Problem 14 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

In a \(\Delta A B C\), the sides \(a, b, c\) are the roots of the equation \(x^{3}-11 x^{2}+38 x-40=0 .\) Then \(\frac{\cos A}{a}+\frac{\cos B}{b}+\) \(\frac{\cos C}{c}\) is (a) 1 (b) \(\frac{3}{4}\) (c) \(\frac{9}{16}\) (d) None

Step-by-Step Solution

Answer

\(\frac{3}{4}\)

Step 1: Explain the Cosine Rule

In any triangle, the cosine rule is obtained from the law of cosines, which states that \(c^2 = a^2 + b^2 – 2ab*cosC\). From here it leads to \(\cos C = \frac{{a^2 + b^2 - c^2}}{{2ab}}\). Similar equations can be obtained for \(\cos A\) and \(\cos B\). For angle A, it is \(\cos A = \frac{{b^2 + c^2 - a^2}}{{2bc}}\). For angle B, it is \(\cos B = \frac{{a^2 + c^2 - b^2}}{{2ac}}\).

Step 2: Substitution into the Asked Expression

Now substitute the cosine values from step 1 into the expression \(\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c}\). Thus it becomes \(\frac{1}{a}*\frac{{b^2 + c^2 - a^2}}{{2bc}} + \frac{1}{b}*\frac{{a^2 + c^2 - b^2}}{{2ac}} + \frac{1}{c}*\frac{{a^2 + b^2 - c^2}}{{2ab}}\). This simplifies to \(\frac{1}{2}(\frac{{b^2 + c^2}}{{abc}} - \frac{2}{bc} + \frac{{a^2 + c^2}}{{abc}} - \frac{2}{ac} + \frac{{a^2 + b^2}}{{abc}} - \frac{2}{ab})\).

Step 3: Identifying the Root-Product Relationships

From the given cubic equation, we notice it's to the standard form \(x^3 - sx^2 + tx - u = 0\). The roots a, b, c of the equation satisfy: \(abc = u = 40\), \(ab + bc + ca = t = 38\), and \(a + b + c = s = 11 ≠ 0\). Here, \(s, t, u\) are the sum of the roots, product of the roots taken two at a time, and the product of the roots respectively.

Step 4: Substitution of the Root-Product Relationships

Substitute these values into the expression found in step 2. The equation thus obtained simplifies to \(\frac{1}{2}(\frac{1}{2} - 1) = \frac{1}{4}\).

Key Concepts

Law of Cosines

The Law of Cosines is a fundamental rule in trigonometry that relates the lengths of the sides of a triangle to the cosine of one of its angles. This rule is especially useful in triangles where we do not have a right angle and need to find an unknown side or an angle. The formula of the law of cosines is given by:

• For side \( c \) in \( \triangle ABC \): \( c^2 = a^2 + b^2 - 2ab\cos(C) \)
• Similarly, for side \( a \): \( a^2 = b^2 + c^2 - 2bc\cos(A) \)
• And for side \( b \): \( b^2 = a^2 + c^2 - 2ac\cos(B) \)

This extension of the Pythagorean theorem allows us to work with oblique triangles and helps to find unknown sides or angles by providing a workable equation. Using these formulas, you can solve pretty much any trigonometry-related question in triangles. Once you understand the direct substitution of known values into the formula, solving for the unknown becomes straightforward.

Roots of a Cubic Equation

Understanding the roots of a cubic equation is essential in algebra when dealing with mathematical modeling and computations. A cubic equation generally has the form \(x^3 + ax^2 + bx + c = 0\). It is a polynomial of degree three and can have up to three real roots. The roots (or solutions) to a cubic equation are the values that satisfy the equation, meaning:

• They make the polynomial equal zero when substituted in place of the variable \( x \).
• A cubic equation will always have three roots, which can be real or complex numbers.

Finding these roots can involve techniques like factorization, synthetic division, or using the cubic formula, depending on the specific equation. These techniques may seem challenging, but they are powerful tools to solve polynomial equations.

Sum of Roots

The sum of the roots of any polynomial equation provides a unique quantitative insight into the equation. For a cubic equation of the form \(x^3 + px^2 + qx + r = 0\), the sum of the roots, traditionally denoted as \( \alpha + \beta + \gamma \), is given by \(-p\) (given by the formula \(-\frac { \text{coefficient of } x^2}{ \text{coefficient of } x^3 } \)). In the specific cubic equation from the exercise, \(x^3 - 11x^2 + 38x - 40 = 0\), the sum of the roots is:

• 11, since it is immediately derived from the negative of the coefficient of \(x^2\).

Knowing the sum of the roots is often the first step in further investigations and simplifications of problems related to polynomial equations.

Product of Roots

The product of roots is another key aspect of polynomial equations, providing valuable information that links directly to the constant term of the polynomial. For the same cubic polynomial \(x^3 + px^2 + qx + r = 0\), the product of the roots \(\alpha\beta\gamma\) is equal to \(- r \), where \( r \) is the constant term, divided by the leading coefficient (for monic polynomials like the one in the example, this is simply \( r \)).In our cubic equation \(x^3 - 11x^2 + 38x - 40 = 0\), the product of the roots is:

• 40, matching the constant term since the leading coefficient is 1.

Understanding the product of the roots along with the sum of roots offers a panoramic view of the equation's intrinsic properties, helping solve algebraic problems with deeper insight.