Question

Ştiind că

$x\in(0,\frac{\pi}{2}) \: \: si \: \: tgx = \frac{1}{2}$

calculați

$\frac{cos2x}{1 + sin2x}$
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Answer 1

Răspuns:

Explicație pas cu pas:

$tg~x=\dfrac{1}{2}\Rightarrow \dfrac{\sin x}{\cos x}=\dfrac{1}{2}\\\texttt{Avem ca:}\\\dfrac{\cos(2x)}{1+\sin(2x)}=\dfrac{\cos^2x-\sin^2x}{\sin^2x+2\sin x\cos x+\cos^2x}=\dfrac{\cos^2x\left(1-\dfrac{\sin^2x}{\cos^2x}\right)}{\cos^2x\left(\dfrac{\sin^2x}{\cos^2x}+2\dfrac{\sin x}{\cos x}+1\right)}\\=\dfrac{1-\left(\dfrac{1}{2}\right)^2}{\left(\dfrac{1}{2}\right)^2+2\cdot\dfrac{1}{2}+1}=\dfrac{1-\dfrac{1}{4}}{\dfrac{1}{4}+1+1}=\dfrac{\dfrac{3}{4}}{\dfrac{9}{4}}=\dfrac{3}{9}=\boxed{\dfrac{1}{3}}$