Question

Найдите cos x, если sin x = 12 /13 , 0 < x < пи/2

Answer 1

$sinx=\dfrac{12}{13}\\\\\\cos^2x=1-sin^2x=1-\dfrac{144}{169}=\dfrac{25}{169}\\\\\\0<x<\dfrac{\pi }{2} \ \ \ \Rightarrow \ \ cosx>0\ \ ,\ \ cosx=+\sqrt{\dfrac{25}{169}}=\dfrac{5}{13}$