Question

suppose that (\frac{\sqrt{2}}{2},y) is a point in quadrant iv lying on the unit circle. find y. write the exact value, not a decimal approximation. y = -\frac{\sqrt{2}}{2}

Explanation:

Step1: Recall unit - circle equation

The equation of the unit circle is $x^{2}+y^{2}=1$. Given $x = \frac{\sqrt{2}}{2}$, we substitute $x$ into the equation: $(\frac{\sqrt{2}}{2})^{2}+y^{2}=1$.

Step2: Simplify the left - hand side

$(\frac{\sqrt{2}}{2})^{2}=\frac{2}{4}=\frac{1}{2}$, so the equation becomes $\frac{1}{2}+y^{2}=1$.

Step3: Solve for $y^{2}$

Subtract $\frac{1}{2}$ from both sides: $y^{2}=1 - \frac{1}{2}=\frac{1}{2}$.

Step4: Determine the sign of $y$

Since the point $(\frac{\sqrt{2}}{2},y)$ is in Quadrant IV, $y\lt0$. Taking the square root of $y^{2}=\frac{1}{2}$, we get $y =-\frac{\sqrt{2}}{2}$.

Answer:

$y =-\frac{\sqrt{2}}{2}$