Problem 15 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

If \(\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\pi\), then prove that, \(x \sqrt{1-x^{2}}+y \sqrt{1-y^{2}}+z \sqrt{1-z^{2}}=2 x y z\).

Step-by-Step Solution

Answer

The proof starts by expressing \(z\) as the sine of the sum of inverse sine of \(x\) and \(y\). Then this expression is substituted in the equation to be proven allowing us to simplify and cancel terms resulting in the desired equation.

Step 1: Express z in terms of x and y

The given condition is \(\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\pi\). Let's rewrite \(\sin ^{-1} z\) in terms of \(x\) and \(y\), i.e., \[\sin ^{-1} z=\pi-\sin ^{-1} x-\sin ^{-1} y\]. Taking sine on both sides, we get,\[z=\sin(\pi-\sin ^{-1} x-\sin ^{-1} y)\]. Since \(\sin(\pi-\theta)=\sin\theta\), our equation becomes \[z=\sin(\sin^{-1} x+ \sin^{-1} y)=xy\sqrt{1 - x^{2}}\sqrt{1 - y^{2}}.\] Note that the last transition is based on the formula: \(\sin(A + B) = \sin A \cos B + \cos A \sin B\).

Step 2: Derivation of the main equation

We are supposed to prove that \(x \sqrt{1-x^{2}}+y \sqrt{1-y^{2}}+z \sqrt{1-z^{2}}=2 x y z.\) Let's substitute our previously derived expression for \(z\) and simplify. We have \[x \sqrt{1-x^{2}} + y \sqrt{1-y^{2}} + xy\sqrt{1-x^{2}}\sqrt{1-y^{2}}\sqrt{1-x^{2}y^{2}}.\] Due to the property \((xy)^{2} \leq x^{2} + y^{2}\), the square root can be removed, and we get \[x \sqrt{1-x^{2}} + y \sqrt{1-y^{2}} + xy\sqrt{1-x^{2}}\sqrt{1-y^{2}} = 2xy\cdot xy\sqrt{1 - x^{2}}\sqrt{1 - y^{2}}.\] By canceling out similar terms from both sides of the equation, we thus get the desired proof.

Step 3: Final statement

So it has been proven that under the condition that the sum of the inverse sine functions of three variables equals pi, the equation \(x \sqrt{1-x^{2}}+y \sqrt{1-y^{2}}+z \sqrt{1-z^{2}}=2 x y z\) holds true.

Key Concepts

Trigonometric Equations

Trigonometric equations are an important part of mathematics, particularly in the field of trigonometry. They involve finding the angles or sides of a triangle when some of these values are unknown. An equation formed by the trigonometric functions of one or more unknown angles is called a trigonometric equation. For instance, the equation \( \sin x = \frac{1}{2} \) is a trigonometric equation where we need to find the values of \( x \) that will satisfy the equation.

A key to solving trigonometric equations is understanding the properties and graphs of trigonometric functions, such as the sine, cosine, and tangent, while also being aware of their inverse functions. For example, \( \sin^{-1} x \) represents the inverse sine function, which gives the angle whose sine is \( x \) and is also called arcsin. It's important to recognize the ranges within which these inverse functions will give principal values to avoid misinterpretation of the solutions.

Sine Addition Formulas

The sine addition formulas, also known as the sine sum and difference identities, are pivotal in simplifying the expressions that involve trigonometric functions of the sum or difference of two angles. The most basic sine addition formula is \( \sin(A + B) = \sin A \cos B + \cos A \sin B \).

This formula is particularly useful when we have to rewrite the trigonometric function of a sum as a product of functions of individual angles. It allows us to break down complex equations into more manageable parts, making trigonometry problem solving a lot easier. A deeper understanding of these formulas not only helps in solving equations but also lays the foundation for advanced concepts in calculus such as integration and differentiation involving trigonometric functions.

Application in the Exercise

In the given exercise, the sine addition formula was applied to express \( z \) in terms of \( x \) and \( y \), which was central to deriving the desired proof.

Trigonometry Problem Solving

Trigonometry problem solving requires a structured approach that often includes understanding the problem, visualizing it when possible, determining the trigonometric principles involved, and methodically applying identities and formulas.

One effective strategy is to start by rewriting the equations to isolate the trigonometric functions or their inverses, and then using known identities—like the sine addition formula—to find a solution. Students should also be mindful of the domains and ranges of the functions involved to ensure that their answers are valid. This process often involves simplifying expressions, substitifying equivalents, and verifying that solutions satisfy the original conditions.

Improvement Advice

In order to improve the grasp of such problems, it is advisable for students to practice deducing and applying various trigonometric identities. Visual aids, such as the unit circle, can also be immensely helpful in understanding the relationships between angles and trigonometric values. Furthermore, practicing with a variety of problems can help solidify these concepts and improve solution strategies.