91Ó°ÊÓ

When working with angles in trigonometry, it's important to know how angles are positioned. In standard position, an angle always starts on the positive x-axis. Imagine drawing a ray starting from the origin (the point where the x-axis and y-axis intersect) outwards.
This initial ray is called the "initial side." When you measure an angle, you'll rotate this ray around the origin. The new direction of the ray after rotation is called the "terminal side." If the angle is positive, the rotation is counterclockwise. If the angle is negative, like in our problem with \(-60^{\circ}\), the rotation is clockwise, bringing our angle into the fourth quadrant (the section of the coordinate plane where both x is positive and y is negative).
A clear understanding of standard position helps when determining which trigonometric ratios to use since it sets the scene for your calculations.

Trigonometric ratios are relationships that help us find unknown side lengths and angles in right triangles. With standard position, any point on the circle created by rotating the angle can have its x and y coordinates determined by these ratios.
The key ratios are sine (\(\sin\theta\)\bullet), cosine (\(\cos\theta\)\bullet), and tangent (\(\tan\theta\)\bullet). However, for a point \(A\) on a circle of radius \(r\), the \(\cos\theta\) ratio gives the x-coordinate: \(x = r \cdot \cos\theta\), and the \(\sin\theta\) ratio provides the y-coordinate: \(y = r \cdot \sin\theta\). This comes in handy when finding exact positions in terms of \(x\) and \(y\) on a circle at a certain angle.
For our problem, \(-60^{\circ}\) means using \(\cos(-60^{\circ}) = \frac{1}{2}\)\bullet) and \(\sin(-60^{\circ}) = -\frac{\sqrt{3}}{2}\)\bullet) tailored to the fourth quadrant to compute coordinates effectively.

Knowing where a point lies based on its coordinates involves using the circle’s radius and an angle in standard position. In our problem, with a radius of \(2\) and a circle centered at the origin, finding the position of point \(A\) involves plugging values into the trigonometric formulas.
The circle is essentially a set of all points \((x, y)\) where \(x^{2} + y^{2} = r^{2}\). To determine where along this circle point \(A\) is, calculate:

• \(x = 2 \cdot \cos(-60^{\circ}) = 2 \cdot \frac{1}{2} = 1\)
• \(y = 2 \cdot \sin(-60^{\circ}) = 2 \cdot \left(-\frac{\sqrt{3}}{2}\right) = -\sqrt{3}\)

Hence, the coordinates are exactly \((1, -\sqrt{3})\), confirming our position on the unit circle in this case.

Simplifying radicals is pivotal in expressing results in their simplest form, especially in trigonometry. Radicals can often involve square roots of numbers, which sometimes aren't whole or perfect square numbers. Our goal is to express the radical in the most straightforward manner.
In our problem, \(-\sqrt{3}\) already represents a simplified state. When simplifying, you often look to:

• Separate the radical to its components, like separating nearest square roots from the number.
• Identify if components can be written as squares, such as \(\sqrt{12}\) becoming \(2\sqrt{3}\).

But in cases like \(-\sqrt{3}\), further simplification isn't necessary. Thus, the coordinates \((1, -\sqrt{3})\) reflect the most reduced form given the initial problem context.