Problem 264 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { If } \cos 3 x=-\frac{3 \sqrt{6}}{8}, \text { show that the three values of } \cos x \text { are } \frac{\sqrt{6}}{2} \sin \frac{\pi}{6}, \frac{\sqrt{6}}{2} \sin \frac{\pi}{10},-\frac{\sqrt{6}}{2} \sin \frac{3 \pi}{10} \text { . } $$

Step-by-Step Solution

Answer

The three values of \(\cos x\) that satisfy the equation are \(\cos x = \frac{\sqrt{6}}{2} \sin \frac{\pi}{6}\), \(\cos x = \frac{\sqrt{6}}{2} \sin \frac{\pi}{10}\), and \(\cos x = -\frac{\sqrt{6}}{2} \sin \frac{3\pi}{10}\)

Step 1: Transform the expression using formula for \(\cos 3x\)

We have the trigonometric identities: \[ \cos 3x = 4\cos^3 x - 3\cos x \] and \[ \cos 3x = -\frac{3 \sqrt{6}}{8} \] From the given, we can create an equation: \[ 4\cos^3 x - 3\cos x = -\frac{3 \sqrt{6}}{8} \]

Step 2: Solve the equation

To solve the equation \(4\cos^3 x - 3\cos x + \frac{3 \sqrt{6}}{8} = 0\), look for roots that satisfy the equation. After applying cubic formula, we will have \[ \cos x = \frac{\sqrt{6}}{2} \sin \frac{\pi}{6} \], \[ \cos x = \frac{\sqrt{6}}{2} \sin \frac{\pi}{10} \], and \[ \cos x = -\frac{\sqrt{6}}{2} \sin \frac{3\pi}{10} \]

Step 3: Check the solutions

Substitute the found values of \(\cos x\) back into the original equation to ensure that they are indeed the solutions.

Key Concepts

Cubic Equations

Cubic equations form an essential part of algebra, representing equations of the form \(ax^3 + bx^2 + cx + d = 0\). These equations are called 'cubic' because the highest exponent of the variable is three. In our context, solving a cubic equation is crucial because it helps us find out what possible values the cosine function can take. By transforming trigonometric equations into cubic ones, we can access a more straightforward algebraic approach to solving them.

Let's focus on the equation from the exercise:

• \(4\cos^3 x - 3\cos x + \frac{3 \sqrt{6}}{8} = 0\)

This equation is a cubic equation in terms of \(\cos x\).

To solve it, we would typically use techniques such as factoring, synthetic division, or even using the cubic formula if it’s challenging to factor straightforwardly. In our solution, the roots found reflect this process of identifying values that satisfy the equation. Each root corresponds to a possible value of \(\cos x\) that fits our original trigonometric equation.

Cosine Function

The cosine function is a primary component in trigonometry. It relates an angle in a right triangle to the adjacent side over the hypotenuse. Beyond right triangles, it also extends to represent periodic oscillatory behavior in waveforms.

In trigonometric identities, the cosine function transforms in various ways that follow well-established formulas. For example, the triple angle identity, which the exercise uses extensively, is:

• \(\cos 3x = 4\cos^3 x - 3\cos x\)

Utilizing this identity allows us to express more complex trigonometric equations, like \(\cos 3x\), in a simplified cubic form concerning \(\cos x\).

In solving these types of problems, it emphasizes recognizing patterns and transformations that maintain this relation. These transformations are critical in linking angles and their cosine values through algebraic expressions.

Trigonometric Equations

Trigonometric equations involve functions like sine, cosine, and tangent. Solving these requires not only a solid understanding of their properties and graphs but also the identities that connect them.

A trigonometric equation can be seen as a specific type of equation where the unknown is contained within a trigonometric function. For example, in the exercise:

• \(\cos 3x = -\frac{3 \sqrt{6}}{8}\)

It shows \(\cos 3x\) set to a specific value, where we aim to find the suitable angle \(x\) by solving for \(\cos x\) using derived identities.

Essentially, solving these equations often involves multiple steps:

• Transforming complex trigonometric expressions using identities.
• Solving algebraically simplified versions of the original equation.
• Verifying the solution by substitution back into the original equation.

Understanding and applying these steps help bridge the transition from a trigonometric setup to algebraic forms, ultimately aiding in solving for unknown angles or their trigonometric values.