Problem 17 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

If \(\alpha\) and \(\beta\) are acute angles and \(\cos 2 \alpha=\frac{3 \cos 2 \beta-1}{3-\cos 2 \beta}\), prove that \(\tan \alpha: \tan \beta=\sqrt{2}: 1\).

Step-by-Step Solution

Answer

\(\tan \alpha: \tan \beta = \sqrt{2}: 1\)

Step 1: Simplify the Given Equation

The given equation is \(\cos 2 \alpha=\frac{3 \cos 2\beta-1}{3-\cos 2 \beta}\). Cross multiply the fractions to simplify it, this gives \[ 3\cos2\alpha - \cos^2 2\alpha = 3\cos 2\beta - 1\].

Step 2: Use Double Angle Formulas

The double-angle formula for cosine is \(\cos 2X = 1- 2\sin^2X\). Replace \(\cos 2\alpha\) and \(\cos 2\beta\) with this identity. The equation becomes \[3- 6\sin^2\alpha - (1 - 2\sin^2\alpha) = 3 - 2\sin^2\beta - 1\]. Simplify this to obtain \[8\sin^2\alpha = 2\sin^2\beta\] or \(\sin^2\alpha = \frac{1}{4}\sin^2\beta\).

Step 3: Use Pythagorean Identity

Then under the condition that both angles are acute we have \( \cos^2\alpha > 0\) and \( \cos^2\beta > 0\). With this condition the equation is possible. We use the pythagorean identity \(\cos^2 X = 1 - \sin^2 X\) and get\[ \cos^2\alpha = 1 - \sin^2\alpha = 1 - \frac{1}{4}\sin^2\beta = 1 - \frac{1}{4}(1-\cos^2\beta)\]. Simplify, we get \[ \cos^2\alpha = \frac{1}{2}\cos^2\beta\].

Step 4: Solve for \(\tan\)

We calculate the ratios of \(\tan \alpha\) and \(\tan \beta\) using the identity \(\tan X = \frac{\sin X}{\cos X}\). Substitute the derived values of \(\sin\alpha\), \(\sin\beta\), \(\cos\alpha\) and \(\cos\beta\) and we find that \(\tan \alpha: \tan \beta = \sqrt{2}: 1\).

Key Concepts

Double Angle Formulas

The double angle formulas are a powerful set of equations in trigonometry. They allow us to express trigonometric functions of twice an angle (\(2\alpha\)and\(2\beta\)) in terms of functions of the angle itself. Let's break it down a bit further:

• The double angle formula for cosine is \(\cos 2X = 1 - 2\sin^2 X\).
• Similarly, \(\sin 2X = 2\sin X \cos X\) for sine, and \(\tan 2X = \frac{2\tan X}{1 - \tan^2 X}\) for tangent.

These formulas are particularly useful when solving problems that involve multiplying angles by two, just like in our problem. They simplify the expressions, making it easier to perform further algebraic operations. By substituting the original expression of \(\cos 2\alpha\) and \(\cos 2\beta\) with their double angle formulas, we were able to progress through the solution efficiently.

Pythagorean Identity

The Pythagorean Identity is another fundamental tool in trigonometry. It is based on the Pythagorean theorem from geometry and relates the squares of sine and cosine for a given angle. The main identity is:

• \(\sin^2 X + \cos^2 X = 1\)

This identity is incredibly useful for converting between sine and cosine. In our specific problem, we used this to find \(\cos^2\alpha\) and \(\cos^2\beta\). Knowing \(\sin X\) from one angle-related identity, \(\cos X\) can be swiftly calculated, given that they're related through this identity.
Utilizing it under the assumption that both \(\alpha\) and \(\beta\) are acute angles (and hence \(\cos^2 X >0\)), we were able to further simplify the expression and derive the necessary ratios of tangents in our problem.

Acute Angles

Understanding acute angles is essential when working with trigonometric identities. Acute angles are angles less than 90 degrees and have unique properties that make calculations with them straightforward in certain contexts:

• Both sine and cosine values are positive for acute angles.
• The sum of sine squared and cosine squared remains equal to one, adhering to the Pythagorean Identity.

Knowing that \(\alpha\) and \(\beta\) are acute angles gives us more than just limits on their size. It ensures that certain trigonometric identities, such as \(\tan X = \frac{\sin X}{\cos X}\), can be applied without negative values causing confusion. This is essential for maintaining the integrity of the derived expressions and makes sure that the solution aligns well with all necessary mathematical rules, as seen in the solution of the exercise where we derived \(\tan \alpha : \tan \beta = \sqrt{2} : 1\). This insight helps guarantee consistency and correctness throughout the solution process.