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Putin ajutor, va rog?

Se considera x, y, z apartin R astfel incat x+y+z=180° (Pi). Sa se demonstreze egalitatea:

[tex]sinx+siny-sinz=4sin\frac{x}{2} *sin\frac{y}{2} *cos\frac{z}{2}[/tex]

Rezolvare completa, dau coroana.


Răspuns :

Explicație pas cu pas:

[tex]x+y+z = \pi = > x+y = \pi - z[/tex]

[tex]\sin(x) + \sin(y) - \sin(z) = 2 \sin(\frac{x + y}{2})\cos(\frac{x - y}{2}) - \sin(\pi - (x + y))= 2 \sin(\frac{x + y}{2})\cos(\frac{x - y}{2}) - \sin(x + y) = 2 \sin(\frac{x + y}{2})\cos(\frac{x - y}{2}) - 2 \sin(\frac{x + y}{2})\cos(\frac{x + y}{2}) = 2 \sin(\frac{x + y}{2})(\cos(\frac{x - y}{2}) - \cos(\frac{x + y}{2})) = - 2 \sin(\frac{x + y}{2}) \times 2 \sin( \frac{x}{2}) \sin( - \frac{y}{2}) = 4 \sin(\frac{x}{2}) \sin( \frac{y}{2}) \sin( \frac{\pi - z}{2}) = 4 \sin(\frac{x}{2}) \sin( \frac{y}{2}) \sin( \frac{\pi}{2} - \frac{z}{2}) = 4 \sin(\frac{x}{2}) \sin( \frac{y}{2})\sin(\frac{z}{2})[/tex]

Răspuns:

Succes! Sper ca este bine!

Vezi imaginea DARIACORINA5