Question

Fie x aparține (0;pi/2), astfel încât tgx+ctgx=3. Arătați ca: 9(sin2x+cos4x)=7

Answer 1

$x\in \left(0,\dfrac{\pi}{2}\right)\,\,\,\text{a.i.}\,\,\,\text{tg}\,x+\text{ctg}\,x = 3 \Rightarrow \\ \\\Rightarrow \dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x}=3 \Rightarrow \dfrac{\sin^2 x+\cos^2 x}{\sin x\cos x} = 3 \Rightarrow \\ \\ \Rightarrow 2\cdot \dfrac{1}{2\sin x\cos x} = 3\Rightarrow \dfrac{2}{\sin 2x} = 3 \Rightarrow \sin 2x = \dfrac{2}{3}$

$\\9(\sin 2x+\cos 4x) =\\ \\ = 9(\sin 2x+\cos^2 2x-\sin^2 2x)\\ \\ = 9\left[\sin 2x+(1-\sin^2 2x)-\sin^2 2x\right]\\ \\ = 9\cdot \left(\dfrac{2}{3}+1-\dfrac{4}{9}-\dfrac{4}{9}\right)\\ \\ = 9\cdot \left(\dfrac{6+9-8}{9}\right)\\ \\ =9\cdot \dfrac{7}{9}\\ \\= \boxed{7}$