Question

on a unit circle, the vertical distance from the x - axis to a point on the perimeter of the circle is twice the horizontal distance from the y - axis to the same point. what is sinθ?

Explanation:

Step1: Define the coordinates of the point

Let the coordinates of the point on the unit - circle be \((x,y)\). The vertical distance from the \(x\) - axis to the point is \(|y|\), and the horizontal distance from the \(y\) - axis to the point is \(|x|\). Given that \(|y| = 2|x|\).

Step2: Use the equation of the unit - circle

The equation of a unit - circle is \(x^{2}+y^{2}=1\). Substitute \(y = 2x\) or \(y=-2x\) into the equation of the unit - circle. If \(y = 2x\), then \(x^{2}+(2x)^{2}=1\), which simplifies to \(x^{2}+4x^{2}=1\), or \(5x^{2}=1\), so \(x^{2}=\frac{1}{5}\) and \(x=\pm\frac{1}{\sqrt{5}}\). Then \(y = 2x=\pm\frac{2}{\sqrt{5}}\).

Step3: Recall the definition of sine

For a point \((x,y)\) on the unit - circle, \(\sin\theta=y\). Since \(y = \pm\frac{2}{\sqrt{5}}=\pm\frac{2\sqrt{5}}{5}\), and considering the four - quadrant cases:

• In the first and second quadrants (\(y>0\)), \(\sin\theta=\frac{2\sqrt{5}}{5}\); in the third and fourth quadrants (\(y < 0\)), \(\sin\theta=-\frac{2\sqrt{5}}{5}\).

Answer:

\(\pm\frac{2\sqrt{5}}{5}\)