The unit circle is defined by the equation:
\[ x^2 + y^2 = 1 \]
We are given that \( y = \frac{\sqrt{2}}{2} \). We can substitute this value of \( y \) into the equation of the unit circle to find \( x \):
\[ x^2 + \left(\frac{\sqrt{2}}{2}\right)^2 = 1 \]
Calculating \( \left(\frac{\sqrt{2}}{2}\right)^2 \):
\[ \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{2}{4} = \frac{1}{2} \]
Now, substituting back into the equation:
\[ x^2 + \frac{1}{2} = 1 \]
To isolate \( x^2 \), subtract \( \frac{1}{2} \) from both sides:
\[ x^2 = 1 - \frac{1}{2} \]
This simplifies to:
\[ x^2 = \frac{1}{2} \]
Taking the square root of both sides gives us:
\[ x = \pm \sqrt{\frac{1}{2}} = \pm \frac{\sqrt{2}}{2} \]
Since we are in Quadrant I, where both \( x \) and \( y \) are positive, we take the positive root:
\[ x = \frac{\sqrt{2}}{2} \]
Thus, the coordinates of the point (x, y) on the unit circle are:
\[ \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) \]
The value of \( x \) is:
\[ \boxed{\frac{\sqrt{2}}{2}} \]
1 answer