Ensino Médio ⇒ Trigonometria - arco metade Tópico resolvido

[janson]

Dada [tex3]tg\frac{x}{2}[/tex3], calcule [tex3]sen\;x, cos\;x \;\;e\;\;tg\;x[/tex3].

[theblackmamba]

Tendo em mãos a fórmula de arco duplo:
[tex3]\cos (2a)=2\cos ^2 a-1[/tex3]

Fazendo [tex3]a=\frac{x}{2}[/tex3] temos,

[tex3]\cos x =2\cos ^2 \(\frac{x}{2}\)-1[/tex3]
[tex3]\cos \(\frac{x}{2}\)=\pm \sqrt{\frac{1+\cos x }{2}}[/tex3]

Como [tex3]\cos ^2 x+\sen ^2x=1[/tex3] podemos reescrever:

[tex3]\cos x =1-2\sen ^2 \(\frac{x}{2}\)[/tex3]
[tex3]\sen \(\frac{x}{2}\)=\pm\sqrt{\frac{1-\cos x }{2}}[/tex3]. Logo,

[tex3]\tg \(\frac{x}{2}\)=\frac{\sqrt{\frac{1-\cos x }{2}}}{\sqrt{\frac{1+\cos x }{2}}}=\sqrt{\frac{1-\cos x }{1+\cos x }}[/tex3]

Eleve ao quadrado:
[tex3]\tg ^2\(\frac{x}{2}\)=\frac{1-\cos x }{1+\cos x }[/tex3]
[tex3]\tg ^2 \(\frac{x}{2}\)+\tg ^2 \(\frac{x}{2}\)\cdot \cos x =1-\cos x [/tex3]
[tex3]\cos x \[\tg ^2 \(\frac{x}{2}\)+1\] = 1-\tg ^2 \(\frac{x}{2}\)[/tex3]
[tex3]\boxed{\cos x =\frac{1-\tg ^2 \(\frac{x}{2}\)}{1+\tg ^2 \(\frac{x}{2}\)}}[/tex3]

Da fórmula fundamental:
[tex3]\sen x =\sqrt{1-\cos ^2x}[/tex3]
[tex3]\sen x =\sqrt{1-\left(\frac{1-\tg ^2 \(\frac{x}{2}\)}{1+\tg ^2 \(\frac{x}{2}\)}\right)^2}[/tex3]
[tex3]\sen x =\sqrt{\frac{\[1+\tg ^2 \(\frac{x}{2}\)\]^2-\[1-\tg ^2 \(\frac{x}{2}\)\]^2}{\[1+\tg ^2 \(\frac{x}{2}\)\]^2}}[/tex3]

Lembre que [tex3](a+b)^2-(a-b)^2=4ab[/tex3]. Sendo [tex3]a=1[/tex3] e [tex3]b=\tg ^2 \(\frac{x}{2}\)[/tex3]. Logo,

[tex3]\sen x =\sqrt{\frac{4\tg ^2 \(\frac{x}{2}\)}{\[1+\tg ^2 \(\frac{x}{2}\)\]^2}}[/tex3]
[tex3]\boxed{\sen x =\frac{2\tg \(\frac{x}{2}\)}{1+\tg ^2 \(\frac{x}{2}\)}}[/tex3]

Por último,
[tex3]\tg x =\frac{\frac{2\tg ^2 \(\frac{x}{2}\)}{\cancel{1+\tg ^2 \(\frac{x}{2}\)}}}{\frac{1-\tg ^2 \(\frac{x}{2}\)}{\cancel{1+\tg ^2 \(\frac{x}{2}\)}}}[/tex3]
[tex3]\boxed{\tg x =\frac{2\tg \(\frac{x}{2}\)}{1-\tg ^2 \(\frac{x}{2}\)}}[/tex3]