Question

se sen x + cos x = 1/3, calcule o cos 2x:

Answer 1

$\mathrm{sen\,}x+\cos x=\dfrac{1}{3}$


Elevando ao quadrado os dois lados da equação acima

$(\mathrm{sen\,}x+\cos x)\,^{2}=\left(\dfrac{1}{3} \right )^{2}\\ \\ \\ \mathrm{sen^{2}\,}x+2\cdot \mathrm{sen\,}x\cdot \cos x+ \cos^{2} x=\dfrac{1}{9}\\ \\ \\ (\mathrm{sen^{2}\,}x+\cos^{2} x)+2\mathrm{\,sen\,}x\cos x=\dfrac{1}{9}\\ \\ \\ 1+\mathrm{sen\,}2x=\dfrac{1}{9}\\ \\ \\ \mathrm{sen\,}2x=\dfrac{1}{9}-1\\ \\ \\ \mathrm{sen\,}2x=\dfrac{1-9}{9}\\ \\ \\ \mathrm{sen\,}2x=-\dfrac{8}{9}$


Relação Trigonométrica Fundamental:

$\cos^{2} 2x+\mathrm{sen^{2}\,}2x=1\\ \\ \cos^{2}2x=1-\mathrm{sen^{2}\,}2x\\ \\ \cos^{2}2x=1-\left(-\dfrac{8}{9} \right )^{2}\\ \\ \\ \cos^{2}2x=1-\dfrac{64}{81}\\ \\ \\ \cos^{2}2x=\dfrac{81-64}{81}\\ \\ \\ \cos^{2}2x=\dfrac{17}{81}\\ \\ \\ \cos 2x=\pm \sqrt{\dfrac{17}{81}}\\ \\ \\ \cos 2x =\pm \dfrac{\sqrt{17}}{9}\\ \\ \\ \begin{array}{rcl} \cos 2x=\dfrac{\sqrt{17}}{9}&\;\text{ ou }\;&\cos 2x=-\dfrac{\sqrt{17}}{9} \end{array}$