Задание:
Sin (a-2π/3) -sin (a+2π/3)
Решение:
sin (a-2π/3) -sin (a+2π/3)=2sin (a-2π/3-a-2π/3) /2)*cos (a-2π/3+a+2π/3) /2)=2sin (-4π/6)*sin (2a/2)=-2sin (2π/3)*cosa=-2*sqrt{3}/2*cosa=-sqrt{3}*cosa
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Sin (a-2π/3) -sin (a+2π/3)
sin (a-2π/3) -sin (a+2π/3)=2sin (a-2π/3-a-2π/3) /2)*cos (a-2π/3+a+2π/3) /2)=2sin (-4π/6)*sin (2a/2)=-2sin (2π/3)*cosa=-2*sqrt{3}/2*cosa=-sqrt{3}*cosa
Знаете другой ответ?