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Magazine Chapter 6: Problem 30
Find a simplified expression for each of the following. $$ \cos \left(\tan ^{-1}(4 x)\right) $$
Short Answer
Expert verified
\( \frac{1}{\sqrt{16x^2 + 1}} \)
Step by step solution
01
Understand the Problem
The problem involves simplifying the expression \( \cos(\tan^{-1}(4x)) \). Our goal is to express this in a simpler form without inverse trigonometric functions.
02
Use the Right Triangle and Definitions
We consider the right triangle definition where \( \theta = \tan^{-1}(4x) \). This implies that \( \tan \theta = 4x \). In a right triangle, if \( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{4x}{1} \), we create a triangle where the opposite side is \( 4x \), and the adjacent side is 1.
03
Calculate the Hypotenuse
Using the Pythagorean Theorem, the hypotenuse \( h \) can be calculated as \( h = \sqrt{(4x)^2 + 1^2} = \sqrt{16x^2 + 1} \).
04
Determine \( \cos \theta \)
In this triangle, \( \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{16x^2 + 1}} \).
05
Simplify Expression
So, \( \cos(\tan^{-1}(4x)) = \frac{1}{\sqrt{16x^2 + 1}} \). This is the simplified form of the expression.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Inverse Trigonometric Functions
Inverse trigonometric functions are used to determine the angles when the ratios of the sides of a right triangle are known. They're essentially the reverse processes of the trigonometric functions like sine, cosine, and tangent. In this exercise, we encounter the inverse tangent, written as \( \tan^{-1} \) or often \( \arctan \). The symbol \( \tan^{-1}(x) \) translates to "the angle whose tangent is \( x \)."
This concept is useful when solving for angles in triangles or converting expressions for those working with complex trigonometric models.
This concept is useful when solving for angles in triangles or converting expressions for those working with complex trigonometric models.
- For instance, if \( \tan \theta = x \), then \( \theta = \tan^{-1}(x) \).
- Inverse trigonometric functions help in expressing angles directly in algebraic terms, rather than through measurements.
Right Triangle
A right triangle is a geometric figure with one angle equal to 90 degrees. The sides connected with the 90-degree angle are the legs, and the side opposite is the hypotenuse. Right triangles lay the foundation for trigonometric functions since each function relates angles to side ratios of right triangles.
For example, in our problem, we used a right triangle:
For example, in our problem, we used a right triangle:
- Label the opposite side to the angle \( \theta \) as \( 4x \) and the adjacent side as 1. This aligns with the equation \( \tan \theta = \frac{4x}{1} \).
- The hypotenuse is then calculated using the Pythagorean theorem. This defines our triangle structure, aiding in identifying the cosine of the angle.
Pythagorean Theorem
The Pythagorean theorem is a fundamental principle in mathematics that is used to find the length of one side of a right triangle when the other two sides are known. The theorem is expressed by the equation:\[ a^2 + b^2 = c^2 \]where \( c \) is the hypotenuse, and \( a \) and \( b \) are the legs of the triangle.
In our problem, the theorem helps calculate the hypotenuse \( h \) of the triangle given by:\[\tan \theta = \frac{4x}{1}\]
In our problem, the theorem helps calculate the hypotenuse \( h \) of the triangle given by:\[\tan \theta = \frac{4x}{1}\]
- Set \( a = 4x \) and \( b = 1 \).
- Then \( h = \sqrt{(4x)^2 + 1^2} = \sqrt{16x^2 + 1} \).
Simplifying Expressions
Simplifying expressions is about reducing mathematical expressions to their simplest form. Often, simplifying involves using algebraic identities or operations to express a function in a more straightforward or more informative way.
In our specific task, modifying \( \cos(\tan^{-1}(4x)) \) into its simplest form involves:
In our specific task, modifying \( \cos(\tan^{-1}(4x)) \) into its simplest form involves:
- Determining \( \cos \theta \) using the triangle: \( \cos \theta = \frac{1}{h} = \frac{1}{\sqrt{16x^2 + 1}} \).
- Thus, simplifying terms and reducing it to a clean expression devoid of inverse trig functions.
