Question

sendo sen x = 2/3 como 0< x < pi/2 determine a) sen 2x b) cos 2x

Answer 1

sen x = 2/3

sen²x + cos²x = 1

(2/3)² + cos²x = 1

cos²x = 1 - 4/9

cos²x = 5/9

cos x = √5/3

(A)

sen 2x = 2*senx*cosx = 2*(2/3)*(√5/3) = 4√5/9

sen 2x = 4√5/9

(B)

cos 2x = cos²x - sen²x = (√5/3)² - (2/3)² = 5/9 - 4/9 = 1/9

cos 2x = 1/9

Answer 2

a)

sin²(x) + cos²(x) = 1

cos = √(1 - 4/9)

cos = (√5)/3

sin(2x) = 2 * sin(x) * cos(x)

2 * 2/3 * (√5)/3

4√5/9

$\boxed{\underline{\mathsf{Resposta\to } { \frac{4\sqrt{5}}{9} }}}$

b)

cos(2x) = 1 -2 * sin²(x)

1 -2 * (2/3)²

1 -2 * 4/9

1 -8/9

(9 -8)/9

1/9

$\boxed{\underline{\mathsf{Resposta\to } { \frac{1}{9} }}}$