Solving For L = Sin²α - Cos²α Given Tan Α = Tan²(π/3)

In this mathematical problem, we are given that $\alpha$ is an acute angle, and the tangent of $\alpha$ is equal to the square of the tangent of $\pi/3$. Our goal is to calculate the value of $L$, which is defined as the difference between the square of the sine of $\alpha$ and the square of the cosine of $\alpha$. This problem involves trigonometric identities and a step-by-step approach to arrive at the solution. In this article, we will go through the necessary steps to solve for the value of $L$. We will delve into the fundamentals of trigonometry, including trigonometric functions and their relationships, to find the sine and cosine of $\alpha$. We will also use trigonometric identities to simplify the expression and get the value of $L$. Solving this problem requires understanding of trigonometric ratios, especially tangent, sine, and cosine, as well as their interrelations. Furthermore, familiarity with the unit circle and special angles such as $\pi/3$ is essential. The application of trigonometric identities, like the Pythagorean identity, helps simplify the equation and arrive at the final answer. This exercise is a great way to practice trigonometric problem-solving, enhancing comprehension of core concepts and their practical application. Through detailed explanations and step-by-step calculations, we aim to make the solution clear and accessible.

Given that $\alpha$ is an acute angle and $\operatorname{tg} \alpha = \operatorname{tg}^2 (\pi/3)$, we need to calculate:

$L = \operatorname{sen}^2 \alpha - \cos^2 \alpha$

We are provided with multiple-choice options: a) 0.2 b) 0.3 c) 0.4 d) 0.6 e) 0.8

Step 1: Calculate $\operatorname{tg}^2(\pi/3)$

First, we need to find the value of $\operatorname{tg}(\pi/3)$. We know that $\pi/3$ radians is equal to 60 degrees. The tangent of 60 degrees ($\operatorname{tg} 60^\circ$) is $\sqrt{3}$.

$\operatorname{tg}(\pi/3) = \sqrt{3}$

Now, we need to square this value:

$\operatorname{tg}^2(\pi/3) = (\sqrt{3})^2 = 3$

Therefore, we have:

$\operatorname{tg} \alpha = 3$

Step 2: Relate $\operatorname{tg} \alpha$ to $\operatorname{sen} \alpha$ and $\cos \alpha$

We know that $\operatorname{tg} \alpha = \frac{\operatorname{sen} \alpha}{\cos \alpha}$. So:

$\frac{\operatorname{sen} \alpha}{\cos \alpha} = 3$

This implies:

$\operatorname{sen} \alpha = 3 \cos \alpha$

Step 3: Use the Pythagorean Identity

The Pythagorean identity states that:

$\operatorname{sen}^2 \alpha + \cos^2 \alpha = 1$

We substitute $\operatorname{sen} \alpha = 3 \cos \alpha$ into the identity:

$(3 \cos \alpha)^2 + \cos^2 \alpha = 1$

$9 \cos^2 \alpha + \cos^2 \alpha = 1$

$10 \cos^2 \alpha = 1$

$\cos^2 \alpha = \frac{1}{10}$

Step 4: Calculate $\operatorname{sen}^2 \alpha$

Using $\operatorname{sen} \alpha = 3 \cos \alpha$, we have:

$\operatorname{sen}^2 \alpha = (3 \cos \alpha)^2 = 9 \cos^2 \alpha$

Substituting $\cos^2 \alpha = \frac{1}{10}$:

$\operatorname{sen}^2 \alpha = 9 \times \frac{1}{10} = \frac{9}{10}$

Step 5: Calculate L

Now we can find the value of $L$:

$L = \operatorname{sen}^2 \alpha - \cos^2 \alpha = \frac{9}{10} - \frac{1}{10} = \frac{8}{10} = 0.8$

Therefore, the value of $L$ is 0.8.

To fully grasp the solution, it's important to understand the underlying trigonometric concepts. Trigonometry deals with the relationships between angles and sides of triangles. The primary trigonometric functions are sine ($\operatorname{sen}$), cosine ($\operatorname{cos}$), and tangent ($\operatorname{tg}$). These functions relate an angle in a right-angled triangle to ratios of its sides.

Trigonometric Functions

• Sine ($\operatorname{sen} \alpha$): In a right triangle, the sine of an angle $\alpha$ is the ratio of the length of the side opposite the angle to the length of the hypotenuse.
• Cosine ($\operatorname{cos} \alpha$): The cosine of an angle $\alpha$ is the ratio of the length of the adjacent side to the length of the hypotenuse.
• Tangent ($\operatorname{tg} \alpha$): The tangent of an angle $\alpha$ is the ratio of the length of the opposite side to the length of the adjacent side. It can also be expressed as the sine of the angle divided by the cosine of the angle, i.e., $\operatorname{tg} \alpha = \frac{\operatorname{sen} \alpha}{\cos \alpha}$.

Special Angles

Certain angles, such as 30°, 45°, and 60° (or $\pi/6$, $\pi/4$, and $\pi/3$ radians, respectively), are considered special angles because their trigonometric function values have simple, exact expressions. In this problem, we encountered the angle $\pi/3$ (60°). The tangent of 60° is $\sqrt{3}$.

Pythagorean Identity

The Pythagorean identity is a fundamental trigonometric identity that relates the sine and cosine of an angle: $\operatorname{sen}^2 \alpha + \cos^2 \alpha = 1$. This identity is derived from the Pythagorean theorem, which states that in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

Acute Angle

An acute angle is an angle that measures less than 90° (or $\pi/2$ radians). Since $\alpha$ is an acute angle in this problem, it falls within the first quadrant of the unit circle, where all trigonometric functions (sine, cosine, tangent) are positive.

Another way to solve this problem involves directly using the trigonometric identity that relates $L$ to $\cos(2\alpha)$.

$L = \operatorname{sen}^2 \alpha - \cos^2 \alpha = -\cos(2\alpha)$

We know that $\operatorname{tg} \alpha = 3$. We can use the identity:

$\cos(2\alpha) = \frac{1 - \operatorname{tg}^2 \alpha}{1 + \operatorname{tg}^2 \alpha}$

Plugging in $\operatorname{tg} \alpha = 3$:

$\cos(2\alpha) = \frac{1 - 3^2}{1 + 3^2} = \frac{1 - 9}{1 + 9} = \frac{-8}{10} = -0.8$

So, $L = -\cos(2\alpha) = -(-0.8) = 0.8$.

This approach uses the double angle formula for cosine and provides a quicker way to the solution.

By following a step-by-step approach and using trigonometric identities, we have successfully calculated the value of $L$ to be 0.8. This problem demonstrates the importance of understanding trigonometric functions, their relationships, and the application of identities in solving mathematical problems. We have explored both a direct method using the Pythagorean identity and an alternative method using the double angle formula, which reinforces the versatility of trigonometric problem-solving techniques. Mastering these concepts is crucial for more advanced topics in mathematics and physics.

The correct answer is: e) 0.8