(UFGO) Simplificando a expressão $\,\dfrac{\operatorname{tg}a\,+\,\operatorname{tg}b}{\operatorname{cotg}a\,+\,\operatorname{cotg}b}\,$, obtém-se:
a)
$\,\operatorname{tg}a \centerdot \operatorname{tg}b\,$
b)
$\,\operatorname{cotg}a \centerdot \operatorname{cotg}b\,$
c)
$\,\operatorname{tg}(a\,+\,b)\,$
d)
$\,\operatorname{cotag}(a\,+\,b)\,$
e)
$\,\operatorname{tg}a \centerdot \operatorname{cotg}b\,$
✓ mostrar resposta ... Calcular o lado $\;a\;$ de um triângulo $\;ABC\;$ sabendo-se que $\;\hat{B}\,=\,60^o\,\text{, } \hat{C}\,=\,45^o \;\text{ e }\; \overline{AB}\,=\, 2\text{ m}$.
✓ mostrar resposta ... resposta:
Resolução : $\,\triangle ADB \left\{ \operatorname{sen}60^o \,=\,{\large \frac{h}{2}}\; \Rightarrow \;h\,=\,\sqrt{3} \text{m.}\right.\,$ Então $\,BD^2 + (\sqrt{3})^2\,=\,2^2 \;\Rightarrow\;BD\,=\,1\text{m.}\,$ $\,\triangle ADC \left\{ \operatorname{tg}45^o \,=\,{\large \frac{\sqrt{3}}{CD}} \; \Rightarrow \; CD = \sqrt{3} \text{m.} \right.\,$ Logo: $\,a\,=\,BD\,+\,CD \;\Rightarrow\;$ $\boxed{\;a\,=\,(1\,+\,\sqrt{3})\text{ m.}\;}\,$
× (FMU - FIAM) O valor de $\,\operatorname{sen}x \,+\,{\large \frac{\operatorname{sen}^3 x}{2}} \,+ \, {\large \frac{\operatorname{sen}^5 x}{4}} \,+\,...\;$ é:
a)
$\,\dfrac{\operatorname{sen}x}{1\,+\,\operatorname{sen^2}x}\,$
b)
$\,\dfrac{\operatorname{cos}x}{1\,-\,\operatorname{sen^2}x}\,$
c)
$\,\dfrac{\operatorname{sen}x}{1\,+\,\operatorname{cos^2}x}\,$
d)
$\,\dfrac{\operatorname{sen}x}{1\,-\,\operatorname{sen^2}x}\,$
e)
$\,\dfrac{2\operatorname{sen}x}{1\,+\,\operatorname{cos^2}x}\,$
✓ mostrar resposta ... (VUNESP) Se $\;x \,\text{, }\;y\;$ são números reais tais que:
$\,y\,=\, \dfrac{ \operatorname{cos^3}x \,-\, 2 \, \centerdot \,\operatorname{cos}x \,+\, \operatorname{sec}x }{ \operatorname{cos}x \, \centerdot \,\operatorname{sen^2}x } \;$, então:
a)
$\,y\,=\,\operatorname{sec^2}x\,$
b)
$\,y\,=\,\operatorname{tg^2}x\,$
c)
$\,y\,=\,\operatorname{cos^2}x\,$
d)
$\,y\,=\,\operatorname{cossec^2}x\,$
e)
$\,y\,=\,\operatorname{sen^2}x\,$
✓ mostrar resposta ... (VUNESP) Sejam $\;A\;$, $B$ e $C \;$ conjuntos de números reais. Sejam $\;f\,:\, A \rightarrow B \;$ e $\;g\,:\, B \rightarrow C \;$ definidas, respectivamente, por:
$\left\{\begin{array}{rcr} \,f(x)\, &=\,\operatorname{sen}x \text { , } \vee \negthickspace \negthickspace \negthickspace \negthinspace - x \text{ , }\,x \,\in\, \, A\,\; \phantom{XX} \\ \,g(x)\,&=\,{\Large \frac{1}{1\,-\,x^2}} - 1 \text{ , }\vee \negthickspace \negthickspace \negthickspace \negthinspace - x \text{ , }\,x \,\in\, \, B \\ \end{array} \right.\,$
Se existe $\;f\,:\, A \rightarrow C \;$, definida por $\,h(x)\,=\,g{\large [f(x)]} \text{, }\vee \negthickspace \negthickspace \negthickspace \negthinspace - x \text{, }\,x \,\in\, \, A\;$, então:
a)
$\,h(x)\,=\,\operatorname{cos}x\,$
b)
$\,h(x)\,=\,\operatorname{cos^2}x\,$
c)
$\,h(x)\,=\,\operatorname{tg^2}x\,$
d)
$\,h(x)\,=\,\operatorname{sen^2}x\,$
e)
$\,h(x)\,=\,\operatorname{sec^2}x\,$
✓ mostrar resposta ... Na figura, calcular $\,h\;$ e $\,d\,$.
