Pré-Vestibular ⇒ (UFRG) Trigonometria

[botelho]

Sinplifique a espressão [[tex3]\frac{cos^{n-1}(x)-cos^{n+1}(x)]}{sen^{n-1}(x)-sen^{n+1}(x)}[/tex3] . [tex3]tan^{n}(x)[/tex3].
a)sec³x
b)tan³x
c)cot³x
d)sen³x
e)cos³x

Resposta

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b

[deBroglie]

Olá, botelho.

=>[tex3]\frac{cos^{n-1}(x)-cos^{n+1}(x)]}{sen^{n-1}(x)-sen^{n+1}(x)}[/tex3].[tex3]tan^{n}(x)[/tex3] =
[tex3]\frac{\frac{cos^n(x)}{cos(x)}-cos^n(x).cos(x)}{\frac{sen^n(x)}{sen(x)}-sen^n(x).sen(x)}.tg^n(x)[/tex3] =
[tex3]\frac{cos^n(x)-cos^n(x).cos^2(x)}{sen^n(x)-sen^n(x).sen^2(x)}.tg(x).tg^n(x)[/tex3] =
[tex3]\frac{cos^n(x).(1-cos^2(x))}{sen^n(x).(1-sen^2(x))}.tg(x).tg^n(x)[/tex3] =
[tex3]\frac{1}{tg^n(x)}.\frac{sen^2(x)}{cos^2(x)}.tg(x).tg^n(x)= \frac{sen^2(x)}{cos^2(x)}.\frac{sen(x)}{cos(x)}=\frac{sen^3(x)}{cos^3(x)}=tg^3(x)[/tex3]