Problem 120 - Algebra and Trigonometry Real Mathematics, Real People

Question

Write the trigonometric expression as an algebraic expression. $$\cos (2 \arctan x)$$

Step-by-Step Solution

Answer

\(\cos (2 \arctan x) = \frac{1 -2x^2}{1 + x^2}\)

Step 1: Apply the double-angle formula

\(\cos (2 \arctan x)\) can be written in terms of sine or cosine using the double-angle formula. Let's go with the version that converts the cosine into sine, \(\cos (2A) = 1 - 2 \sin^2 (A)\). So, we get \(\cos (2 \arctan x) = 1 - 2 \sin^2 (\arctan x)\).

Step 2: Express sin(A) in terms of tan(A)

Recall the Pythagorean identity, \(\sin^2(A) + \cos^2(A) = 1\). Expressing sine in terms of tangent using this identity gives us \(\sin(A) = \frac{\tan(A)}{\sqrt{1+\tan^2(A)}}\). Substituting A with \(\arctan x\) and simplifying, we get \(\sin (\arctan x) = \frac{x}{\sqrt{1+x^2}}\). Squaring it gives us \(\sin^2 (\arctan x) = \frac{x^2}{1+x^2}\).

Step 3: Substitute sin^2(A) in the expression found in Step 1

Revisiting our expression from Step 1, replace sin^2 of \(\arctan x\) with the algebraic equivalent found in Step 2, we get \(\cos (2 \arctan x) = 1 - 2 \cdot \frac{x^2}{1+x^2} = \frac{1 -2x^2}{1 + x^2}\).

Key Concepts

Double-Angle Formulas

The double-angle formulas are used to simplify expressions involving trigonometric functions multiplied by two. Particularly, the cosine double-angle formula is helpful in transforming expressions like \( \cos(2A) \) into simpler algebraic forms. Here’s how it looks:

• \( \cos(2A) = \cos^2(A) - \sin^2(A) \)
• Alternatively, it can be adjusted to \( \cos(2A) = 1 - 2\sin^2(A) \) or \( \cos(2A) = 2\cos^2(A) - 1 \)

For the step-by-step exercise, \( \cos(2\arctan x) \) was rewritten using the version \( \cos(2A) = 1 - 2\sin^2(A) \). By substituting \( A \) with \( \arctan x \), the expression is transformed into one involving \( \sin^2(\arctan x) \), setting the stage for further simplification.

Pythagorean Identity

The Pythagorean identity is a fundamental relation in trigonometry providing a direct link between sine and cosine functions: \( \sin^2(A) + \cos^2(A) = 1 \). This identity is crucial when we need to express sine or cosine in terms of tangent.

• In our exercise, to find \( \sin(A) \) using \( \tan(A) \), the formula is rearranged to \( \sin(A) = \frac{\tan(A)}{\sqrt{1+\tan^2(A)}} \).

For \( A = \arctan x \), the Pythagorean identity helps to express \( \sin(\arctan x) \) in a simpler algebraic form: \( \frac{x}{\sqrt{1+x^2}} \). This expression allows us to replace \( \sin^2(\arctan x) \) in the larger equation efficiently.

Inverse Trigonometric Functions

Inverse trigonometric functions are used to find the angle whose trigonometric function is known. In this exercise, \( \arctan x \) is the inverse tangent function, returning the angle \( A \) such that \( \tan A = x \).

• The function \( \arctan x \) specifically provides an angle in the range \(-\frac{\pi}{2} \) to \( \frac{\pi}{2} \).

Understanding inverse functions is key because they allow us to switch between different forms of trigonometric expressions. Here, knowing \( \arctan x \) means we can derive expressions involving \( \sin(\arctan x) \) and \( \cos(2\arctan x) \), eventually solving the original problem by leveraging relationships like the Pythagorean identity.