大师作文网作业帮 分类:数学作业 时间:2024/09/21 22:37:08
化简:√[1/2-1/2√(1/2-1/2cos2α)],270
√[1/2-1/2√(1/2-1/2cos2α)]
=√[1/2-1/2√(1/2-1/2(1- 2sin²α))]
=√[1/2-1/2√(1/2+sin²α-1/2)]
=√[1/2-1/2√sin²α]
=√[1/2-1/2sinα]
=(√2/2)√(sin(α/2)-cos(α/2))²
=(√2/2)(sin(α/2)-cos(α/2))
=(√2/2)(sin(α/2)-(√2/2)(cos(α/2))
=cos(π/4)(sin(α/2)-sin(π/4)(cos(α/2))
=sin(α/2-π/4)
=√[1/2-1/2√(1/2-1/2(1- 2sin²α))]
=√[1/2-1/2√(1/2+sin²α-1/2)]
=√[1/2-1/2√sin²α]
=√[1/2-1/2sinα]
=(√2/2)√(sin(α/2)-cos(α/2))²
=(√2/2)(sin(α/2)-cos(α/2))
=(√2/2)(sin(α/2)-(√2/2)(cos(α/2))
=cos(π/4)(sin(α/2)-sin(π/4)(cos(α/2))
=sin(α/2-π/4)
求证:cos2αcos2β=1/2{cos2(α+β)+cos2(α-β)}
(sin2α-cos2α+1)/(1+tanα)=2sin2αcos2α 为什么
(2sin2α/1+cos2α)*(cosα)^2/cos2α=?
化简(sinα-cosα)^2-1/-cos2α
化简:sin^2αsin^2β+cos^2αcos^2β-1/2cos2αcos2β
化简(sinα)^2*(sinβ)^2+(cosα)^2(cosβ)^2-1/2cos2αcos2β
化简sinα^2sinβ^2+cos^2cosβ^2-1/2cos2αcos2β
化简:sin²αsin²β+cos²αcos²β-1/2cos2αcos2β
化简:sin²αsin²β+cos²αcos²β—1/2cos2αcos2β
化简 sin²αsin²β+cos²αcos²β-1/2(cos2αcos2β)
求证cos2acos2β=1/2{cos2(a+β)+cos2(a-β)
化解sin^2αsin^2β+cos^2αcos^2β-1/2cos2αcos2β