Question

Exercícios de "arcos duplos" para achar a tangente, cosseno e seno. 39 PONTOS

Sendo cos x = -4/5 e pi/2 < x < pi, calcule sen 2x, cos 2x e tg 2x.

Sendo sen x = - \/3/2, pi < x < 3pi/2, escreva sen 2x, cos 2x e tg 2x.

Dado sen a = cos a - 0,2, a E QI, calcule 2a

etc.

Todos os exercícios que precisam ser resolvidos (7), (COM RESOLUÇÃO), na imagem. Por favor, faça o máximo que puder! Se possível todos, mas só um já ajudaria bastante. (ou uma explicação de como resolver.)

Answer 1

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$\boxed{\begin{array}{l}\tt 51)~\sf cos(x)=-\dfrac{4}{5}\implies cos^2(x)=\dfrac{16}{25}\\\sf sen^2(x)=\dfrac{25}{25}-\dfrac{16}{25}=\dfrac{9}{25}\\\sf sen(x)=\sqrt{\dfrac{9}{25}}=\dfrac{3}{5}\\\sf sen(2x)=2\cdot sen(x)\cdot cos(x)\\\sf sen(2x)=2\cdot\dfrac{3}{5}\cdot\bigg(-\dfrac{4}{5}\bigg)=-\dfrac{24}{25}\\\sf cos(2x)=cos^2(x)-sen^2(x)\\\sf cos(2x)=\dfrac{16}{25}-\dfrac{9}{25}=\dfrac{7}{25}\\\sf tg(2x)=\dfrac{sen(2x)}{cos(2x)}\\\sf tg(2x)=\dfrac{-\frac{24}{25}}{\frac{7}{25}}=-\dfrac{24}{7}\end{array}}$

$\boxed{\begin{array}{l}\tt 52)~\sf sen(x)=-\dfrac{\sqrt{3}}{2}\implies sen^2(x)=\dfrac{3}{4}\\\sf cos^2(x)=\dfrac{4}{4}-\dfrac{3}{4}=\dfrac{1}{4}\\\sf cos(x)=-\sqrt{\dfrac{1}{4}}=-\dfrac{1}{2}\\\sf sen(2x)=2\cdot sen(x)\cdot cos(x)\\\sf sen(2x)=\diagup\!\!\!2\cdot\bigg(-\dfrac{\sqrt{3}}{\diagup\!\!\!2}\bigg)\cdot\bigg(-\dfrac{1}{2}\bigg)=\dfrac{\sqrt{3}}{2}\\\sf cos(2x)=cos^2(x)-sen^2(x)\\\sf cos(2x)=\dfrac{1}{4}-\dfrac{3}{4}=-\dfrac{2}{4}=-\dfrac{1}{2}\\\sf tg(2x)=\dfrac{sen(2x)}{cos(2x)}\end{array}}$

$\large\boxed{\begin{array}{l}\sf tg(2x)=\dfrac{\frac{\sqrt{3}}{\diagup\!\!\!2}}{-\frac{1}{\diagup\!\!\!2}}\\\sf tg(2x)=-\sqrt{3}\end{array}}$

$\large\boxed{\begin{array}{l}\tt 53)~\sf sen(a)=cos(a)-0,2\\\sf sen(a)-cos(a)=-0,2\\\sf (sen(a)-cos(a))^2=(-0,2)^2\\\sf sen^2(a)-2sen(a)cos(a)+cos^2(a)=0,04\\\sf 1-sen(2a)=0,04\\\sf sen(2a)=1-0,04\\\sf sen(2a)=0,96\end{array}}$

$\large\boxed{\begin{array}{l}\tt 54)~\sf tg(x)=\dfrac{3}{5}\\\sf tg(2x)=\dfrac{2tg(x)}{1-tg^2(x)}\\\sf tg(2x)=\dfrac{2\cdot\frac{3}{5}}{1-\bigg(\frac{3}{5}\bigg)^2}\\\sf tg(2x)=\dfrac{\frac{6}{5}}{1-\frac{9}{25}}\\\sf tg(2x)=\dfrac{\frac{6}{5}}{\frac{16}{25}}\\\sf tg(2x)=\dfrac{\diagdown\!\!\!\!6^3}{\diagdown\!\!\!\!5}\cdot\dfrac{\diagdown\!\!\!\!\!\!25^5}{\diagdown\!\!\!\!\!\!16_8}\\\sf tg(2x)=\dfrac{15}{8}\end{array}}$

$\large\boxed{\begin{array}{l}\tt 55)~\sf cossec(x)=2\\\sf sen(x)=\dfrac{1}{2}\implies sen^2(x)=\dfrac{1}{4}\\\sf cos^2(x)=\dfrac{4}{4}-\dfrac{1}{4}=\dfrac{3}{4}\\\sf cos(2x)=cos^2(x)-sen^2(x)\\\sf cos(2x)=\dfrac{3}{4}-\dfrac{1}{4}\\\sf cos(2x)=\dfrac{2\div2}{4\div2}\\\sf cos(2x)=\dfrac{1}{2}\end{array}}$

$\boxed{\begin{array}{l}\tt 56)~\sf sen(2x)=-0,6=-\dfrac{3}{5}\implies sen^2(2x)=\dfrac{9}{25}\\\sf cos^2(2x)=\dfrac{25}{25}-\dfrac{9}{25}=\dfrac{16}{25}\\\sf cos(2x)=-\sqrt{\dfrac{16}{25}}=-\dfrac{4}{5}\\\tt a)~\sf cos(2x)=2cos^2(x)-1\\\sf-\dfrac{4}{5}=2cos^2(x)-1\\\sf 2cos^2(x)=1-\dfrac{4}{5}\\\sf 2cos^2(x)=\dfrac{1}{5}\\\sf cos^2(x)=\dfrac{1}{10}\\\sf cos(x)=-\sqrt{\dfrac{1}{10}}\\\sf cos(x)=-\dfrac{\sqrt{10}}{10}\end{array}}$

$\large\boxed{\begin{array}{l}\tt b)~\sf cos^2(x)=\dfrac{10}{100}=\dfrac{1}{10}\\\sf sen^2(x)=\dfrac{10}{10}-\dfrac{1}{10}\\\sf sen^2(x)=\dfrac{9}{10}\\\sf sen(x)=-\sqrt{\dfrac{9}{10}}\\\sf sen(x)=-\dfrac{3}{\sqrt{10}}\\\sf sen(x)=-\dfrac{3\sqrt{10}}{10}\end{array}}$

$\large\boxed{\begin{array}{l}\tt c)~\sf tg(x)=\dfrac{sen(x)}{cos(x)}\\\sf tg(x)=\dfrac{-\frac{3\sqrt{10}}{10}}{-\frac{\sqrt{10}}{10}}\\\sf tg(x)=-\dfrac{3\diagdown\!\!\!\!\!\!\sqrt{10}}{\diagdown\!\!\!\!\!\!10}\cdot-\dfrac{\diagdown\!\!\!\!\!\!10}{\diagdown\!\!\!\!\!\!\sqrt{10}}=3\end{array}}$