Question

ALGUÉM BOM EM MATEMÁTICA ME AJUDA POR FAVOR!!!

1- Indique os valores de sen, cos, tg, cotg, sec, cosec de um arco de 210°

2- Dado cos x = - 4/5, com π/2 < x < π, calcule sen x, tg x, sec x cosec x

Answer 1

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⠀⠀⠀☞ 1) sen(210º) = -1/2; cos(210º) = -√3/2; tg(210º) = √3/3; cotg(210º) = √3/3; sec(210º) = -2√3/3; cosec(210º) = -2; 2) sen(x) = 3/5; tg(x) = -3/4; sec(x) = -5/4; cosec(x) = 5/3. ✅  

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⠀⠀⠀⭐⠀Para realizar este exercício vamos rever o círculo trigonométrico.⠀⭐⠀

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⠀⠀⠀☔⠀Oi novamente, Georgie ✌. Vamos inicialmente identificar em qual quadrante o arco de 210º se encontra:

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   $\setlength{\unitlength}{0.95cm}\begin{picture}(6,5)\thicklines\put(0,0){\vector(1,0){5}}\put(0,0){\vector(0,1){5}}\put(0,0){\vector(-1,0){5}}\put(0,0){\vector(0,-1){5}}\put(4.8,0.2){x}\put(4.1,-0.4){cos(x)}\put(0.2,4.8){y}\put(-1.15,4.75){sen(x)}\bezier(-3,0)(-2.77,2.77)(0,3)\bezier(3,0)(2.77,2.77)(0,3)\bezier(-3,0)(-2.77,-2.77)(0,-3)\bezier(3,0)(2.77,-2.77)(0,-3)\put(-1.8,1.5){\Huge\bf{Q_2}}\put(0.9,1.5){\Huge\bf{Q_1}}\put(-1.7,-1.8){\Huge\bf{Q_3}}\put(0.9,-1.8){\Huge\bf{Q_4}}\put(-6,3.5){\dashbox{0.1}(3.4,1){ \sf 90^{\circ} < Q_2 < 180^{\circ} }}\put(3,3.5){\dashbox{0.1}(3.4,1){ \sf 0^{\circ} < Q_1 < 90^{\circ} }}\put(3,-4.2){\dashbox{0.1}(3.4,1){ \sf 270^{\circ} < Q_4 < 360^{\circ} }}\put(-6,-4.2){\dashbox{0.1}(3.4,1){ \sf 180^{\circ} < Q_3 < 270^{\circ} }}\bezier(0.7,0)(0.65,0.65)(0,0.7)\put(0.0,0.67){\line(7,28){0.38}}\bezier(-0.7,0)(-0.65,0.65)(0,0.7)\put(-0.29,0.4){\line(-4,-40){0.38}}\bezier(-0.7,0)(-0.65,-0.65)(0,-0.7)\put(0,-0.3){\line(-4,-22){0.38}}\bezier(0.7,0)(0.65,-0.65)(0,-0.7)\put(1.07,0){\line(-4,-31){0.38}}\put(5.7,0.2){\LARGE\text{ \mathbf {0^{\circ}} }}\put(5.4,-0.4){\LARGE\text{ \mathbf {360^{\circ}} }}\put(-0.4,5.5){\LARGE\text{ \mathbf {90^{\circ}} }}\put(-6.3,-0.1){\LARGE\text{ \mathbf {180^{\circ}} }}\put(-0.5,-5.8){\LARGE\text{ \mathbf {270^{\circ}} }}\put(-3.6,-0.45){\bf{-1}}\put(3.2,-0.45){\bf{1}}\put(0.2,-3.5){\bf{-1}}\put(0.2,3.4){\bf{1}}\end{picture}$  

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⠀⠀⠀⠀➡️⠀Sabendo, portanto, que 210º está no terceiro quadrante (sen<0 e cos<0) então temos que, seu ângulo correspondente no primeiro quadrante é de 210 - 180 = 30º. Desta forma sabemos que:

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⠀⠀⠀⠀☃️⠀sen(210º) = -sen(30º) = -1/2 ✅

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⠀⠀⠀⠀☃️⠀cos(210º) = -cos(30º) = -√3/2 ✅

