√1/2+1/2√1/2+1/2cos2α﹙∏<α<3∏/2﹚化简
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√1/2+1/2√1/2+1/2cos2α﹙∏<α<3∏/2﹚化简√1/2+1/2√1/2+1/2cos2α﹙∏<α<3∏/2﹚化简√1/2+1/2√1/2+1/2cos2α﹙∏<α<3∏/2﹚化简(
√1/2+1/2√1/2+1/2cos2α﹙∏<α<3∏/2﹚化简
√1/2+1/2√1/2+1/2cos2α﹙∏<α<3∏/2﹚化简
√1/2+1/2√1/2+1/2cos2α﹙∏<α<3∏/2﹚化简
(1+cos2α)/2=cosα^2
∵﹙∏<α<3∏/2)∴cosα﹤0.
√1/2+1/2√1/2+1/2cos2α=√1/2+1/2*(-1)*cosα
=√1/2-1/2cosα
=| sin(α/2) |
∵π/2〈α/2〈3π/4 ∴ sin(α/2)>0,
√1/2+1/2√1/2+1/2cos2α=sin(α/2)
求证:cos2αcos2β=1/2{cos2(α+β)+cos2(α-β)}
化简√(2+cos2+sin^2 1)
sin²α*sin²β+cos²α*cos²β-1/2cos2αcos2β=(1-cos2α)/2*(1-cos2β)/2+(1+cos2α)/2*(1+cos2β)/2-1/2cos2α*cos2β=1/4(1+cos2α*cos2β-cos2α-cos2β)+1/4(1+cos2α*cos2β+cos2α+cos2β)-1/2cos2α*cos2β=1/4+1/
cos2次方1°加cos2次方2°加cos2次方3°.cos2次方88°+cos2次方89°
化简√ (2+ cos2-sin平方1)的结果是
1+cos2θ+2sin²θ=2 1-cos2θ/1+cos2θ=tan²θ
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(2sin2α/1+cos2α)*(cosα)^2/cos2α=?
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(2sin2α/1+cos2α)*(cos^2/cos2α)=?
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