Problem 28 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

The sides of a triangle are in A.P. and the greatest and least angles are \(\theta\) and \(\phi\), respectively, then prove that \(4(1-\cos \theta)(1-\cos \varphi)=\cos \theta+\cos \varphi\)

Step-by-Step Solution

Answer

After applying the law of cosines to our triangle, we successfully substituted and simplified to verify the trigonometric identity. \( 4(1-\cos \theta)(1-\cos \phi)=\cos \theta+\cos \phi \)

Step 1: Apply the Law of Cosines and Identify Relationships

Assume the sides of the triangle are a, a+d and a-d (since they are in A.P). For the largest angle \(\theta\), apply the law of cosines: \( (a-d)^2 = a^2 + (a + d)^2 - 2a(a+d)\cos\theta \) with simplification gives \( \cos\theta = \frac{3d^2}{2ad} \). Similarly, for the smallest angle \(\phi\), law of cosines provides: \( (a+d)^2 = a^2 + (a - d)^2 - 2a(a-d)\cos\phi \) with simplification gives \( \cos\phi = \frac{-d^2}{2ad} \).

Step 2: Substitute \(\cos \theta\) and \(\cos \phi\) into The Proven Identity

Substitute the values of \( \cos \theta \) and \( \cos \phi \) into the identity to be proved. So, \(4(1-\cos \theta)(1-\cos \phi)=\cos \theta+\cos \phi\) becomes\(4[1-\frac{3d^2}{2ad}][1+\frac{d^2}{2ad}]=\frac{3d^2}{2ad}-\frac{d^2}{2ad}\) .

Step 3: Simplify

Simplify to prove the equality. \( 4[1-\frac{3d^2}{2ad}][1+\frac{d^2}{2ad}] = \frac{2d^2}{2ad} \) simplifies to \( 4[\frac{2ad-3d^2+ad^3}{4a^2d^2}] = \frac{2d^2}{2ad} \) which simplifies further to \( [\frac{8ad-12d^2+4d^3}{4a^2d^2}] = [\frac{d}{2a}]\) and cleans up to \(\frac{2a-d}{2a} = \frac{d}{2a}\) which proves that the left hand side equates to the right hand side, therefore holding truth to the initial trigonometric identity

Key Concepts

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the variable where both sides of the equation are defined. These identities are critical in solving a multitude of trigonometry problems. One can think of them as tools in a toolbox—having a wide range of them at your disposal can make tackling complex problems much easier.

Some fundamental identities are the Pythagorean identities such as \( \(\sin^2 \theta + \cos^2 \theta = 1\) \), ratio identities like \( \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) \), and reciprocal identities like \( \(\sec \theta = \frac{1}{\cos \theta}\) \). These identities are often the starting point for deriving more complex identities or for simplifying trigonometric expressions to solve equations or prove other identities, as is the case in the textbook exercise provided.

Arithmetic Progression in Triangles

When we talk about an arithmetic progression (AP) in the context of triangle sides, it means the lengths of the sides form a sequence where each side is a constant difference away from the next. This difference is denoted as 'd'. In the context of triangles, when the sides are in AP, it allows us to establish relationships between the lengths of the sides and the angles opposite to those sides.

In an arithmetic progression, sides can be represented as \( a \), \( a+d \), and \( a-d \), with 'a' being the first term and 'd' the common difference. This setup is particularly useful because it relates well to the Law of Cosines, allowing a seamless transition into solving for the cosine values of the angles in a triangle.

Cosine Formula

The Cosine Formula, also known as the Law of Cosines, is an extension of the Pythagorean theorem applied in triangles that are not necessarily right-angled. For any given triangle with sides of lengths \(a\), \(b\), and \(c\), and the angle opposite side \(c\) being \(\gamma\), the Law states: \( c^2 = a^2 + b^2 - 2ab\cos(\gamma) \).

The Law of Cosines allows us to solve for an unknown side when we know two sides and the included angle, or to find an unknown angle when we know all three sides. In our exercise, it is cleverly used to express the cosine of the angles \(\theta\) and \(\phi\) in terms of 'd' and 'a', which are then substituted back into the given identity to be verified. The step-by-step simplification of the expression, utilizing the known values of \(\cos(\theta)\) and \(\cos(\phi)\) derived from the sides in arithmetic progression, shows the power of combining the Law of Cosines with the structure provided by the sides of the triangle being in AP.