Problem 70 - Precalculus Essentials

Question

Use a right triangle to write each expression as an algebraic expression. Assume that \(x\) is positive and that the given inverse trigonometric function is defined for the expression in \(x\). $$\cot \left(\tan ^{-1} \frac{x}{\sqrt{2}}\right)$$

Step-by-Step Solution

Answer

The algebraic expression for \( \cot(\tan^{-1} \frac{x}{\sqrt{2}}) \) is \( \frac{\sqrt{2}}{x} \)

Step 1: Simplify Inside the Parentheses

Given: \( \cot(\tan^{-1} \frac{x}{\sqrt{2}}) \). We are considering \(\tan^{-1}\) as an angle. We can assume that \(\tan^{-1} \frac{x}{\sqrt{2}} = \theta \), then \( \theta \) is an angle whose tangent is \( \frac{x}{\sqrt{2}} \).

Step 2: Create a Right Triangle

We can draw a right triangle such that the opposite side 'O' corresponds to 'x' (from the numerator of the expression) and the adjacent side 'A' corresponds to \(\sqrt{2}\) (from the denominator). Consequently, the hypotenuse 'H' will be \(sqrt{x^2+2}\) using Pythagoras’ theorem as 'H' = \(\sqrt{O^2+A^2}\). The angle opposite to the side 'O' will be \( \theta \)

Step 3: Substitute cotangent Equation

Cotangent is the reciprocal of the tangent i.e, \(\cot{\theta} = \frac{A}{O}\). Substituting the values of 'A' and 'O' from the triangle, we get \(\cot{\theta} = \frac{\sqrt{2}}{x}\). Since we assumed \(\tan^{-1} \frac{x}{\sqrt{2}} = \theta \), we can substitute this to our main equation.

Key Concepts

Right Triangle

A right triangle is a type of triangle that has one of its angles equal to 90 degrees. This makes it easy to apply certain mathematical principles, such as trigonometric ratios and the Pythagorean theorem. In our exercise, we visualized a right triangle to represent the angle \( \theta \) as \( \tan^{-1} \frac{x}{\sqrt{2}} \).
Understanding the sides of a right triangle is crucial:

• Opposite Side: This is the side opposite the angle of interest (here represented by \( x \)).
• Adjacent Side: This is the side adjacent to the angle \( \theta \) (here represented by \( \sqrt{2} \)).
• Hypotenuse: The longest side of the triangle, opposite the right angle.

By positioning \( x \) as the opposite and \( \sqrt{2} \) as the adjacent side, we set the stage to calculate the hypotenuse using the Pythagorean theorem. This setup helps us to evaluate trigonometric functions like cotangent effectively.

Pythagorean Theorem

The Pythagorean theorem is a fundamental relation in geometry that relates the three sides of a right triangle. It states that the sum of the squares of the two shorter sides equals the square of the hypotenuse. The formula can be expressed as:
\[ c^2 = a^2 + b^2 \] where \( c \) is the hypotenuse and \( a \) and \( b \) are the other two sides.
In our problem, when we set \( O = x \) and \( A = \sqrt{2} \), we applied the theorem to find the hypotenuse:
\[ H = \sqrt{x^2 + 2} \]
This allows us to fully understand the spatial relationships within our triangle and provides the necessary ratios for further trigonometric calculations. Deriving the hypotenuse ensures that we can calculate cotangent correctly, as the problem demands.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the variables. Two well-known functions related to right triangles are tangent and cotangent.
The tangent of an angle is the ratio of the opposite side to the adjacent side, while cotangent is its reciprocal. Knowing that,
\( \tan{\theta} = \frac{O}{A} \) implies \( \cot{\theta} = \frac{A}{O} \).
For our problem, \( \theta = \tan^{-1} \frac{x}{\sqrt{2}} \). Therefore, the cotangent becomes:
\[ \cot{\theta} = \frac{\sqrt{2}}{x} \]
This expression stems from reversing the tangent ratio. By understanding these basic identities, students can manipulate and simplify expressions involving inverse trigonometric functions, analyzing them step by step. These identities form the backbone of solving problems involving angles and sides in right triangles.