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Questões Comentadas: Trigonometria – Operações com Arcos

Exercícios:

1. (Fuvest – SP) A tangente do ângulo 2x é dada em função da tangente de x pela seguinte fórmula:

<a href=”http://www.codecogs.com/eqnedit.php?latex=tg2x=\frac{2tgx}{1-tg^2x}” target=”_blank”><img src=”http://latex.codecogs.com/gif.latex?tg2x=\frac{2tgx}{1-tg^2x}” title=”tg2x=\frac{2tgx}{1-tg^2x}” /></a>

Calcule um valor aproximado da tangente do ângulo 22°30′.
a) 0,22
b) 0,41
c) 0,50
d) 0,72
e) 1,00

2. (UF-RN) Na representação a seguir, EF é o diâmetro da circunferência; EG e FG são catetos do triângulo retângulo FGE, inscrito na circunferência trigonométrica; e FG é perpendicular a Ox para qualquer α. O raio da circunferência é unitário.

Resumo 4. Geometria - 13-04-2015 010

Nessas condições, podemos afirmar que, para qualquer α (0° < α < 90°):
<a href=”http://www.codecogs.com/eqnedit.php?latex=a)&space;\,&space;\frac{FG}{EG}=2tg\alpha” target=”_blank”><img src=”http://latex.codecogs.com/gif.latex?a)&space;\,&space;\frac{FG}{EG}=2tg\alpha” title=”a) \, \frac{FG}{EG}=2tg\alpha” /></a>
<a href=”http://www.codecogs.com/eqnedit.php?latex=b)&space;\,&space;sen^2\alpha&plus;cos^2\alpha=EF” target=”_blank”><img src=”http://latex.codecogs.com/gif.latex?b)&space;\,&space;sen^2\alpha&plus;cos^2\alpha=EF” title=”b) \, sen^2\alpha+cos^2\alpha=EF” /></a>
<a href=”http://www.codecogs.com/eqnedit.php?latex=c)&space;\,&space;OH&space;=&space;cos(90^{\circ}-\alpha)” target=”_blank”><img src=”http://latex.codecogs.com/gif.latex?c)&space;\,&space;OH&space;=&space;cos(90^{\circ}-\alpha)” title=”c) \, OH = cos(90^{\circ}-\alpha)” /></a>
<a href=”http://www.codecogs.com/eqnedit.php?latex=d)&space;FG&space;=&space;2sen\alpha” target=”_blank”><img src=”http://latex.codecogs.com/gif.latex?d)&space;FG&space;=&space;2sen\alpha” title=”d) FG = 2sen\alpha” /></a>

 

3. (UCDB-MS) Se cosx + senx.tgx = 3, x pertencente ao 1° quadrante, o valor da cotgx é igual a:
<a href=”http://www.codecogs.com/eqnedit.php?latex=a)&space;\,&space;1″ target=”_blank”><img src=”http://latex.codecogs.com/gif.latex?a)&space;\,&space;1″ title=”a) \, 1″ /></a>
<a href=”http://www.codecogs.com/eqnedit.php?latex=b)&space;\,&space;\sqrt3″ target=”_blank”><img src=”http://latex.codecogs.com/gif.latex?b)&space;\,&space;\sqrt3″ title=”b) \, \sqrt3″ /></a>
<a href=”http://www.codecogs.com/eqnedit.php?latex=c)&space;\,&space;\sqrt2″ target=”_blank”><img src=”http://latex.codecogs.com/gif.latex?c)&space;\,&space;\sqrt2″ title=”c) \, \sqrt2″ /></a>
<a href=”http://www.codecogs.com/eqnedit.php?latex=d)&space;\,&space;\frac{\sqrt2}{4}” target=”_blank”><img src=”http://latex.codecogs.com/gif.latex?d)&space;\,&space;\frac{\sqrt2}{4}” title=”d) \, \frac{\sqrt2}{4}” /></a>
<a href=”http://www.codecogs.com/eqnedit.php?latex=d)&space;\,&space;\frac{\sqrt3}{9}” target=”_blank”><img src=”http://latex.codecogs.com/gif.latex?d)&space;\,&space;\frac{\sqrt3}{9}” title=”d) \, \frac{\sqrt3}{9}” /></a>

