Problem 62

Question

Show that \(\cos (2 \arccos (x))\) is the restriction to the interval [-1,1] of a polynomial in \(x\).

Step-by-Step Solution

Verified
Answer
The expression \( \cos(2 \arccos(x)) \) is \( 2x^2 - 1 \), a polynomial for \( x \in [-1, 1] \).
1Step 1: Understand the Problem
We need to find the expression for \( \cos(2 \arccos(x)) \) and show that it simplifies to a polynomial in \( x \) that is valid for \( x \) in the interval \([-1, 1]\).
2Step 2: Use Double Angle Formula for Cosine
Recall the double angle formula for cosine: \( \cos(2\theta) = 2\cos^2(\theta) - 1 \). Here, set \( \theta = \arccos(x) \), so \( \cos(\theta) = x \).
3Step 3: Substitute Values Into the Formula
Substitute \( \cos(\theta) = x \) into the double angle formula to get \( \cos(2\arccos(x)) = 2\cos^2(\arccos(x)) - 1 = 2x^2 - 1 \).
4Step 4: Verify Polynomial Form and Domain
The expression \( 2x^2 - 1 \) is a polynomial in \( x \). Since \( x \) is in \([-1, 1]\), this polynomial is valid for our problem.

Key Concepts

Double Angle FormulaArccosine FunctionPolynomial ExpressionInterval Restriction
Double Angle Formula
The Double Angle Formula is a key identity in trigonometry that allows us to express trigonometric functions of double angles in terms of single angles. This is particularly useful when simplifying expressions or solving equations involving trigonometric functions. The formula for the cosine of a double angle is given by:
  • \( \cos(2\theta) = 2\cos^2(\theta) - 1 \)
  • Alternative forms include \( \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) \) and \( \cos(2\theta) = 1 - 2\sin^2(\theta) \)
In our problem, we use the form \( \cos(2\theta) = 2\cos^2(\theta) - 1 \). This becomes especially powerful when combined with the concept of arccosines, as it enables transforming a more complex expression into a simpler polynomial form.
Arccosine Function
The Arccosine Function, also known as the inverse cosine function, is a fundamental inverse trigonometric function denoted by \( \arccos(x) \). It answers the question: "For what angle \( \theta \) does the cosine equal \( x \)?" Here are some critical points about the arccosine:
  • Its range is \([0, \pi]\), meaning it gives angles in radians between 0 and \( \pi \) (0 to 180 degrees).
  • The domain of arccosine is \([-1, 1]\) because cosine values only exist within this interval.
In the exercise, we set \( \theta = \arccos(x) \), which implies \( \cos(\theta) = x \). This substitution bridges the trigonometric function with our polynomial transformation. Using arccosine in this way brings us to the polynomial form after applying the Double Angle Formula.
Polynomial Expression
Polynomial expressions are algebraic expressions consisting of variables raised to whole number powers. They typically take the form \( a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \). These expressions are vital due to their simplicity and the fact that they can be defined for any real number input.

In the context of our exercise, the expression \( 2x^2 - 1 \) is a polynomial in \( x \). It arises after substituting \( \cos(\theta) = x \) into the Double Angle Formula and simplifies the trigonometric function \( \cos(2 \arccos(x)) \) into a form that fits a polynomial function, revealing the utility and beauty of these algebraic transformations.
Interval Restriction
An Interval Restriction involves limiting the domain or range of a function to ensure all resultant values are valid for a given context. In problems involving trigonometric identities, ensuring the domain meets the function's constraints is vital for accuracy.
  • The domain of \( \arccos(x) \) is \([-1, 1]\), which becomes the restriction for valid \( x \) values in the problem.
  • Since \( \cos(2 \arccos(x)) \) needs to be evaluated correctly, its restricted interval ensures this transformation to a polynomial remains valid.
This careful consideration of intervals helps align the polynomial form of \( 2x^2 - 1 \) with the restriction \( x \in [-1, 1] \). This insight ensures that all transformations and simplifications comply with the original trigonometric and inverse functions' constraints.