证明(sinα)^2+(sin(120°-α))^2-2sinαsin(120°-α)cos60°=(sin60°)^2证明(sinα)^2+(sin(120°-α))^2-2sinαsin(120°-α)cos60°=(sin60°)^2

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证明(sinα)^2+(sin(120°-α))^2-2sinαsin(120°-α)cos60°=(sin60°)^2证明(sinα)^2+(sin(120°-α))^2-2sinαsin(120°

证明(sinα)^2+(sin(120°-α))^2-2sinαsin(120°-α)cos60°=(sin60°)^2证明(sinα)^2+(sin(120°-α))^2-2sinαsin(120°-α)cos60°=(sin60°)^2
证明(sinα)^2+(sin(120°-α))^2-2sinαsin(120°-α)cos60°=(sin60°)^2
证明(sinα)^2+(sin(120°-α))^2-2sinαsin(120°-α)cos60°=(sin60°)^2

证明(sinα)^2+(sin(120°-α))^2-2sinαsin(120°-α)cos60°=(sin60°)^2证明(sinα)^2+(sin(120°-α))^2-2sinαsin(120°-α)cos60°=(sin60°)^2
左边=sin²a+sin²(a-60°)-2sinasin(a-60°)cos60°
=〔(sina-sin(a-60°)〕²+sinasin(a-60°)
=cos²(a-30°)-½〔cos(2a-60°)-cos60°〕
=½〔(cos2a-60°)+1〕-½cos(2a-60°) +½cos60°=¾=(sin60°)²=右边.
证毕.

步骤如下:
因为:sin(120°-α))与sin(120°-α)抵消则得出
(sinα)^2+sin-2sinα=(sin60°)^2