Question

Sendo sen x=1/4, determine: (x ∈ III Q)
a)Tg x=

b)Sec x=

c)Cossec x=

Answer 1

$\Huge \bf Resolucao:$

Bem, podemos dizer que uma razão trigonométrica advêm de um triângulo retângulo. Como pode perceber, não temos um valor tabelado para "seno=1/4", dessa forma façamos um triângulo retângulo que corresponda a razão trigonométrica "seno=1/4".

$\huge \begin{array}{l} \bf Sendo~a~razao~trigonometrica~seno= \frac{cat. op}{hip}\end$

$\Huge \begin{array}{l} \boxed{\boxed{\frac{cat.op}{hip.} = \frac{1}{4}}} \\ \\ Dessa~forma~: \\ \\ Cat.op = 1 \\ \\ Hip=4 \end$

$\Large \begin{array}{l} ~~~| \backslash \\~~~ | ~~\backslash \\~~~ |~~~~ \backslash \\~~~ |~~~~~~ \backslash \\~1| ~~~~~~~~\backslash ~~4 \\~~~ | ~~~~~~~~~~\backslash \\ ~~~| ~~~~~~~~~ ~~ ~\backslash \\~~ ~ |.................\backslash \\~~~~~~ ~~~ ~ X}\end$


$\huge \bf Para descobrir o lado "x", basta aplicar pitagoras. \\ \\ 4^{2}=1^{2}+x^{2} \\16=1+x^2 \\ x^{2}=15 \\ x =\sqrt{15}$

$\Large \begin{array}{l} ~~~| \backslash \\~~~ | ~~\backslash \\~~~ |~~~~ \backslash \\~~~ |~~~~~~ \backslash \\~1| ~~~~~~~~\backslash ~~4 \\~~~ | ~~~~~~~~~~\backslash \\ ~~~| ~~~~~~~~~ ~~ ~\backslash \\~~ ~ |.................\backslash \\~~~~~~ ~~ \sqrt{15}\end$

$\Large \begin{array}{l} Sendo~assim:\end$

$\Huge \begin{array}{l} Tgx = \frac{1}{ \sqrt{15} } \\ \\ \boxed{Tgx = \frac{ \sqrt{15} }{15}} \end$
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$\Huge \begin{array}{l} secx = \frac{1}{cosx } \\ \\ secx = \frac{1}{\frac{cat.adj.}{hip.}} \\ \\ secx = \frac{hip.}{cat.adj} \\ \\ secx = \frac{4}{\sqrt{15} } \\ \\ \boxed{secx = \frac{4 \sqrt{15} }{15}} \end$

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$\Huge \begin{array}{l} cossecx = \frac{1}{senx} \\ \\ cossecx = \frac{1}{\frac{1}{4}} \\ \\ \boxed{cossecx = 4}\end$