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⠀⠀⠀⠀☃️⠀tg(210º) = sen(210º)/cos(210º) = (-1/2)/(-√3/2) = 1/√3 = √3/3 ✅

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⠀⠀⠀⠀☃️⠀cotg(210º) = 1/tg(210º) = 3/√3 = √3 ✅

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⠀⠀⠀⠀☃️⠀sec(210º) = 1/cos(210º) = -2/√3 = -2√3/3 ✅

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⠀⠀⠀⠀☃️⠀cosec(210º) = 1/sen(210º) = -2/1 = -2 ✅

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2)⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀✍

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⠀⠀⠀⠀➡️⠀Temos, pela relação fundamental da trigonometria, que:

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                               $\Large\red{\boxed{\pink{\boxed{\begin{array}{lcr}\green{\star}&&\green{\star}\\&\!\!\orange{\bf sen^2(x) + cos^2(x) = 1}\!\!&\\\green{\star}&&\green{\star}\\\end{array}}}}}$  

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⠀⠀⠀⠀➡️⠀Ou seja:  

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$\LARGE\blue{\sf sen^2(x) + \left(\dfrac{-4}{5}\right)^2 = 1}$

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$\LARGE\blue{\sf sen^2(x) + \dfrac{16}{25} = 1}$

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$\LARGE\blue{\sf sen^2(x) = 1 - \dfrac{16}{25}}$

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$\LARGE\blue{\sf sen^2(x) = \dfrac{25}{25} - \dfrac{16}{25}}$

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$\LARGE\blue{\sf sen^2(x) = \dfrac{9}{25}}$

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⠀⠀⠀⠀➡️⠀Aplicando a radiciação em ambos os lados da igualdade temos:  

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$\LARGE\blue{\text{ \sf \sqrt{sen^2(x)} = \pm \sqrt{\dfrac{9}{25}} }}$

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$\LARGE\blue{\sf sen(x) = \pm \dfrac{3}{5}}$

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⠀⠀⠀⠀➡️⠀Como x pertence ao segundo quadrante (ou seja, sen>0 e cos<0) então assumiremos somente a solução positiva desta radiciação:  

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⠀⠀⠀⠀☃️⠀sen(x) = 3/5 ✅

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⠀⠀⠀⠀☃️⠀tan(x) = (3/5)/(-4/5) = -3/4 ✅

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⠀⠀⠀⠀☃️⠀sec(x) = 1/cos(x) = -5/4 ✅

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⠀⠀⠀⠀☃️⠀cosec(x) = 1/sen(x) = 5/3 ✅  

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                             $\bf\large\red{\underline{\quad\quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}$☁

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                                     https://brainly.com.br/tarefa/37962049 ✈  

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                                     $\huge\blue{\text{\bf\quad Bons~estudos.}}$ ☕

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                                          $\quad\qquad(\orange{D\acute{u}vidas\ nos\ coment\acute{a}rios})$ ☄

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                             $\bf\large\red{\underline{\qquad \qquad \qquad \qquad \qquad \qquad \quad }\LaTeX}$✍

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                                                          $\Huge\green{\text{ \underline{\red{\mathbb{S}}\blue{\mathfrak{oli}}~}~\underline{\red{\mathbb{D}}\blue{\mathfrak{eo}}~}~\underline{\red{\mathbb{G}}\blue{\mathfrak{loria}}~} }}$ ✞

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Answer 2

Vamos là.

1)

sen(210) = sen(180 + 30) = -sen(30) = -1/2

cos(210) = cos(180 + 30) = -cos(30) = -√3/2

tg(210) = -sen(30)/-cos(30) = tg(30) = √3/3

cotg(210) = cotg(30) = √3

sec(210) = 1/-cos(30) = -2√3/3

cosec(210) = 1/-sen(30) = -2

2)

cos²(x) + sen²(x) = 1

16/25 + sen²(x) = 25/25

sen²(x) = 9/25

sen(x) = 3/5

tg(x) = sen(x)/cos(x) = -3/4

sec(x) = 1/cos(x) = -5/4

cosec(x) = 1/sen(x) = 5/3