Problem 44 - Algebra and Trigonometry

Question

\(23-44=\) Find the exact value of the expression, if it is defined. \(\sin \left(\tan ^{-1}(-\sqrt{3})\right)\)

Step-by-Step Solution

Answer

\(-\frac{\sqrt{3}}{2}\)

Step 1: Understanding the Problem

We need to evaluate the expression \( \sin \left( \tan^{-1}(-\sqrt{3}) \right) \). This involves understanding the relationship between tangent and sine in a right triangle.

Step 2: Express in Terms of Right Triangle

\( \tan^{-1}(-\sqrt{3}) \) represents an angle \( \theta \) whose tangent is \(-\sqrt{3}\). We visualize a right triangle where the opposite side is \(-\sqrt{3}\) and the adjacent side is \(1\).

Step 3: Calculate Hypotenuse using Pythagorean Theorem

Apply the Pythagorean theorem: the hypotenuse \( h = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2 \).

Step 4: Find Sine of the Angle

Sine of an angle is the ratio of the opposite side to the hypotenuse. Therefore, \( \sin(\theta) = \frac{-\sqrt{3}}{2} \).

Step 5: Final Evaluation

Thus, the value of the expression \( \sin \left( \tan^{-1}(-\sqrt{3}) \right) \) is \(-\frac{\sqrt{3}}{2}\).

Key Concepts

Sine Function

The sine function is a primary trigonometric function. It relates the angle of a right triangle to the ratio of the length of the opposite side over the hypotenuse. This function is significant because it allows you to calculate unknown sides or angles in right triangles. When dealing with problems involving sine, remember:

• The range of the sine function is from -1 to 1.
• It is periodic with a period of \(2\pi\).

For example, in our given exercise, we calculated \( \sin(\theta) \) by using the formula \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \). This resulted in a value of \(-\frac{\sqrt{3}}{2}\) given the side lengths derived from the triangle for \( \theta = \tan^{-1}(-\sqrt{3}) \).

Tangent Function

The tangent function in trigonometry is vital for understanding angle relationships in right triangles. It is defined as the ratio of the length of the opposite side to the adjacent side. In symbols, this is expressed as \( \tan(\theta) = \frac{\text{opposite side}}{\text{adjacent side}} \).
For angles where tangent values are known, we can reverse the process using the inverse tangent, often written as \( \tan^{-1} \). This function gives us the angle when the tangent value is known. In our exercise, we used \( \tan^{-1}(-\sqrt{3}) \) to find \( \theta \), based on the understanding that the opposite side is \(-\sqrt{3}\) and the adjacent side is \(1\).
This relationship is crucial in constructing the triangle that helped us solve the expression.

Right Triangle

A right triangle is a triangle where one of its angles is 90 degrees. This type of triangle has specific properties that are useful in trigonometry:

• One angle is always 90 degrees.
• The other two angles must add up to 90 degrees.
• The side opposite the 90-degree angle is the longest and is called the hypotenuse.

In our example, we visualized a right triangle where the tangent of the angle \( \theta \) is known. The opposite side was \(-\sqrt{3}\), and the adjacent side was \(1\). This configuration helps us apply trigonometric functions to find the hypotenuse using the Pythagorean theorem, and ultimately, to find the sine of \( \theta \).

Pythagorean Theorem

The Pythagorean theorem remains a fundamental principle in trigonometry and geometry. It states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Mathematically, it is written as \( a^2 + b^2 = c^2 \), where \( c \) is the hypotenuse.
In the context of our exercise, we wanted to find the hypotenuse of the triangle where \( a = -\sqrt{3} \) and \( b = 1 \). Applying the theorem, we computed the hypotenuse \( c \) as follows:\[ h = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2 \]
Using this result, we then determined the sine of \( \theta \) by using the length of the opposite side over the calculated hypotenuse, which was essential in solving our original problem.