Given:
The ratio is given as,
[tex]\sin x=\frac{3}{7}[/tex]The x is in the second quadrant.
The objective is to find the value of sin2x, cos2x and tan2x.
Explanation:
To find cos x :
Using the trigonometric identity,
[tex]\begin{gathered} \sin ^2x+\cos ^2x=1 \\ \cos ^2x=1-\sin ^2x \\ \cos x=\sqrt[]{1-\sin^2x}\text{ . . . . .(1)} \end{gathered}[/tex]On plugging the given values in equation (1),
[tex]\begin{gathered} \cos x=\sqrt[]{1-(\frac{3}{7})^2} \\ =\sqrt[]{1-\frac{9}{49}} \\ =\sqrt[]{\frac{49-9}{49}} \\ =\pm\frac{\sqrt[]{40}}{7} \\ =\pm\frac{2\sqrt[]{10}}{7} \end{gathered}[/tex]Since x lies in the second quadrant,
[tex]\cos x=-\frac{2\sqrt[]{10}}{7}[/tex]a)
To find sin(2x):
The general formula to find sin(2x) is,
[tex]\sin 2x=2\sin x\cos x\text{ . . . . .(2)}[/tex]On plugging the obtained values in equation (2),
[tex]\begin{gathered} \sin 2x=2(\frac{3}{7})(-\frac{2\sqrt[]{10}}{7}) \\ =-\frac{12\sqrt[]{10}}{49} \end{gathered}[/tex]Hence, the value of sin(2x) is (-12√10)/49.
b)
To find cos (2x):
The general formula of cos(2x) is,
[tex]\cos (2x)=\cos ^2x-\sin ^2x\text{ . . . . . .(3)}[/tex]On plugging the obtained values in equation (3),
[tex]\begin{gathered} \cos (2x)=(-\frac{2\sqrt[]{10}}{7})^2-(\frac{3}{7})^2 \\ =\frac{4\times10}{49}-\frac{9}{49} \\ =\frac{40-9}{49} \\ =\frac{31}{49} \end{gathered}[/tex]Hence, the value of cos(2x) is 31/49.
c)
To find tan(2x):
The general formula of tan(2x) is,
[tex]\text{tan}(2x)=\frac{\sin2x}{\cos2x}\text{ . . . . . .(4)}[/tex]On plugging the obtained values in equation (4),
[tex]\begin{gathered} \text{tan}(2x)=\frac{\frac{-12\sqrt[]{10}}{49}}{\frac{31}{49}} \\ =-\frac{12\sqrt[]{10}}{31} \end{gathered}[/tex]Hence, the value of tan(2x) is (-12√10/31).
