sin20°+sin40°-sin100°. Created by irayoeywaw. algebra-ru

sin20°+sin40°-sin100°...

$Sin20^{0}+Sin40^{0}-Sin100^{0}=2Sin\frac{20^{0}+40^{0}}{2}Cos\frac{20^{0} -40^{0}}{2}-Sin(90^{0}+10^{0})=2Sin30^{0}Cos10^{0}-Cos10^{0}=2*\frac{1}{2}Cos10^{0}-Cos10^{0}=Cos10^{0}-Cos10^{0}=0$

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Ответ:

Объяснение:

2sin(30°)cos(-10°)-sin(90°+10°)=2*1/2*cos(10°)-cos(10°)=cos(10°)-cjs(10°)=0

1/2 потрібно записати дрібю.

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