[tex]sin \frac{x-y}{2} sin \frac{y-z}{2} sin \frac{z-x}{2} = \frac{1}{4} sin(x-y)+sin(y-z)+sin(z-x)=?[/tex]. Created by Ilyasssssss. algebra-ru

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July 2022 1 3 Report

[tex]sin \frac{x-y}{2} sin \frac{y-z}{2} sin \frac{z-x}{2} = \frac{1}{4}

sin(x-y)+sin(y-z)+sin(z-x)=?[/tex]

Answers & Comments

Удачник66 Есть такие формулы преобразования произведения в сумму:
sin a*sin b = 1/2*(cos(a-b) - cos(a+b))
sin a*cos b = 1/2*(sin(a+b) + sin(a-b))
Умножаем по порядку
1) $sin \frac{x-y}{2}sin \frac{y-z}{2}= \frac{1}{2}(cos \frac{x-y-y+z}{2} -cos \frac{x-y+y-z}{2} )=$
$= \frac{1}{2}(cos \frac{x-2y+z}{2}-cos \frac{x-z}{2} )$
2) $sin \frac{z-x}{2}*\frac{1}{2}(cos \frac{x-2y+z}{2}-cos \frac{x-z}{2} )=\frac{1}{2}sin \frac{z-x}{2}cos \frac{x-2y+z}{2}-$
$-\frac{1}{2}sin \frac{z-x}{2}cos \frac{x-z}{2}= \frac{1}{4}(sin \frac{z-x+x-2y+z}{2}+sin \frac{z-x-x+2y-z}{2}) - \frac{1}{4}sin(z-x)$
$= \frac{1}{4}(sin(z-y)+sin(y-x)-sin(z-x))= \frac{1}{4}$
sin(z-y) + sin(y-x) - sin(z-x) = -sin(x-y) - sin(y-z) - sin(z-x) = 1
sin(x-y) + sin(y-z) + sin(z-x) = -1

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