Question

Найдите tgα, если sinα = 1/6 и 0 < < π/2

Answer 1

Ответ:

$sina=\dfrac{1}{6}\\\\cos^2a=1-sin^2a=1-\dfrac{1}{36}=\dfrac{35}{36}\\\\0<a<\dfrac{\pi}{2}\ \ \ \Rightarrow \ \ \ cosa>0\ ,\ \ cosa=+\dfrac{\sqrt{35}}{6}\\\\tga=\dfrac{sina}{cosa}=\dfrac{\frac{1}{6} }{\frac{\sqrt{35}}{6}}=\dfrac{1}{\sqrt{35}}=\dfrac{\sqrt{35}}{35}$