Question

Запишите выражения в виде произведения:

1+cos 9 градусов =
1+-ctb =
корень3 + tga=

Answer 1

Ответ:

$1+cos9^\circ =2\cdot cos^2\, (4,5^\circ )\\\\\\\\1+ctga=1+\dfrac{cosa}{sina}=\dfrac{sina+cosa}{sina}=\dfrac{1}{sina}\cdot (\, sina+sin(\dfrac{\pi}{2}-a)\, )=\\\\=\dfrac{1}{sina}\cdot 2\, sin\dfrac{\pi}{4}\cdot cos(a-\dfrac{\pi}{4})=\dfrac{\sqrt2}{sina}\cdot cos(a-\dfrac{\pi}{4})$

$\sqrt3+tga=tg\dfrac{\pi}{3} +tga=\dfrac{sin(\frac{\pi }{3}+a)}{cos\frac{\pi}{3} \cdot cosa}=\dfrac{sin(\frac{\pi }{3}+a)}{cos\frac{\pi}{3} \cdot cosa}=\dfrac{sin(\frac{\pi }{3}+a)}{\frac{1}{2} \cdot cosa}=\\\\\\=2\, sin(\dfrac{\pi }{3}+a)\cdot cosa$