Question

sendo x um arco do intervalo {0,X/2} e sen x=5/13 calcule tg2x a)tg2x=120/119 b)tg2x=5/12,c)tg2x=25/27,d)tg2x=8raiz de 2/3 e)tg2x=133/13​

Answer 1

Resposta:

A) 120/119

Explicação passo-a-passo:

Relações trigonométricas

Intervalo [0°,90°]= (+)

Sen x = 5/13

sen^2x + cos^2x = 1

(5/13)^2 + cos ^2 x = 1

25/169 + cos^2 x = 1

cos^2 x = 1 - 25/169

cos^2 x = (169-25)/169

Cos^2 x = 144/169

cos x = 12/13

Tg = sen x / cos x

Tg x = 5/13 : 12/13

Tg x = 5/13 . 13/12

Tg x = 5/12

Tg 2x = 2 tg x /(1- tg^2 x)

Tg 2x = 2 . 5/12 / [1 - (5/12)^2]

Tg 2x = 10/12 : (1 - 25/144]

Tg 2x = 5/6 : (144 - 25)/144

Tg 2x = 5/6 : 119/144

Tg 2x = 5/6 . 144/119

Tg 2x = 144/6 . 5/119

Tg 2x = 24.5/119

Tg 2x = 120/119

Resp.: a) tg 2x = 120/119