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Sagot :
Given sin(x)=5/13
First, lets find cos(x).
It is known that:
[tex]\begin{gathered} \sin ^2(x)+\cos ^2(x)=1 \\ (\frac{5}{13})^2+\cos ^2(x)=1 \\ \cos ^2(x)=1-\frac{25}{169} \\ \cos ^2(x)=\frac{169-25}{169}=\frac{144}{169} \\ \cos (x)=\pm\sqrt[]{\frac{144}{169}}\text{ = }\frac{\sqrt[]{144}}{\sqrt[]{169}} \\ \cos (x)=\pm\frac{12}{13} \end{gathered}[/tex]Since π/2 < x < π, we are in 2nd quadrant. Then, cos(x) is negative.
[tex]\cos (x)=-\frac{12}{13}[/tex]Since we know the values for sin and cos, we can find tan(x):
[tex]\begin{gathered} \tan (x)=\frac{\sin(x)}{\cos(x)} \\ \tan (x)=\frac{\frac{5}{13}}{-\frac{12}{13}} \\ \tan (x)==-\frac{5}{12} \end{gathered}[/tex]Now, lets work with the expression tan(2x)
It is known that:
[tex]\tan (2x)=\frac{2\tan(x)}{1-\tan^2(x)}[/tex]
Since we know tan(x), we can substitute in the expression above and find the value of tan(2x):
[tex]\begin{gathered} \tan (2x)=\frac{2\tan(x)}{1-\tan^2(x)} \\ \tan (2x)=\frac{2\cdot(-\frac{5}{12}_{})}{1-(-\frac{5}{12})^2} \\ \tan (2x)=\frac{-\frac{10}{12}}{1-\frac{25}{144}}=\frac{-\frac{10}{12}}{\frac{144-25}{144}}=\frac{-\frac{10}{12}}{\frac{119}{144}}=-\frac{10}{12}\cdot\frac{144}{119} \\ \tan (2x)=-\frac{120}{119} \end{gathered}[/tex]Answer: -120/119
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