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on a unit circle, the vertical distance from the x - axis to a point on…

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 sin e?

Explanation:

Step1: Define coordinates

Let 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 unit - circle equation

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

Step3: Recall the definition of sine

For a point $(x,y)$ on the unit - circle, $\sin\theta=y$. Since $|y| = 2|x|$ and $x^{2}+y^{2}=1$, we have two cases: when $y=\frac{2}{\sqrt{5}}$ and when $y =-\frac{2}{\sqrt{5}}$. So $\sin\theta=\pm\frac{2}{\sqrt{5}}=\pm\frac{2\sqrt{5}}{5}$.

Answer:

$\pm\frac{2\sqrt{5}}{5}$