Question

sen 18º = √5-1 /4 , utilizando a relação fundamental da trigonometria, qual o cosseno de 18º?

Answer 1

Resposta:  $\mathsf{cos(18^\circ)=\dfrac{\sqrt{10+2\sqrt{5}}}{4}.}$

Explicação passo a passo:

Pela relação fundamental trigonométrica, temos

    $\mathsf{sen^2(18^\circ)+cos^2(18^\circ)=1}\\\\ \mathsf{\Longrightarrow\quad cos^2(18^\circ)=1-sen^2(18^\circ)}$

Subsituindo acima o valor de sen(18°), temos

    $\mathsf{\Longrightarrow\quad cos^2(18^\circ)=1-\Big(\dfrac{\sqrt{5}-1}{4}\Big)^{\!2}}\\\\\\ \mathsf{\Longrightarrow\quad cos^2(18^\circ)=1-\dfrac{(\sqrt{5}-1)^2}{4^2}}\\\\\\ \mathsf{\Longrightarrow\quad cos^2(18^\circ)=1-\dfrac{(\sqrt{5})^2-2\cdot \sqrt{5}\cdot 1+1^2}{16}}\\\\\\ \mathsf{\Longrightarrow\quad cos^2(18^\circ)=1-\dfrac{5-2\sqrt{5}+1}{16}}\\\\\\ \mathsf{\Longrightarrow\quad cos^2(18^\circ)=1-\dfrac{6-2\sqrt{5}}{16}}$

    $\mathsf{\Longrightarrow\quad cos^2(18^\circ)=\dfrac{16-(6-2\sqrt{5})}{16}}\\\\\\ \mathsf{\Longrightarrow\quad cos^2(18^\circ)=\dfrac{16-6+2\sqrt{5}}{16}}\\\\\\ \mathsf{\Longrightarrow\quad cos^2(18^\circ)=\dfrac{10+2\sqrt{5}}{16}}\\\\\\ \mathsf{\Longrightarrow\quad cos(18^\circ)=\sqrt{\dfrac{10+2\sqrt{5}}{16}}}\\\\\\ \mathsf{\Longrightarrow\quad cos(18^\circ)=\dfrac{\sqrt{10+2\sqrt{5}}}{4}\quad\longleftarrow\quad resposta.}$

Dúvidas? Comente.

Bons estudos!