Question

Sendo cotg x = 12/5 e 0 < x < pi/2, calcule cos 2x

Answer 1

Resposta:

Explicação passo-a-passo:

cotgx= 12/5

cotgx²=144/25

cos2x= cos²x -sen²x

1+ctg²x= cossec²x

cossec²x= 1+ 144/25

cossec²x=25/25 + 144/25

cossec²x=169/25

sen²x= 1/ cossec²x

sen²x= 1:169/25

sen²x=25/169

senx= 5/13

cosx=senx * cotgx

cosx=5/13 * 12/5

cosx=5*12/13*5

cosx=60/65

cosx=12/13

cos²x=144/169

cos2x=cos²x - sen²x

cos2x=144/169 - 25/169

cos2x=119/169

cos2x=0,704142

ok? espero ter ajudado.

Answer 2

Olá Deisi!!

Resposta:

$\boxed{\mathtt{\frac{119}{169}}}$

Explicação passo-a-passo:

De acordo com o enunciado,

$\displaystyle \mathtt{cotg \ x = \frac{12}{5}}$

Desse modo,

$\displaystyle \mathtt{\exists \ k \in \mathbb{Z}; \frac{\cos x}{\sin x} = \frac{12k}{5k}}$

Com efeito, temos:

$\\ \displaystyle \bullet \qquad \mathtt{\cos x = 12k \qquad \qquad \qquad (i)} \\\\ \bullet \qquad \mathtt{\sin x = 5k \qquad \qquad \qquad \ (ii)}$

Uma vez que $\displaystyle \boxed{\mathtt{\sin^2 x + \cos^2 x = 1}}$, fazemos...

$\\ \displaystyle \mathsf{\sin^2 x + \cos^2 x = 1} \\\\ \mathsf{\left ( 5k \right )^2 + \left ( 12k \right )^2 = 1} \\\\ \mathsf{25k^2 + 144k^2 = 1} \\\\ \boxed{\mathsf{k^2 = \frac{1}{169}}}$

Por fim,

$\\ \displaystyle \mathsf{\cos (2x) = \cos^2 x - \sin^2 x} \\\\ \mathsf{\qquad \quad \ = \left ( 12k \right )^2 - \left ( 5k \right )^2} \\\\ \mathsf{\qquad \quad \ = 144k^2 - 25k^2} \\\\ \mathsf{\qquad \quad \ = 119\underbrace{\mathsf{k^2}}_{1/169}} \\\\\\ \mathsf{\qquad \quad \ = 119 \cdot \frac{1}{169}} \\\\ \boxed{\boxed{\mathsf{\cos(2x) = \frac{119}{169}}}}$