Solução Passo-a-Passo:

1. Primeiro Passo: Substituir x por 22°30′ na fórmula.
<a href=”http://www.codecogs.com/eqnedit.php?latex=tg(2.22^{\circ}30′)=\frac{2tg(22^{\circ}30′)}{1-tg^2(22^{\circ}30′)}\Rightarrow&space;tg45^{\circ}=\frac{2tg(22^{\circ}30′)}{1-tg^2(22^{\circ}30′)}” target=”_blank”><img src=”http://latex.codecogs.com/gif.latex?tg(2.22^{\circ}30′)=\frac{2tg(22^{\circ}30′)}{1-tg^2(22^{\circ}30′)}\Rightarrow&space;tg45^{\circ}=\frac{2tg(22^{\circ}30′)}{1-tg^2(22^{\circ}30′)}” title=”tg(2.22^{\circ}30′)=\frac{2tg(22^{\circ}30′)}{1-tg^2(22^{\circ}30′)}\Rightarrow tg45^{\circ}=\frac{2tg(22^{\circ}30′)}{1-tg^2(22^{\circ}30′)}” /></a>

Segundo Passo: Chamar 22°30′ de x, resolver a equação e encontrar o valor aproximado da tangente de x.
<a href=”http://www.codecogs.com/eqnedit.php?latex=1=\frac{2tgx}{1-tg^2x}\Rightarrow&space;1-tg^2x=2tgx\Rightarrow&space;0&space;=&space;tg^2x&plus;2tgx-1″ target=”_blank”><img src=”http://latex.codecogs.com/gif.latex?1=\frac{2tgx}{1-tg^2x}\Rightarrow&space;1-tg^2x=2tgx\Rightarrow&space;0&space;=&space;tg^2x&plus;2tgx-1″ title=”1=\frac{2tgx}{1-tg^2x}\Rightarrow 1-tg^2x=2tgx\Rightarrow 0 = tg^2x+2tgx-1″ /></a>
Teremos uma equação do segundo grau com incógnita tgx, logo:
<a href=”http://www.codecogs.com/eqnedit.php?latex=a=1,&space;b=2,&space;c=-1;&space;\Delta&space;=&space;b^2-4ac\Rightarrow&space;\Delta&space;=&space;2^2-4.1.(-1)=4&plus;4\Rightarrow&space;\Delta=8″ target=”_blank”><img src=”http://latex.codecogs.com/gif.latex?a=1,&space;b=2,&space;c=-1;&space;\Delta&space;=&space;b^2-4ac\Rightarrow&space;\Delta&space;=&space;2^2-4.1.(-1)=4&plus;4\Rightarrow&space;\Delta=8″ title=”a=1, b=2, c=-1; \Delta = b^2-4ac\Rightarrow \Delta = 2^2-4.1.(-1)=4+4\Rightarrow \Delta=8″ /></a>
<a href=”http://www.codecogs.com/eqnedit.php?latex=\sqrt{\Delta}=\sqrt{8}\Rightarrow&space;\sqrt{\Delta}=2\sqrt2;&space;x=\frac{-b\pm&space;\sqrt{\Delta}}{2a}=\frac{-2\pm2\sqrt2}{2.1}\Rightarrow&space;x=-1\pm\sqrt2″ target=”_blank”><img src=”http://latex.codecogs.com/gif.latex?\sqrt{\Delta}=\sqrt{8}\Rightarrow&space;\sqrt{\Delta}=2\sqrt2;&space;x=\frac{-b\pm&space;\sqrt{\Delta}}{2a}=\frac{-2\pm2\sqrt2}{2.1}\Rightarrow&space;x=-1\pm\sqrt2″ title=”\sqrt{\Delta}=\sqrt{8}\Rightarrow \sqrt{\Delta}=2\sqrt2; x=\frac{-b\pm \sqrt{\Delta}}{2a}=\frac{-2\pm2\sqrt2}{2.1}\Rightarrow x=-1\pm\sqrt2″ /></a>
<a href=”http://www.codecogs.com/eqnedit.php?latex=Como&space;\,&space;x=tg(22^{\circ}30′)\,&space;e\,&space;0^{\circ}\leq22^{\circ}30’\leq90^{\circ},\,&space;x>0\Rightarrow&space;x=-1&plus;\sqrt2″ target=”_blank”><img src=”http://latex.codecogs.com/gif.latex?Como&space;\,&space;x=tg(22^{\circ}30′)\,&space;e\,&space;0^{\circ}\leq22^{\circ}30’\leq90^{\circ},\,&space;x>0\Rightarrow&space;x=-1&plus;\sqrt2″ title=”Como \, x=tg(22^{\circ}30′)\, e\, 0^{\circ}\leq22^{\circ}30’\leq90^{\circ},\, x>0\Rightarrow x=-1+\sqrt2″ /></a>
<a href=”http://www.codecogs.com/eqnedit.php?latex=Considerando&space;\,&space;\sqrt2\approx&space;1,41,&space;\,&space;x=-1&plus;1,41\Rightarrow&space;x=&space;0,41.” target=”_blank”><img src=”http://latex.codecogs.com/gif.latex?Considerando&space;\,&space;\sqrt2\approx&space;1,41,&space;\,&space;x=-1&plus;1,41\Rightarrow&space;x=&space;0,41.” title=”Considerando \, \sqrt2\approx 1,41, \, x=-1+1,41\Rightarrow x= 0,41.” /></a>

