Problem 13 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

In a \(\triangle A B C\), prove that, \(2\left[a \sin ^{2}\left(\frac{C}{2}\right)+c \sin ^{2}\left(\frac{A}{2}\right)\right]=c+a-b\)

Step-by-Step Solution

Answer

The original equation is proven to be true by manipulating both sides of the equation, using trigonometric identities and the sine rule until the expressions on both sides are identical, which means the given equation holds true for a triangle ABC.

Step 1: Express sides in terms of angles

Use the sine rule to express sides a and c in terms of angles A and C, respectively. By the sine rule, we know that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is same for all three sides of the triangle. Hence, \(a=\frac{b \sin A}{\sin B}\) and \(c=\frac{b \sin C}{\sin B}\).

Step 2: Substitute expressions from step 1

Substitute the expressions obtained in step 1 into the given equation. Resulting equation will be \(2\left[\frac{b \sin A \sin^{2}\left(\frac{C}{2}\right)}{\sin B}+\frac{b \sin C \sin^{2}\left(\frac{A}{2}\right)}{\sin B}\right]=\frac{b \sin C}{\sin B}+\frac{b \sin A}{\sin B}-b\)

Step 3: Simplify

Factor out common terms and simplify, which will give \(2b \left(\sin A \sin^{2}\left(\frac{C}{2}\right)+\sin C \sin^{2}\left(\frac{A}{2}\right)\right) = b(\sin A+\sin C) - b\)

Step 4: Use the sine addition identity

Use the sine addition identity to transform \(\sin A + \sin C\). Since \(\sin A + \sin C = 2\sin \left(\frac{A+C}{2}\right)\cos \left(\frac{A-C}{2}\right)\) and \(A+C = 180 - B\), we have: \(2\sin \left(90-\frac{B}{2}\right)\cos \left(\frac{A-C}{2}\right)\), which simplifies to \(2\cos \left(\frac{B}{2}\right)\cos \left(\frac{A-C}{2}\right)\)

Step 5: Use the double-angle identity

Use the double-angle identity for sine squared to transform \(\sin^{2}\left(\frac{C}{2}\right)\) and \(\sin^{2}\left(\frac{A}{2}\right)\), which leads to \(\frac{1-\cos C}{2}\) and \(\frac{1-\cos A}{2}\), respectively. Substituting into the given expression yields: \(2b \left[\sin A\frac{1-\cos C}{2}+\sin C\frac{1-\cos A}{2}\right] = 2b \cos\left(\frac{B}{2}\right)\cos \left(\frac{A-C}{2}\right) - b\)

Step 6: Make further simplification

Simplify both sides. Left side simplifies to \(b(\sin A+\sin C-\sin A\cos C-\sin C\cos A)\). Using double angle identity for sine and the angle sum identity for cosine, right side becomes \(b(2\cos\left(\frac{A-C}{2}\right)\cos \left(\frac{B}{2}\right) - 1)\). The expression on both sides becomes identical, therefore, the original equation is proven true.

Key Concepts

Double-Angle Identity

The double-angle identity is a crucial tool in trigonometry, especially when dealing with expressions involving squares of trigonometric functions. For sine and cosine, the double-angle identities are expressed as:

• For sine: \(\sin 2\theta = 2\sin \theta \cos \theta\)
• For cosine: \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta = 2\cos^2 \theta - 1 = 1 - 2\sin^2 \theta\)

These identities allow us to express functions of double angles in terms of single angles, simplifying many types of algebraic and trigonometric manipulations. In the exercise, we see the application of these identities by transforming \(\sin^2\left(\frac{C}{2}\right)\) and \(\sin^2\left(\frac{A}{2}\right)\) using the identity for sine, helping in the simplification of the given expression. By using the double-angle identities, complex trigonometric expressions are broken down into more manageable parts, which aids in arriving at the solution or proving relations simpler.

Sine Squared Identity

The sine squared identity is an expression derived from the Pythagorean identity, which is fundamental in understanding relations between trigonometric functions. It states that:

• \(\sin^2 \theta = \frac{1-\cos 2\theta}{2}\)
• \(\cos^2 \theta = \frac{1+\cos 2\theta}{2}\)

These are particularly useful for substituting squared terms to evaluate or transform equations, as seen in the original exercise. By substitution of \(\sin^2\left(\frac{C}{2}\right)\) and \(\sin^2\left(\frac{A}{2}\right)\), the exercise employs the transformation to enable further simplifications later on. The sine squared identity helps make calculations involving sine and cosine in squared forms more straightforward by expressing them in terms of cosines of double angles. This simplification is a key step in deriving the proof for the equation provided.

Angle Sum Identity

The angle sum identity refers to rules that expand trigonometric functions of sums and differences of angles into more basic trigonometric functions. The identities are given by:

• \(\sin(A + C) = \sin A \cos C + \cos A \sin C\)
• \(\cos(A + C) = \cos A \cos C - \sin A \sin C\)

In the exercise, the sine addition identity is particularly useful for transforming expressions involving the sum of angles, such as \( \sin A + \sin C \). By knowing the sum and difference identities, one can break down expressions into basic sine and cosine components. This is what happens in the solution when the identity \(\sin A + \sin C = 2 \sin \left( \frac{A+C}{2} \right) \cos \left( \frac{A-C}{2} \right)\) is used, thus facilitating the proof process by further reducing the complexity of terms.

Triangle Angle Relationships

Understanding the relationships between angles in a triangle is foundational knowledge in trigonometry and geometry. In any triangle, the sum of the angles is always 180 degrees, which can be expressed as:

• \(A + B + C = 180^\circ\)

This relationship is pivotal when applying the sine rule and other trigonometric identities. For example, knowing that \(A + C = 180^\circ - B\) aids in transformations using the angle sum identity. In the given problem, recognizing the sum of the triangle’s angles enables the conversion of angle expressions to a format that can be used in identities such as \( \sin(180^\circ - B) = \sin B \) and related transformations. Grasping these basic angle relationships helps students understand the interplay of various trigonometric identities within a triangle and provides a framework for both proving equations and solving geometric problems effectively.