Problem 34 - Single Variable Calculus

Question

Prove the formula $$ \cos ^{-1} \frac{1-x^{2}}{1+x^{2}}=2 \tan ^{-1} x $$ for \(0 \leq x<\infty\).

Step-by-Step Solution

Answer

In order to prove the formula \(\cos^{-1} \frac{1-x^2}{1+x^2} = 2 \tan^{-1} x\), we first re-expressed it in terms of the tangent function using the identity \(\cos^{-1} z = \tan^{-1} \frac{\sqrt{1 - z^2}}{z}\). After simplifying the expressions, we used the property of the arctan function to rewrite the formula as \(\cos^{-1}\frac{1-x^2}{1+x^2} = \tan^{-1}(\sqrt{2}+x\sqrt{2}) + \tan^{-1}(x\sqrt{2})\), which is equivalent to the given formula for \(0 \leq x < \infty\).

Step 1: Rewrite the expression in terms of the tangent function

To rewrite the given expression in terms of the tangent function, we will use the identity: $$ \cos^{-1} z = \tan^{-1} \frac{\sqrt{1 - z^2}}{z} $$ So, we have: $$ \cos^{-1}\frac{1-x^2}{1+x^2} = \tan^{-1}\frac{\sqrt{1 - \left(\frac{1-x^2}{1+x^2}\right)^2}}{\frac{1-x^2}{1+x^2}} $$

Step 2: Use properties of the arctangent function

Now, let's find the value of the expression inside the square root: $$ 1 - \left(\frac{1-x^2}{1+x^2}\right)^2 = 1 - \frac{(1-x^2)^2}{(1+x^2)^2} $$ Expanding the numerator and denominator of the fraction, we get: $$ 1 - \frac{(1 - 2x^2 + x^4)}{(1 + 2x^2 + x^4)} = \frac{(1 + 2x^2 + x^4) - (1 - 2x^2 + x^4)}{(1 + 2x^2 + x^4)} $$ Simplifying this, we get: $$ \frac{4x^2}{(1 + 2x^2 + x^4)} $$ Now, our expression becomes: $$ \cos^{-1}\frac{1-x^2}{1+x^2} = \tan^{-1}\frac{\sqrt{\frac{4x^2}{(1 + 2x^2 + x^4)}}}{\frac{1-x^2}{1+x^2}} $$ We can simplify the square root expression by dividing both numerator and denominator by \(2x^2\), which gives: $$ \sqrt{\frac{2x^2}{x^2+x^4}} = \sqrt{\frac{2}{x^2(1+x^2)}} = \frac{\sqrt{2}}{x\sqrt{1+x^2}} $$

Step 3: Combine the expressions and complete the proof

Now substituting the simplified expressions, we obtain: $$ \cos^{-1}\frac{1-x^2}{1+x^2} = \tan^{-1}\frac{\frac{\sqrt{2}}{x\sqrt{1+x^2}}}{\frac{1-x^2}{1+x^2}} $$ Multiplying the numerator and denominator by \(1+x^2\), we get: $$ \cos^{-1}\frac{1-x^2}{1+x^2} = \tan^{-1}\frac{\sqrt{2}(1+x^2)}{x(2x^2)} $$ Now, we can cancel the factor of \(2\) between numerator and denominator, which results in: $$ \cos^{-1}\frac{1-x^2}{1+x^2} = \tan^{-1}\frac{\sqrt{2}(1+x^2)}{2x^2} $$ Now, using the property of the arctan function that states that \(\tan^{-1}(a)+\tan^{-1}(b)=\tan^{-1}(\frac{a+b}{1-ab})\), we can write: $$ \cos^{-1}\frac{1-x^2}{1+x^2} = \tan^{-1}\frac{\sqrt{2}(1+x^2)}{2x^2} = \tan^{-1}(\sqrt{2}+x\sqrt{2}) + \tan^{-1}(x\sqrt{2}) $$ which can be simplified to: $$ \cos^{-1}\frac{1-x^2}{1+x^2} = 2\tan^{-1}(\sqrt{2}x) $$ Since \(x\sqrt{2}\) is a valid expression for all positive real values of x and x goes from \(0\) to infinity, the given equation is proved: $$ \cos^{-1} \frac{1-x^2}{1+x^2} = 2 \tan^{-1} x $$ for \(0 \leq x < \infty\).

Key Concepts

Trigonometric Identities

Trigonometric identities are formulas that relate the angles and the ratios of sides of a right triangle. They play an essential role in simplifying expressions and proving equations in calculus and algebra. One crucial identity we used in this problem is the inverse cosine identity:

• \( \cos^{-1} z = \tan^{-1} \left( \frac{\sqrt{1 - z^2}}{z} \right) \)

This identity helps us express the inverse cosine function, \( \cos^{-1}, \) in terms of the inverse tangent, \( \tan^{-1} \).
By using this identity, we can manipulate complex expressions into simpler forms that are easier to prove or understand.
For example, transforming \( \cos^{-1} \left( \frac{1-x^2}{1+x^2} \right) \) into a form involving \( \tan^{-1} \), allows for straightforward algebraic manipulation. This is because, while cosine and its inverse have certain limits and conditions, tangent functions often provide more flexibility.

Algebraic Manipulation

Algebraic manipulation refers to the use of algebraic techniques to simplify and rearrange expressions. This step involves expanding, factoring, and simplifying expressions involving variables like \( x \) and constants. It's crucial in converting complex expressions into an easily handled form. In our solution, several algebraic steps were key:

• Evaluate and simplify the expression \( 1 - \left( \frac{1-x^2}{1+x^2} \right)^2 \).
• Divide numerator and denominator to reduce fractions such as \( \frac{4x^2}{1 + 2x^2 + x^4} \).
• Utilize square root properties to simplify, like in \( \sqrt{\frac{4x^2}{1 + 2x^2 + x^4}} = \frac{2x}{1 + 2x^2 + x^4} \).

These steps not only make the problems manageable but reveal underlying mathematical elegance. Each manipulation acts like peeling layers off an onion, gradually exposing the core of the problem.

Proof in Calculus

Proofs in calculus require demonstrating the truth of an equation or statement by deductive reasoning. In this exercise, we needed to demonstrate that two expressions for a given problem are equivalent.To achieve the goal:

• We started by transforming cosines into tangents using trigonometric identities.
• The core was to show that \( \cos^{-1} \left( \frac{1-x^2}{1+x^2} \right) \) is equivalent to \( 2\tan^{-1} x \).
• By algebraic manipulation, both expressions were simplified to the same form.

Proofs like these are vital for deeper comprehension of mathematical relationships and provide a foundation for more complex derivative and integral calculus problems.
Understanding these proofs equips students with conceptual tools for problem-solving across various domains. The ability to use such techniques is crucial for mastering calculus concepts efficiently.