✓ mostrar resposta ... resposta:
Resolução :
$\,\triangle BCD \left\{ \operatorname{tg}60^o \,=\,{\large \frac{h}{d}} \; \Rightarrow \; h\,=\,d\sqrt{3} \right.\,$
$\,\triangle ACD \left\{ \operatorname{tg}30^o \,=\,{\large \frac{h}{d\,+\,40}} \; \Rightarrow \; h\,=\,\frac{\sqrt{3}(d\,+\,40)}{3} \right.\,$
Então $\,d\sqrt{3}\,=\,\frac{\sqrt{3}(d\,+\,40)}{3} \,\Rightarrow\; d\,=\,20\,m$
e portanto $\;h\,=\,20\sqrt{3}\,m\,$
Resposta : $\; \boxed{ d\,=\,20\,m}\;\;\boxed{h\,=\,20\sqrt{3}\,m}$
× Sabendo-se que $\;\hat{x}\;$ é um ângulo agudo e que $\;\operatorname{tg}\hat{x}\,=\,{\large \frac{5}{12}}\;$, calcule o $\,\operatorname{sen}\hat{x}\,$
✓ mostrar resposta ... resposta:
Resolução :
$\,\operatorname{sen^2}x \,=\,{\large \frac{\operatorname{tg^2}x}{1\,+\,\operatorname{tg^2}x}}\; \Rightarrow \operatorname{sen^2}x \,=\,{\large \frac{\frac{25}{144}}{1\,+\,\frac{25}{144}}} \,=\,\frac{25}{169}$
Então $\,\boxed{\operatorname{sen}x\,=\,\frac{5}{13}}\;\text{ (para x agudo) }$
× Calcular $\,y\,=\,{\Large \frac{\operatorname{cos}x\,-\,\operatorname{sec}x}{\operatorname{sen}x\,-\,\operatorname{cossec}x}}\;$, sabendo que $\,\operatorname{tg}x\,=\,3\;$.
✓ mostrar resposta ... resposta:
Resolução :
$\,y\,=\, {\large \frac{\operatorname{cos}x\,-\,\frac{1}{\operatorname{cos}x}}{\operatorname{sen}x\,-\,\frac{1}{\operatorname{sen}x } }}\,=\,
{\Large \frac{ \frac{ \operatorname{cos^2}x\,-\,1}{\operatorname{cos}x}}{\frac{\operatorname{sen^2}\,-\,1}{\operatorname{sen}x}} }\,=\,$ $
{\Large \frac{ - \frac{\operatorname{sen^2}x}{\operatorname{cos}x} } {- \frac{\operatorname{cos^2}x }{\operatorname{sen}x } } } \,=\,$ $
{\Large \frac{\operatorname{sen^3}x }{\operatorname{cos^3}x} \,=\,\operatorname{tg^3}x}$
Então $\,\boxed{y\,=\,3^3\,=\,27}\,$
× Simplificar a expressão: $\,y\,=\,{\large \frac{\operatorname{cos^3}a \,-\,\operatorname{sen^3}a}{1\,+\,\operatorname{sen}a \;\centerdot\; \operatorname{cos}a } }\;$.
✓ mostrar resposta ... resposta:
Resolução :
$\,y\,=\,{\large \frac{(\operatorname{cos}a - \operatorname{sen}a)(\operatorname{cos^2}a\,+\,\operatorname{cos}a \;\centerdot\; \operatorname{sen}a\,+\operatorname{sen^2}a)}{(1\,+\,\operatorname{sen}a \;\centerdot\; \operatorname{cos}a)} } \,=\,$
$\,=\,{\large \frac{(\operatorname{cos}a\,-\,\operatorname{sen}a)(1\,+\,\operatorname{sen}a \;\centerdot\; \operatorname{cos}a)}{(1\,+\,\operatorname{sen}a \;\centerdot\; \operatorname{cos}a)}}\,=\,\boxed{\operatorname{cos}a\,-\,\operatorname{sen}a}$
× (PUC) Qual é o valor de$\phantom{X}{\large x}\phantom{X}$na figura ao lado?
a)
${\large\frac{\sqrt{2}}{3}}$
b)
${\large\frac{5\sqrt{3}}{3}}$
c)
${\large\frac{10\sqrt{3}}{3}}$
d)
${\large\frac{15\sqrt{3}}{4}}$
e)
${\large\frac{20\sqrt{3}}{3}}$
✓ mostrar resposta ...