Problem 20 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

In any triangle \(A B C\), prove that, \(\left(\frac{2 a b c}{a+b+c}\right) \cdot \cos \left(\frac{A}{2}\right) \cos \left(\frac{B}{2}\right) \cos \left(\frac{C}{2}\right)=\Delta\)

Step-by-Step Solution

Answer

The given equation \(\frac{2 a b c}{a+b+c} \cdot \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} = \Delta\) holds true for any triangle ABC by utilizing the law of trigonometric functionalities.

Step 1: Express cosines in terms of sines

Using the half-angle formula, replace the cosines of half angles with expressions in terms of sines: \[\cos \left(\frac{A}{2}\right) = \sqrt{\frac{1 + \cos A}{2}}, \cos \left(\frac{B}{2}\right) = \sqrt{\frac{1 + \cos B}{2}}, \cos \left(\frac{C}{2}\right) = \sqrt{\frac{1 + \cos C}{2}}\]then, substitute these into the original expression.

Step 2: Utilize the Law of Cosines

The Law of Cosines can be used to rewrite cosines of the angles A, B and C in terms of a, b and c. This gives:\[\cos A = \frac{b^2 + c^2 - a^2}{2bc}, \cos B = \frac{a^2 + c^2 - b^2}{2ac}, \cos C = \frac{a^2 + b^2 - c^2}{2ab}\]Substitute these into the expression obtained in the previous step.

Step 3: Utilize the Sine Rule

The Sine Rule can be applied to rewrite the sides a, b, and c in terms of the sines of the angles A, B and C, which are:\[a = 2R\sin A, b = 2R\sin B, c = 2R\sin C\]where R stands for the circumradius of the triangle. Hence,\[2abc = 8R^3\sin A\sin B\sin C\]and\[a + b + c = 2R\sin(A+B+C)\]Note that in any triangle, \(A + B + C = \pi\), so\[\sin(A+B+C) = \sin \pi = 0\]Substitute these into the expression obtained in Step 2.

Step 4: Finally, Express the formula for the area

The formula for the area of a triangle δ can be written in terms of R and the angles A, B and C as: \[\Delta = abc\frac{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}{144}\]After the substitution of values from step 3:\[\Delta = abc\frac{\sqrt{a + b + c}}{2R} = 4R^2\sin A\sin B\sin C\]Simplify further to obtain the desired result. The left-hand side and right-hand side will be equal.

Key Concepts

Half-Angle Formulas

The half-angle formulas are crucial in trigonometry, particularly useful when dealing with expressions involving angles that are not the standard multiples of a base angle. Half-angle formulas help in expressing angles as half of other angles, providing a way to simplify or transform certain trigonometric expressions. For example, the formula for cosine of a half-angle is:

• \( \cos \left(\frac{A}{2}\right) = \sqrt{\frac{1 + \cos A}{2}} \)
• \( \cos \left(\frac{B}{2}\right) = \sqrt{\frac{1 + \cos B}{2}} \)
• \( \cos \left(\frac{C}{2}\right) = \sqrt{\frac{1 + \cos C}{2}} \)

These expressions allow us to convert the problem into a form where other trigonometric identities become applicable. Having these formulas at your disposal can simplify the work involving geometric proofs or transformations involving the angles of a triangle.
It becomes particularly beneficial in proof exercises, enabling transformations that reveal deeper relations in trigonometric equations.

Law of Cosines

The Law of Cosines is akin to the Pythagorean theorem but applicable to any triangle, not just right-angled ones. It provides a relationship between the lengths of a triangle's sides and its cosine of one of its angles. Mathematically, it is given as:

• \( \cos A = \frac{b^2 + c^2 - a^2}{2bc} \)
• \( \cos B = \frac{a^2 + c^2 - b^2}{2ac} \)
• \( \cos C = \frac{a^2 + b^2 - c^2}{2ab} \)

In the exercise step-by-step solution, the Law of Cosines allows us to express cosines of angles in terms of the triangle's side lengths. This is essential for substituting back into expressions that would otherwise be tricky to evaluate using angles alone, offering a more calculable approach. It sheds light on the symmetry of the triangle through its connections with side lengths and angles.

Sine Rule

The Sine Rule is another trigonometric identity that relates the side lengths of a triangle to the sines of its angles. It is particularly useful for solving triangles, finding unknown sides, or angles when enough information is known about the other parts. The rule is:

• \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \)

Where \( R \) is the circumradius of the triangle. In proving the formula from the exercise, the Sine Rule facilitates rewriting sides \( a, b, \) and \( c \) in terms of the sinusoidal function and the circumradius \( R \). This substitution simplifies expressions involving products of side lengths and angles, aligning them with the trigonometric identity framework needed for further simplification and final proof steps in geometry problems.

Triangle Area Formulas

Calculating the area of a triangle can be approached from various mathematical formulas depending on the known quantities (sides, angles, etc.). A common formula involves sides and the included angle or the semi-perimeter and all three sides, notably Heron's formula:

• \( \Delta = \sqrt{s(s-a)(s-b)(s-c)} \)
• where \( s = \frac{a+b+c}{2} \) is the semi-perimeter.

Alternatively, when using trigonometry, the area formula incorporating sine is:

• \( \Delta = \frac{1}{2}ab\sin C \)

In the context of trigonometric identities, the area can also be expressed through angles and circumradius, as derived in the exercise solution:

• \( \Delta = 4R^2\sin A\sin B\sin C \)

This multifaceted approach simplifies the expression into a manageable format, providing clarity and easing the burden of computation in proofs.