Question

Preciso urgentemente das respostas disso:

Answer 1

a)

1620| 360

-1440 4

180

Tg(1620°) =tg(180°) =0

b)

$\cot( \frac{9\pi}{4} ) = \cot(2\pi + \frac{7\pi}{4} ) \\ = \cot( \frac{7\pi}{4} ) = - 1$

c) 1710 |360

-1440 4

270

$\sec(1720) = \sec(270)$

∄ sec(270°) → ∄ sec(1710°)

d) 1755|360

-1440 4

315

-1755=4×-360°-315

Calculando a primeira determinação positiva temos

$\alpha = - 315 + 360 = 45$

Daí

$\csc( - 1755) = \csc( - 315) \\ = \csc(45) = \frac{2}{ \sqrt{2} } = \sqrt{2}$

2)

$\sin(x) = \frac{ \sqrt{6} }{3} \\ { \sin }^{2}x = \frac{6}{9} = \frac{2}{3} \\ 1 - \frac{2}{3} = \frac{1}{3}$

a)

$\cos(x) = - \sqrt{ \frac{1}{3} } = - \frac{ \sqrt{3} }{3}$

b)

$\tan(x) = \frac{ \sin(x) }{ \cos(x)} = \frac{ \frac{ \sqrt{6} }{3} }{ \frac{ - \sqrt{3} }{3} } \\ \tan(x) = \frac{ \sqrt{6} }{ - \sqrt{3} } = - \sqrt{2}$

c)

$\cot(x) = \frac{1}{ \tan(x) } = - \frac{1}{ \sqrt{2} } = - \frac{ \sqrt{2} }{2}$

d)

$\sec(x) = \frac{1}{ \cos(x)} = - \frac{3}{ \sqrt{3} } = \\ \sec(x) = - \frac{3 \sqrt{3} }{3} = - \sqrt{3}$

e)

$\csc(x) = \frac{3}{ \sqrt{6} } = \frac{ 3\sqrt{6} }{6} = \frac{ \sqrt{6} }{2}$

3)

$k = \frac{ { \sin }^{2}x}{ { \tan }^{2}x } + \frac{ { \cos }^{2}x}{ { \cot }^{2}x } \\ k = { \sin }^{2}x. \frac{ { \cos}^{2}x }{ { \sin}^{2}x} + { \cos}^{2}x. \frac{ { \sin}^{2}x}{ { \cos}^{2}x}$

$k = { \cos }^{2}x + { \sin}^{2}x \\ k = 1$

4) vou mandar por foto

5)

$\cot(x) = - \sqrt{ { \csc }^{2}x - 1 } \\ \cot(x) = - \sqrt{( { \frac{3}{2}) }^{2} - 1 } \\ \cot(x) = - \sqrt{ \frac{9}{4} - 1 }$

$\cot(x) = - \sqrt{ \frac{9 - 4}{4} } = - \sqrt{ \frac{5}{4} } \\ \cot(x) = - \frac{ \sqrt{5} }{2}$

6)

$k = \sin(2\pi + x) - 3 \cos(x + \frac{3\pi}{2} )$

$\sin(2\pi + x) \\ = \sin(2\pi) \cos(x) + \sin(x) \cos(2\pi)$

$0. \cos(x) \sin(x).(1) =\sin(x)$

$\cos(x + \frac{3\pi}{2} ) \\ \cos(x) \cos( \frac{3\pi}{2} ) - \sin(x). \sin( \frac{3\pi}{2} )$

$k = \sin(x) + \sin(x) = 2 \sin(x)$

$\cos(x).0 - \sin(x) .( - 1) = \sin(x)$

7)

${ \sin}^{2}x = {( \frac{3}{5} )}^{2} = \frac{9}{25} \\ 1 - \frac{9}{25} = \frac{16}{25}$

$\cos(x) = \sqrt{ \frac{16}{25} } = \frac{4}{5}$

a)

$\sin(2x) = 2. \sin(x) . \cos(x) \\ \sin(2x) = 2. \frac{3}{5}. \frac{4}{5} = \frac{24}{25}$

b)

$\cos(2x) = { \cos}^{2}x - { \sin }^{2}x \\ \cos(2x) = {( \frac{4}{5} )}^{2} - {( \frac{3}{5} )}^{2} \\ = \frac{16}{25 } - \frac{9}{25} = \frac{7}{25}$

c)

$\tan(2x) = \frac{ \sin(2x) }{ \cos(2x) } = \frac{ \frac{24}{25} }{ \frac{7}{25} } = \frac{24}{7}$

8)

a) senx=-½←→x=7π/6 ou x=11π/6

b) cosx=√3/2 ←→x=π/3 ou x=5π/3