Question

Calculați cos 2x știind că x ∈ {0;$\frac{n}{2}$} și sinx = $\frac{ \sqrt{3} }{5}$

Answer 1

[tex]cox 2x = cox^{2}x - sin^{2}x \\ cox^{2}x = 1 - sin^2{x} = 1 - \frac{3}{25} = \frac{22}{25} \\ [/tex]

$cox 2x = \frac{22}{25} - \frac{3}{25} = \frac{19}{25}$


Cred ca asta e, nu sunt 100% sigur 

Answer 2

$\cos 2x = \cos^2 x-\sin^2x \\ \\ \cos 2x = (1-\sin^2x)-\sin^2 x \\ \\ \cos 2x = 1-\sin^2 x-\sin^2 x \\ \\ \cos 2x = 1-2\sin^2 x \\ \\ \cos 2x = 1 - 2\cdot \Big(\dfrac{\sqrt3}{5}\Big)^2 \\ \\ \cos 2x = 1 - 2\cdot \dfrac{3}{25} \\ \\ \cos 2x = 1-\dfrac{6}{25} \\ \\ \cos 2x = \dfrac{25-6}{25} \\ \\ \boxed{\cos 2x = \dfrac{19}{25}}$