• Matéria: Matemática
  • Autor: gordonnnw
  • Perguntado 3 anos atrás

qual o sen, cos e tg de 1590°

Respostas

respondido por: CyberKirito
0

\Large\boxed{\begin{array}{l}\sf\, 1590^\circ|\underline{360^\circ}\\\sf\underline{\!\!\!\!-1440^\circ}~~4\\\sf~~150^\circ\\\sf 1590^\circ=4\cdot(360^\circ)+150^\circ\\\sf sen(1590^\circ)=sen(150^\circ)=sen(30^\circ)=\dfrac{1}{2}\\\\\sf cos(1590^\circ)=cos(150^\circ)=-cos(30^\circ)=-\dfrac{\sqrt{3}}{2}\\\\\sf tg(1590^\circ)=tg(150^\circ)=-tg(30^\circ)=-\dfrac{\sqrt{3}}{3}\end{array}}

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