化简:√[1/2-1/2√(1/2-1/2cos2α)],270

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化简:√[1/2-1/2√(1/2-1/2cos2α)],270化简:√[1/2-1/2√(1/2-1/2cos2α)],270化简:√[1/2-1/2√(1/2-1/2cos2α)],270√[1/

化简:√[1/2-1/2√(1/2-1/2cos2α)],270
化简:√[1/2-1/2√(1/2-1/2cos2α)],270

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)

用倍角余弦公式代入呗!最后根据角的取值确定正负号

(sinA)^2=(1-cos2A)/2