Question

求高手：求证三角形和差化积公式，只写其中一个即可

如，sinθ+sinφ=2sin[(θ+φ)/2]cos[(θ-φ)/2] ，写证明过程，简写的就不麻烦您动手了，采纳有奖

Answer 1

证明θ=(θ+φ)/2+(θ-φ)/2
φ=(θ+φ)/2-(θ-φ)/2
sinθ+sinφ=sin[(θ+φ)/2+(θ-φ)/2]+sin[(θ+φ)/2-(θ-φ)/2]
=sin(θ+φ)/2*cos(θ-φ)/2+cos(θ+φ)/2*sin(θ-φ)/2+sin(θ+φ)/2*cos(θ-φ)/2-cos(θ+φ)/2*sin(θ-φ)/2
=sin(θ+φ)/2*cos(θ-φ)/2+sin(θ+φ)/2*cos(θ-φ)/2+cos(θ+φ)/2*sin(θ-φ)/2-cos(θ+φ)/2*sin(θ-φ)/2
=sin(θ+φ)/2*cos(θ-φ)/2+sin(θ+φ)/2*cos(θ-φ)/2
=2sin(θ+φ)/2*cos(θ-φ)/2

Answer 2

θ=[(θ+φ)/2]+[(θ-φ)/2]
φ=[(θ+φ)/2]-[(θ-φ)/2]
sinθ+sinφ=sin{[(θ+φ)/2]+[(θ-φ)/2]}+sin{[(θ+φ)/2]-[(θ-φ)/2]}
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