Estude! Com Prof. Caju

Calcular [tex3]\text{sen} 2x[/tex3] sabendo que [tex3]\text{tg}x + \text{cotg}x = 3.[/tex3]

No membro esquerdo:

• [tex3]\text{sen}\,2x\,=\,2\,\text{sen}\,x\,\cos\,x\,=\,2\,\text{tg}\,x\,\cos^2\,x\,=\,\Large\frac{2\text{tg}\,x}{\text{sec}^2\,x}\large\,=\,\Large\frac{2\,\text{tg}\,x}{1\,+\,\text{tg}^2\,x}\large[/tex3]

No membro direito:

• [tex3]\text{tg}\,x\,+\,\Large\frac{1}{\text{tg}\,x}\large\,=\,\Large\frac{\text{tg}^2\,x\,+\,1}{\text{tg}\,x}\large\,\,\Longleftrightarrow\,\,\Large\frac{\text{tg}^2\,x\,+\,1}{2\text{tg}\,x}\large\,=\,\Large\frac{3}{2}[/tex3]

Assim o valor procurado é [tex3]\Large\frac{2}{3}[/tex3]

Obrigado Triple!

Abraço!

Opa,

• [tex3]\text{sen} 2x\Rightarrow \,\text{sen}(x+x)=2. cos x.\text{sen}x\text{ } \boxed{i}[/tex3]

Como [tex3]\text{tg} x=\frac{\text{sen} x}{cos x},\,\text{cotg} x=\frac{cos x}{\text{sen} x} \text{ e } \text{sen}^{2}x+cos^{2}x=1[/tex3]

temos,

• [tex3]\frac{\text{sen} x}{cos x}+\frac{cos x}{\text{sen} x}=3\Rightarrow\,\frac{\text{sen}^{2}x + cos^{2}}{\cancel{\text{sen} x . cos x}}=3.\text{sen}x. cos x\Rightarrow\,\text{sen}x . cos x=\frac{1}{3}\text{ }\boxed{ii}[/tex3]

Agora substituindo [tex3](ii)[/tex3] em [tex3](i)[/tex3] vem que,

• [tex3]\text{sen} 2x=2.\frac{1}{3}=\frac{2}{3}[/tex3]

Abraço e t+