Question

Draw a right triangle to simplify the given expressions. $$\cos \left(\tan ^{-1}\left(\frac{x}{\sqrt{9-x^{2}}}\right)\right)$$

Short answer

Question: Find the cosine of the angle whose tangent is equal to \(\frac{x}{\sqrt{9-x^2}}\) in a right triangle. Answer: The cosine of the angle is \(\frac{\sqrt{9-x^2}}{3}\).

Step-by-step solution

1. Draw the right triangle and label sides

Draw a right triangle ABC with angle \(\theta\) at vertex B. Since the tangent of the angle is given, let side AB (opposite to angle \(\theta\)) be x, and side BC (adjacent to angle \(\theta\)) be \(\sqrt{9 - x^2}\).

2. Determine the hypotenuse using the Pythagorean theorem

In the right triangle ABC, we apply the Pythagorean theorem to find side AC (the hypotenuse). The theorem states that the sum of the squares on the legs of the right triangle is equal to the square on the hypotenuse: $$AB^2 + BC^2 = AC^2$$ $$x^2 + (\sqrt{9-x^2})^2 = AC^2$$ $$(x^2 + (9 - x^2)) = AC^2$$ $$AC = \sqrt{9}$$

3. Find the cosine of the right triangle

By definition, the cosine of an angle in a right triangle is the ratio of the adjacent side to the hypotenuse. In our case, the adjacent side is BC (length \(\sqrt{9-x^2}\)) and the hypotenuse is AC (length \(\sqrt{9}\)): $$\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{9-x^2}}{\sqrt{9}}$$

4. Simplify the expression

The expression we need to evaluate is: $$\cos(\tan^{-1}(\frac{x}{\sqrt{9-x^2}}))$$ We found that \(\cos(\theta) = \frac{\sqrt{9-x^2}}{\sqrt{9}}\). Since the cosine is the same as the angle whose tangent is \(\frac{x}{\sqrt{9-x^2}}\), we can substitute our result into the expression: $$\cos(\tan^{-1}(\frac{x}{\sqrt{9-x^2}})) = \frac{\sqrt{9-x^2}}{\sqrt{9}} = \frac{\sqrt{9-x^2}}{3}$$ So the given expressions simplifies to: $$\frac{\sqrt{9-x^2}}{3}$$

Key concepts

Inverse Trigonometric Functions

Inverse trigonometric functions help us find angles from trigonometric ratios. They are the inverse operations of the regular trigonometric functions, such as sine, cosine, and tangent. These functions include \( \arcsin \), \( \arccos \), and \( \arctan \). They answer questions like: "What angle has a given sine, cosine, or tangent value?"

In the exercise, the inverse tangent, \( \tan^{-1} \), is used. This helps us determine the angle \( \theta \) in a right triangle when the ratio \( \frac{x}{\sqrt{9-x^2}} \) is known.

• Function: \( \tan^{-1} \left( \frac{x}{\sqrt{9-x^2}} \right) \) gives angle \( \theta \).
• Application: Helps to find \( \theta \) so other trigonometric values can be determined.

Inverse functions are essential because they translate trigonometric ratios back into practical angles. This allows us to work backwards from known side lengths to discover important angle measures.

Pythagorean Theorem

The Pythagorean Theorem is a cornerstone of geometry. It relates the sides of a right triangle, providing a way to find the length of a side if the other two are known. It is written as \( a^2 + b^2 = c^2 \), where \( c \) is the hypotenuse—the longest side opposite the right angle.

In the problem, the theorem calculates the hypotenuse of the triangle, side AC. Knowing that \( AB = x \) and \( BC = \sqrt{9-x^2} \):

• \( AB^2 + BC^2 = AC^2 \)
• \( x^2 + (9-x^2) = 9 \)
• \( AC = \sqrt{9} = 3 \)

This theorem is instrumental for solving right triangles; it allows finding the unknown side length, ensuring precise calculations as seen in the triangle applied here.

Right Triangle

A right triangle is a triangle where one of the angles is exactly 90 degrees. It has special properties allowing the use of trigonometric functions to solve for unknown sides and angles. These triangles are critical because of how they simplify complex problems.

In the context of the exercise, a right triangle is constructed with an angle \( \theta \) based on the tangent ratio \( \frac{x}{\sqrt{9-x^2}} \). Understanding a right triangle involves:

• Identifying the right angle and labeling sides (opposite, adjacent, hypotenuse).
• Using trigonometric definitions to relate sides and angles.
• Applying the Pythagorean theorem for precise calculations.

The simplicity of right triangles makes them useful for teaching the relationship between angles and side lengths, as highlighted in this exercise.