Question

Como montar o gráfico da função cosseno y=2-3cos(2x-pi/3) Calculando cada um dos cinco ângulos?

Answer 1

Resposta:

Eu faço da seguinte forma:

1) Construo a tabela. O arco (2x-π/3) tem que variar entre 0 e 2π para o cosseno completar um arco.

$\begin{tabular}{|c|c|c|}x&2x-\pi/3&y=2 - 3cos(2x-\pi/3)\\&0&\\&\pi/2&\\&\pi&\\&3\pi/2&\\&2\pi&\\\end{tabular}$

2) Agora vamos calcular o valor de x e completar a tabela na primeira coluna:

$Para{:}\\2x-\pi/3=0\Rightarrow 2x=\pi/3\Rightarrow x=\pi/6\\\\2x-\pi/3=\pi/2\Rightarrow 2x=\pi/2+\pi/3\Rightarrow x=5\pi/12\\\\2x-\pi/3=\pi\Rightarrow 2x=\pi+\pi/3\Rightarrow x=4\pi/6\\\\2x-\pi/3=3\pi/2\Rightarrow 2x=3\pi/2+\pi/3\Rightarrow x=11\pi/12\\\\2x-\pi/3=2\pi\Rightarrow 2x=2\pi+\pi/3\Rightarrow x=7\pi/6\\\\\\\begin{tabular}{|c|c|c|}x&2x-\pi/3&y=2 - 3cos(2x-\pi/3)\\\pi/6&0&\\5\pi/12&\pi/2&\\4\pi/6&\pi&\\11\pi/12&3\pi/2&\\7\pi/6&2\pi&\\\end{tabular}$

3) E agora vamos completar a tabela e incluir mais uma coluna

$\begin{tabular}{|c|c|c|c|}x&2x-\pi/3&y=2 - 3cos(2x-\pi/3)&Ponto (x,y)\\\pi/6&0&2-3(1)=-1&A(\pi/6,-1)\\5\pi/12&\pi/2&2-3(0)=2&B(5\pi/12,2)\\4\pi/6&\pi&2-3(-1)=5&C(4\pi/6,5)\\11\pi/12&3\pi/2&2-3(0)=2&D(11\pi/12,2)\\7\pi/6&2\pi&2-3(1)=-1&E(7\pi/6,-1)\\\end{tabular}$