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Sagot :
When a point (x,y) lies on a unit circle, the following equation holds true:
[tex]x^2+y^2=1[/tex]We are given
[tex]y=\frac{1}{3}[/tex]and need to find x.
Let's put it into the equation and figure out x. Shown below:
[tex]\begin{gathered} x^2+y^2=1 \\ x^2+(\frac{1}{3})^2=1 \\ x^2+\frac{1}{9}=1 \\ x^2=1-\frac{1}{9} \\ x^2=\frac{8}{9} \\ x=\sqrt[]{\frac{8}{9}} \\ x=\frac{\sqrt[]{8}}{\sqrt[]{9}} \\ x=\frac{\sqrt[]{8}}{3} \end{gathered}[/tex]We can simplify the square root of 8 by using the radical property:
[tex]\sqrt[]{a\cdot b}=\sqrt[]{a}\sqrt[]{b}[/tex]Thus, square root of 8 becomes:
[tex]\sqrt[]{8}=\sqrt[]{4\cdot2}=\sqrt[]{4}\sqrt[]{2}=2\sqrt[]{2}[/tex]Thus, the simplest form of x is:
[tex]x=\frac{2\sqrt[]{2}}{3}[/tex]
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