Question

calcule sen2x cos2x e tg2x, sabendo que senx=1/3 e que π/2 < x < π.

Answer 1

Resposta:

sen2x = -4√2/9

cos2x = 7/9

tg2x = -4√2/7

Explicação passo-a-passo:

senx = 1/3

cos²x = 1 - sen²x

cos²x = 1 - (1/3)²

cos²x =1 - 1/9

cos²x = 8/9

cosx = -2√2/3 (x ∈ 2°Q)

tgx = senx : cosx

tgx = 1/3 : (-2√2/3

tgx = 1/3 .(-3/2√2)

tgx = -1/2√2

tgx = -√2/4

sen2x = 2senx cosx

sen2x = 2.1/3 . (-2√2/3)

sen2x = -4√2/9

cos2x = cos²x - sen²x

cos2x = 1 - sen²x - sen²x

cos2x = 1 - 2sen²x

cos2x = 1 - 2(1/3)²

cos2x = 1 - 2/9

cos2x = 7/9

tg2x = 2tgx/(1 - tg²x)

tg2x = 2(-√2/4) :[1 - (-√2/4)²]

tg2x = -√2/2 : (1 - 2/16)

tg2x = -√2/2 :  14/16

tg2x = -√2/2 .8/7

tg2x = -4√2/7

Answer 2

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

$\cos(x)=-\sqrt{1-\frac{1}{9}}\\ \cos(x)=-\frac{2\sqrt{2}}{3}$

$\tan(x)=\frac{\sin(x)}{\cos(x)} \\ \tan(x) =\frac{\frac{1}{3}}{-\frac{2\sqrt{2}}{3}}$

$\tan(x)=-\frac{\sqrt{2}}{4}$

$\sin(2x)=2.\sin(x).\cos(x) \\ \sin(2x)=2.\frac{1}{3}.(-\frac{2\sqrt{2}}{3})$

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

$\cos(2x)={(\cos(x))}^{2}-{(\sin(x)) </p><p>}^{2}$

$\cos(2x) =\frac{8}{9}-\frac{1}{9} \\ \cos(2x)=\frac{7}{9}$

O cálculo de tg2x esta anexo.