91Ó°ÊÓ

Understanding the Pythagorean Identity is crucial for solving trigonometric problems like the one in this exercise. The Pythagorean Identity for secant and tangent is expressed as: \[\mathrm {sec}^2(\theta) - \tan^2(\theta) = 1 \].
This equation showcases a fundamental relationship between the secant and tangent functions, which parallels the more familiar identity involving sine and cosine: \[ \sin^2(\theta) + \cos^2(\theta) = 1 \].
In simpler terms, it states that for any angle \( \theta \), the square of the secant of \( \theta \) minus the square of the tangent of \( \theta \) always equals 1.
This identity is extremely powerful in trigonometry, as it can simplify complex expressions and aid in solving equations. If you commit this identity to memory, it will make various trigonometric computations much easier.

The secant function, abbreviated as \( \sec \), is one of the six fundamental trigonometric functions. It is the reciprocal of the cosine function:
\[ \sec(\theta) = \frac{1}{\cos(\theta)} \].
Understanding secant can provide deeper insights into trigonometric relationships in both theoretical and practical contexts.
Let’s break it down further:

• When \( \cos(\theta) \) equals zero, \( \sec(\theta) \) is undefined because you cannot divide by zero.
• The secant function can tell us about the length of the hypotenuse relative to the adjacent side of a right triangle when \( \cos(\theta) \) is known.

In our exercise, we specifically work with \( \sec^2 \left( \frac{\pi}{15} \right) \).
Because of the Pythagorean Identity, we can say that \[ \sec^2 \left( \frac{\pi}{15} \right) - \tan^2 \left( \frac{\pi}{15} \right) = 1 \].

The tangent function, abbreviated as \( \tan \), is another fundamental trigonometric function. It is defined as the ratio of the sine and cosine functions:
\[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \].
Here are some key points about the tangent function:

• It is periodic with a period of \( \pi \).
• The tangent function is undefined when \( \cos(\theta) = 0 \), causing vertical asymptotes in its graph.
• In a right triangle, it represents the ratio of the length of the opposite side to the length of the adjacent side.

In the context of our exercise, we look at \( \tan^2 \left( \frac{\pi}{15} \right) \). By using the Pythagorean identity, \[ \sec^2 \left( \frac{\pi}{15} \right) - \tan^2 \left( \frac{\pi}{15} \right) \], we see that these terms interact closely.
Thanks to this identity, even without knowing the exact values of these functions at \( \frac{\pi}{15} \), we can confidently state that their relationship yields 1, simplifying our calculations significantly.