Question

exercitiul 6 ! puncte multe si coroana ​

Answer 1

Explicație pas cu pas:

$\tan(x) = \sqrt{3} ,0 < x < \frac{\pi}{2}$

sin²(x) + cos²(x) = 1 => sin²(x) = 1 - cos²(x)

cos(2x) = 2cos²(x) - 1

$\frac{ \sin(x) }{ \cos(x) } = \sqrt{3} = > (\frac{ \sin(x) }{ \cos(x) })^{2} = {( \sqrt{3} )}^{2} \\ \frac{ \sin^{2} (x) }{ \cos^{2} (x) } = 3 = > 1 - \cos^{2} (x) = 3 \cos^{2} (x) \\ 4\cos^{2} = 1 = > 4\cos^{2} - 2 = 1 - 2 \\ 2(2\cos^{2} - 1) = - 1 = > 2 \cos(2x) = - 1 \\ = > \cos(2x) = - \frac{1}{2}$