二倍角公式

[(cot A)^2 - 1] sin 4A / [(1 + cos 4A) cot A]
= [(cot A)^2 - 1] (2 sin 2A cos 2A) / {[1 + 2(cos 2A)^2 - 1] cot A}
= [(cot A)^2 - 1] (4 sin A cos A cos 2A) / [2(cos 2A)^2 cot A]
= 2 [(cot A)^2 - 1] sin A cos A / (cos 2A cot A)
= 2 [(cot A)^2 - 1] (sin A)^2 / cos 2A
= 2 [(cos A)^2 - (sin A)^2] / cos 2A
= 2 cos 2A / cos 2A
= 2

(cot^2 A-1)sin4A / (1+cos4A)cotA
=(cos^2 A - sin^2 A)sin4A / (1+cos4A)sinAcosA [上下乘sin^2 A]
=(cos2A)[2sin2Acos2A] / (1+cos4A)[sin2A/2]
=4cos^2 2A / (1+cos4A)
=4cos^2 2A / (2cos^2 2A) [半角公式之一 : 1+cos2X=2cos^2 X]
=2
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你補充果部分既1/2係
(1) (1/2)(sin2θ+√3cos2θ)
定係
(2) 1/ [2(sin2θ+√3cos2θ)]