Question

Calcule o valor de cos 3TT/5.​

Answer 1

$\text{Seja}\ \alpha=\dfrac{3\pi}{5}$

$\Longrightarrow\beta=\dfrac{2\pi}{5}\ \mathrm{\acute{e}\ suplementar\ de}\ \alpha\Longrightarrow \cos{\alpha}=-\cos{\beta}$

$\Longrightarrow \phi=\dfrac{\pi}{10}\ \mathrm{\acute{e}\ complementar\ de}\ \beta\Longrightarrow \cos{\alpha}=-\cos{\beta}=-\sin{\phi}$

$\Longrightarrow\cos{\dfrac{3\pi}{5}}=-\cos{\dfrac{2\pi}{5}}=\boxed{-\sin{\dfrac{\pi}{10}}}$

$\text{Seja}\ \phi=\dfrac{\pi}{10}\Longrightarrow 5\phi=\dfrac{\pi}{2}\Longrightarrow 2\phi=\dfrac{\pi}{2}-3\phi$

$\Longrightarrow\sin{2\phi}=\sin{\bigg(\dfrac{\pi}{2}-3\phi\bigg)}\Longrightarrow\sin{2\phi}=\cos{3\phi}$

$\Longrightarrow2\sin{\phi}\cos{\phi}=4\cos^3{\phi}-3\cos{\phi}\Longrightarrow2\sin{\phi}=4\cos^2{\phi}-3$

$\Longrightarrow2\sin{\phi}=4\left(1-\sin^2{\phi}\right)-3\Longrightarrow\boxed{4\sin^2{\phi}+2\sin{\phi}-1=0}$

$\Longrightarrow\sin{\phi}=\dfrac{-2\pm\sqrt{2^2-4(4)(-1)}}{2(4)}\Longrightarrow\sin{\phi}=\dfrac{-1\pm\sqrt{5}}{4}$

$\Longrightarrow\sin{\phi}=\dfrac{-1+\sqrt{5}}{4}\ \because\ \dfrac{\pi}{2}>\phi>0$

$\cos{\alpha}=-\sin{\phi}\ \therefore\ \boxed{\cos{\dfrac{3\pi}{5}}=\dfrac{1-\sqrt{5}}{4}}$