Question

Cum calculez $\cos( \frac{45}{2} )$ ?

Answer 1

Salut,

cos(2x) = 2cos²x -- 1, deci cos²x = (1+cos(2x))/2.

$cosx=\pm\sqrt{\dfrac{1+cos(2x)}{2}}.\ Pentru\ x=\dfrac{45^{\circ}}2<90^{\circ}\ avem\ c\breve{a}:\\\\\\cos\left( \dfrac{45^{\circ}}2\right)=+\sqrt{\dfrac{1+cos45^{\circ}}{2}}=\sqrt{\dfrac{1+\dfrac{\sqrt2}2}{2}}=\dfrac{\sqrt{2+\sqrt2}}{2}.$

Green eyes.

Answer 2

$\cos(2x) = \cos^2 x - \sin ^2 x = \cos^2 x + \cos^2 x - \cos^2 x - \sin^2 x = \\ = 2\cos^2 x-(\cos^2 x + \sin^2 x) = 2\cos^2 x - 1 \\ \\ \\ \\ \cos(2x) = 2\cos^2 x - 1\Rightarrow 2\cos^2(x) = \cos(2x) + 1 \Rightarrow \\ \\ \Rightarrow \cos^2 (x) = \dfrac{\cos(2x)+1}{2} \Rightarrow \cos^2\Big(\dfrac{x}{2}\Big) = \dfrac{\cos\Big(2\cdot \dfrac{x}{2}\Big)+1}{2} \Rightarrow$

$\Rightarrow \cos^2\Big(\dfrac{x}{2}\Big) = \dfrac{\cos(x) + 1}{2} \Rightarrow \boxed{\cos\Big(\dfrac{x}{2}\Big) = \pm \sqrt{\dfrac{\cos(x)+1}{2}}} \\ \\ \\ \dfrac{45^{\circ}}{2} =22.5^{\circ} \in (0,90^\circ) \Rightarrow \cos\Big(\dfrac{45^{\circ}}{2}\Big) \ \textgreater \ 0 \\ \cos\Big(\dfrac{45^{\circ}}{2}\Big) = + \sqrt{\dfrac{\cos(45^\circ) + 1}{2}} = \sqrt{\dfrac{\dfrac{\sqrt 2}{2} + 1}{2}}} = \\ \\ = \sqrt{\dfrac{\dfrac{\sqrt 2 + 2}{2}}{2}} = \sqrt{\dfrac{\sqrt 2 + 2}{4}} = \boxed{\dfrac{\sqrt{2+\sqrt 2}}{2}}$