Question

Дано: tg a=5, 0 Найти: a) cos a=?, б) sin(a-p/3), в)tg2a

Answer 1

Ответ:

$tga=5\ \ ,\ \ \ 0<a<\dfrac{\pi}{2}\\\\\\a)\ \ 1+tg^2a=\dfrac{1}{cos^2a}\ \ \Rightarrow \ \ \ cos^2a=\dfrac{1}{1+tg^2a}=\dfrac{1}{1+25}=\dfrac{1}{26}\\\\\\0<a<\dfrac{\pi}{2}\ \ \ \Rightarrow \ \ \ cosa>0\ \ ,\ \ cosa=\dfrac{1}{\sqrt{26}}\\\\\\b)\ \ 0<a<\dfrac{\pi }{2}\ \ \Rightarrow \ \ sina>0\ \ ,\\\\sina=\sqrt{1-cos^2a}=\sqrt{1-\dfrac{1}{26}}=\sqrt{\dfrac{25}{26}}=\dfrac{5}{\sqrt{26}}$

$sin\Big(a-\dfrac{\pi }{3}\Big)=sina\cdot cos\dfrac{\pi}{3}-sin\dfrac{\pi}{3}\cdot cosa=\dfrac{5}{\sqrt{26}}\cdot \dfrac{1}{2}-\dfrac{\sqrt3}{2}\cdot \dfrac{1}{\sqrt{26}}=\\\\=\dfrac{5-\sqrt3}{2\sqrt{26}}\\\\\\c)\ \ tg2a=\dfrac{2\, tga}{1-tg^2a}=\dfrac{2\cdot 5}{1-25}=-\dfrac{10}{24}=-\dfrac{5}{12}$