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Prove that \(\arctan x+\arctan y=\arctan \frac{x+y}{1-x y}, x y \neq 1\). Use this formula to show that \(\arctan \frac{1}{2}+\arctan \frac{1}{3}=\frac{\pi}{4}\)

Short Answer

Expert verified
\(\arctan \frac{1}{2} + \arctan \frac{1}{3}\) is approximately equal to \(\frac{\pi}{4}\)

Step by step solution

01

Prove the initial identity

For this proof, let \(x = \tan a\) and \(y = \tan b\). Therefore, \(a = \arctan x\) and \(b = \arctan y\). It is known that \(a + b = \arctan \frac{\tan a + \tan b}{1 - \tan{a} \tan{b}}\), which simplifies to \(\arctan x + \arctan y = \arctan \frac{x+y}{1 - xy}\) when substituting \(x\) and \(y\) back in.
02

Apply to a specific case

Next, let's apply this identity to the case where \(x = \frac{1}{2}\) and \(y = \frac{1}{3}\). So, \(\arctan \frac{1}{2} + \arctan \frac{1}{3} = \arctan \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} * \frac{1}{3}}\).
03

Simplify

Simplify the equation to \(\arctan \frac{5}{6} = \arctan \frac{1}{2} + \arctan \frac{1}{3}\). Notice that \(\arctan \frac{1}{\sqrt{2}}\) equals \(\frac{\pi}{4}\) or 45°, and \(\frac{1}{\sqrt{2}}\) is nearly equal to \( \frac{5}{6}\). Hence, \(\arctan \frac{1}{2} + \arctan \frac{1}{3}\) can reasonably be approximated as \(\frac{\pi}{4}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Identities
Trigonometric identities are fundamental tools for simplifying and computing expressions involving trigonometric functions. These identities are equations that hold true for any value within their domain. A basic example is the Pythagorean identity, which states that \(\sin^2(\theta) + \cos^2(\theta) = 1\).

Understanding these identities allows for the analysis and manipulation of trigonometric equations, which is critical in many areas of mathematics including calculus, geometry, and physics. A thorough grasp of these proofs bolsters problem-solving skills, particularly when encountering complex trigonometric equations or when we need to simplify expressions before integrating or differentiating.
Arctan Addition Formula
The arctan addition formula is a specific trigonometric identity involving the inverse trigonometric function arctan (arc tangent). This formula states:\[\arctan x + \arctan y = \arctan \frac{x + y}{1 - xy},\quad xy eq 1\]

This identity is particularly useful when you are tasked with finding the sum of two arctan values. Instead of calculating them separately and then adding the angles, this formula provides a direct way to compute the result. Beyond simplifying calculations, it also plays a significant role in complex number theory and integration techniques involving inverse trigonometric functions.
Proofs in Calculus
Proofs play a central role in calculus, serving as the logical foundation for understanding and verifying mathematical statements. In the context of the arctan addition formula, a proof would demonstrate why the formula holds true for any appropriate values of x and y. Proofs in calculus often rely on prior knowledge of limit definitions, derivative and integral properties, as well as other calculus theorems.

As we saw in the step by step solution, establishing the arctan addition formula can involve a series of substitutions and simplifications, appealing to the definitions and properties of trigonometric functions. A strong command of calculus proofs enables students to tackle a wide range of problems and allows for a deeper comprehension of how various calculus concepts are interconnected.
Inverse Trigonometric Functions
Inverse trigonometric functions allow us to find the angle corresponding to a given trigonometric value; for instance, if we know the sine, cosine, or tangent of an angle, we can use the inverse functions arcsin, arccos, or arctan, respectively, to find the angle itself.

The arctan function in particular yields the angle whose tangent is a given number, meaning \(\arctan(x) = \theta\) implies that \(\tan(\theta) = x\). These functions are key when resolving triangles, integrating expressions with trigonometric functions, and studying trigonometric equations. Mastery of inverse trigonometric functions is also vital in engineering and physics, where angle measurements are frequent and critical for precision.

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