91Ó°ÊÓ

A reference angle is the smallest angle between the terminal side of the given angle and the x-axis. It helps to simplify the calculations of trigonometric functions for non-acute angles.

In this exercise, the angle given is \(\frac{3 \pi}{4}\), which is in the second quadrant. Reference angles are always measured from the x-axis, so now we will subtract the given angle from \(\pi\) to find the reference angle.

To find the reference angle of \(\frac{3 \pi}{4}\):
- Subtract the angle from \(\pi\): \(\pi - \frac{3 \pi}{4} = \frac{\pi}{4}\).

Thus, the reference angle is \(\frac{\pi}{4}\).

Using the reference angle helps because trigonometric functions of an angle and its reference angle have the same absolute values but may differ in sign depending on the quadrant.

Understanding the four quadrants of the coordinate plane is essential in trigonometry. Each quadrant has distinctive properties that affect the signs of the trigonometric functions.

Here is a brief overview:

• Quadrant I (angles between 0 to \(\pi/2\)): All trigonometric functions are positive.
• Quadrant II (angles between \(\pi/2\) to \(\pi\)): Sine is positive, and cosine and tangent are negative.
• Quadrant III (angles between \(\pi\) to \(3\pi/2\)): Tangent is positive, and sine and cosine are negative.
• Quadrant IV (angles between \(3\pi/2\) to \(2\pi\)): Cosine is positive, and sine and tangent are negative.

In our problem, the angle \(\frac{3 \pi}{4}\) is in the second quadrant, where sine is positive. This is why the solution \(\frac{1}{\sqrt{2}}\) is positive.

The sine function is one of the basic trigonometric functions that relates the angle in a right triangle to the ratio of the length of the opposite side to the hypotenuse.

In terms of the unit circle, which simplifies understanding of angles and their trigonometric values:
- The sine of an angle represents the y-coordinate of the point on the unit circle corresponding to that angle.

For the given angle \(\frac{3 \pi}{4}\):
- Using the reference angle \(\frac{\pi}{4}\):
\(\text{Since } \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}, \text{ and } \sin \) is positive in the second quadrant:
\(\therefore \sin \frac{3 \pi}{4} = \frac{1}{\sqrt{2}} \).

Understanding the sine function and reference angles helps not only in this problem but in many other trigonometric problems.