Gabarito: B

2. Primeiro Passo: Identificar todas as semelhanças na figura.

Resumo 4. Geometria - 13-04-2015 011

 

Segundo Passo: Escrever as conclusões tiradas da figura.
<a href=”http://www.codecogs.com/eqnedit.php?latex=tg\alpha&space;=&space;\frac{FH}{OH}=\frac{OD}{ED}=&space;\frac{FG}{EG}” target=”_blank”><img src=”http://latex.codecogs.com/gif.latex?tg\alpha&space;=&space;\frac{FH}{OH}=\frac{OD}{ED}=&space;\frac{FG}{EG}” title=”tg\alpha = \frac{FH}{OH}=\frac{OD}{ED}= \frac{FG}{EG}” /></a>
Logo, a primeira alternativa está errada.
<a href=”http://www.codecogs.com/eqnedit.php?latex=sen^2\alpha&space;&plus;&space;cos^2\alpha&space;=&space;1&space;\,&space;e&space;\,&space;EF=2\Rightarrow&space;sen^2\alpha&plus;cos^2\alpha=\frac{EF}{2}” target=”_blank”><img src=”http://latex.codecogs.com/gif.latex?sen^2\alpha&space;&plus;&space;cos^2\alpha&space;=&space;1&space;\,&space;e&space;\,&space;EF=2\Rightarrow&space;sen^2\alpha&plus;cos^2\alpha=\frac{EF}{2}” title=”sen^2\alpha + cos^2\alpha = 1 \, e \, EF=2\Rightarrow sen^2\alpha+cos^2\alpha=\frac{EF}{2}” /></a>
Portanto, a segunda alternativa também está errada.
<a href=”http://www.codecogs.com/eqnedit.php?latex=cos\alpha=OH&space;\,&space;e&space;\,&space;sen\alpha=FH\Rightarrow&space;cos(90^{\circ}-\alpha)=FH&space;\,&space;e&space;\,&space;sen(90^{\circ}-\alpha)=OH.” target=”_blank”><img src=”http://latex.codecogs.com/gif.latex?cos\alpha=OH&space;\,&space;e&space;\,&space;sen\alpha=FH\Rightarrow&space;cos(90^{\circ}-\alpha)=FH&space;\,&space;e&space;\,&space;sen(90^{\circ}-\alpha)=OH.” title=”cos\alpha=OH \, e \, sen\alpha=FH\Rightarrow cos(90^{\circ}-\alpha)=FH \, e \, sen(90^{\circ}-\alpha)=OH.” /></a>
Assim, a terceira alternativa também está errada.
<a href=”http://www.codecogs.com/eqnedit.php?latex=FH=OD=HG=sen\alpha&space;\,&space;e&space;\,&space;FG=FH&plus;HG\Rightarrow&space;FG=2sen\alpha” target=”_blank”><img src=”http://latex.codecogs.com/gif.latex?FH=OD=HG=sen\alpha&space;\,&space;e&space;\,&space;FG=FH&plus;HG\Rightarrow&space;FG=2sen\alpha” title=”FH=OD=HG=sen\alpha \, e \, FG=FH+HG\Rightarrow FG=2sen\alpha” /></a>

