Respuesta :
To solve this, we can use the following trigonometric identity:
[tex]\sin ^2\theta+\cos ^2\theta=1[/tex]By replacing 0 for θ, we get:
[tex]\sin ^2(0)+\cos ^2(0)=1[/tex]by replacing -1/sqrt5 for cos(0), we get:
[tex]\begin{gathered} \sin ^2(0)+(-\frac{1}{\sqrt[]{5}})^2=1 \\ \sin ^2(0)+\frac{(-1)^2}{(\sqrt[]{5)^2}}^{}=1 \\ \sin ^2(0)+\frac{1^{}}{5^{}}^{}=1 \end{gathered}[/tex]From this equation, we can solve for the sin(0), like this:
[tex]\begin{gathered} \sin ^2(0)+\frac{1^{}}{5^{}}^{}=1 \\ \sin ^2(0)+\frac{1^{}}{5^{}}^{}-\frac{1}{5}=1-\frac{1}{5} \\ \sin ^2(0)+0=1\times\frac{5}{5}-\frac{1}{5} \\ \sin ^2(0)=\frac{5}{5}-\frac{1}{5} \\ \sin ^2(0)=\frac{5-1}{5} \\ \sin ^2(0)=\frac{4}{5} \\ \sqrt[]{\sin ^2(0)}=\sqrt[]{\frac{4}{5}} \\ \sin (0)=\pm\sqrt[]{\frac{4}{5}} \\ \sin (0)=\pm\frac{\sqrt[]{4}}{\sqrt[]{5}}=\pm\frac{2}{\sqrt[]{5}}= \end{gathered}[/tex]Then, the posible values of sin(0) are 2/sqrt(5) and -2/sqrt(5)
