Problem 38 - Precalculus: A Concise Course

Question

Evaluate the trigonometric function using its period as an aid. $$ \sin \frac{9 \pi}{4} $$

Step-by-Step Solution

Answer

\(\sin \frac{9 \pi}{4} = \frac{\sqrt{2}}{2}\)

Step 1: Recognizing period of sine function

The sine function, \(\sin(x)\), is a periodic function with a period of \(2\pi\). This means that every \(2\pi\) the function will repeat its values, or in other words, \(\sin(x + 2n\pi) = \sin(x)\), where \(n\) is any integer.

Step 2: Calculation of an equivalent angle in range [0, 2π]

The angle \(\frac{9 \pi}{4}\) is more than \(2\pi\) so it is out of the primary period. An equivalent angle within the range [0, \(2\pi\)] can be found by subtracting \(2\pi\) (or multiples thereof) from the original angle until it falls within the desired range. In terms of quarters of Pi, \(2\pi\) is the same as \(8\pi/4\), so \(\frac{9 \pi}{4} - 2\pi = \frac{9 \pi}{4} - \frac{8 \pi}{4} = \frac{\pi}{4}\).

Step 3: Evaluating sine of the calculated angle

Now the problem is reduced to the well known trigonometric value of \(\sin(\frac{\pi}{4})\). This value is well known and equals \(\frac{\sqrt{2}}{2}\).

Key Concepts

Period of Sine Function

The sine function is one of the most fundamental concepts in trigonometry, repeating its pattern in a predictable cycle known as its period. The period of the sine function, denoted as \( 2\pi \), is the length of one complete cycle of the wave. In simpler terms, it's just like the circular track of a Ferris wheel where, after one full rotation, the seats return to their initial position.

Understanding this concept is crucial for evaluating trigonometric functions, especially when dealing with angles greater than \( 2\pi \) or less than \( -2\pi \). By leveraging the period, we can \