IME / ITA ⇒ (IME) Trigonometria Tópico resolvido

[MJ14]

Prove a identidade

[tex3]\frac {cos^2x}{cot\frac {x}{2}-tan\frac{x}{2}}= \frac{1}{4}sen2x[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac {cos^2x}{cot\frac {x}{2}-tan\frac{x}{2}}= \frac{1}{4}sen2x[/tex3]

[PedroCunha]

Olá, Mariana.

Código: Selecionar tudo

[tex3]\frac{\cos ^2x}{\cot \frac{x}{2} - \tan \frac{x}{2}} = \frac{\sin (2x)}{4} \therefore \frac{\cos^2x}{\frac{\cos \frac{x}{2}}{\sin \frac{x}{2}} - \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}} = \frac{\sin (2x)}{4} \therefore \frac{\cos ^2x}{\frac{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}}{\sin \frac{x}{2} \cdot \cos \frac{x}{2}}} = \frac{\sin (2x)}{4} \therefore \\\\ \frac{\frac{\sin x}{2} \cdot \cos ^2x}{\cos^2 \frac{x}{2} - (1 - \cos^2 \frac{x}{2})} = \frac{\sin (2x)}{4} \therefore \frac{2\sin x \cdot \cancelto{\cos x}{\cos ^2x}}{\cancel{\cos x}} = \sin (2x)[/tex3]

Código: Selecionar tudo

[tex3]\boxed{\boxed{ \sin (2x) = \sin(2x), C.Q.P. }}[/tex3]

Qualquer dúvida é só falar.

Abraços,
Pedro