Gabarito: D

 

3. Primeiro Passo: Transformar tgx.
<a href=”http://www.codecogs.com/eqnedit.php?latex=cosx&plus;senx.tgx=3\Rightarrow&space;cosx&plus;senx.\frac{senx}{cosx}=3\Rightarrow&space;cosx&plus;\frac{sen^2x}{cosx}” target=”_blank”><img src=”http://latex.codecogs.com/gif.latex?cosx&plus;senx.tgx=3\Rightarrow&space;cosx&plus;senx.\frac{senx}{cosx}=3\Rightarrow&space;cosx&plus;\frac{sen^2x}{cosx}=3″ title=”cosx+senx.tgx=3\Rightarrow cosx+senx.\frac{senx}{cosx}=3\Rightarrow cosx+\frac{sen^2x}{cosx}=3″ /></a>
<a href=”http://www.codecogs.com/eqnedit.php?latex=\frac{cos^2x&plus;sen^2x}{cosx}=3\Rightarrow&space;\frac{1}{cosx}=3\Rightarrow&space;secx=3″ target=”_blank”><img src=”http://latex.codecogs.com/gif.latex?\frac{cos^2x&plus;sen^2x}{cosx}=3\Rightarrow&space;\frac{1}{cosx}=3\Rightarrow&space;secx=3″ title=”\frac{cos^2x+sen^2x}{cosx}=3\Rightarrow \frac{1}{cosx}=3\Rightarrow secx=3″ /></a>

Segundo Passo: Descobrir o valor da tgx.
<a href=”http://www.codecogs.com/eqnedit.php?latex=tg^2x&plus;1=sec^2x\Rightarrow&space;tg^2x&plus;1=3^2\Rightarrow&space;tg^2x=9-1=8\Rightarrow&space;tgx=\sqrt8″ target=”_blank”><img src=”http://latex.codecogs.com/gif.latex?tg^2x&plus;1=sec^2x\Rightarrow&space;tg^2x&plus;1=3^2\Rightarrow&space;tg^2x=9-1=8\Rightarrow&space;tgx=\sqrt8″ title=”tg^2x+1=sec^2x\Rightarrow tg^2x+1=3^2\Rightarrow tg^2x=9-1=8\Rightarrow tgx=\sqrt8″ /></a>

Terceiro Passo: Encontrar o valor de cotgx.
<a href=”http://www.codecogs.com/eqnedit.php?latex=cotgx=\frac{1}{tgx}\Rightarrow&space;cotgx=\frac{1}{\sqrt8}=\frac{1}{2\sqrt2}\Rightarrow&space;cotgx=\frac{\sqrt2}{4}” target=”_blank”><img src=”http://latex.codecogs.com/gif.latex?cotgx=\frac{1}{tgx}\Rightarrow&space;cotgx=\frac{1}{\sqrt8}=\frac{1}{2\sqrt2}\Rightarrow&space;cotgx=\frac{\sqrt2}{4}” title=”cotgx=\frac{1}{tgx}\Rightarrow cotgx=\frac{1}{\sqrt8}=\frac{1}{2\sqrt2}\Rightarrow cotgx=\frac{\sqrt2}{4}” /></a>

Gabarito: D