Trigonometri İspatları-1

Question

Bir ABC üçgeninde    a+b+c=180  olmak üzere

$sin\dfrac{a}{2}+sin\dfrac{b}{2}+sin\dfrac{c}{2}-1=4.sinH.sinJ.sinK$      oluyorsa

$H,J,K$  nedir?

Comments

0 olması lazım ^^

Answer 1

Sol taraftan başlayalım, $$[sin(a/2)+sin(b/2)]+[sin(c/2)-sin(\pi/2)]=$$

$$2sin(\frac{a/2+b/2}{2}).cos(\frac{a/2-b/2}{2})+2sin(\frac{c/2-\pi/2}{2}).cos(\frac{c/2+\pi/2}{2})=$$ $$2sin(\frac{\pi/2-c/2}{2}).cos(\frac{a/2-b/2}{2})+2sin(\frac{c/2-\pi/2}{2}).cos(\frac{c/2+\pi/2}{2})=$$

$$2sin(\frac{\pi/2-c/2}{2}).\left[cos(\frac{a/2-b/2}{2})-cos(\frac{c/2+\pi/2}{2})\right]=$$

$$2sin(\frac{\pi/2-c/2}{2}).\left[-2.sin(\frac{a/2-b/2+c/2+\pi/2}{4}).sin(\frac{a/2-b/2-c/2-\pi/2}{4})\right]=$$

$$-4sin(\frac{\pi/2-c/2}{2}).\left[sin(\frac{\pi-b}{4}).sin(\frac{-b-c}{4})\right]=$$

$$4sin(\frac{\pi-c}{4}).sin(\frac{\pi-b}{4}).sin(\frac{b+c}{4})=4sin(\frac{\pi-c}{4}).sin(\frac{\pi-b}{4}).sin(\frac{\pi-a}{4})$$ olacaktır.

DİKKAT İşlem hatası içerebilir.

Comments

hata yok, tebrikler hocam.