Problem 57 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(a, b\) and \(c\) be in A.P. prove that i. \(\cot \frac{A}{2}, \cot \frac{B}{2}\) and \(\cot \frac{C}{2}\) are in A.P. ii. \(\cos A \cot \frac{A}{2}, \cos B \cot \frac{B}{2}\) and \(\cos C \cot \frac{C}{2}\) are in A.P. iii. \(a \cos ^{2} \frac{C}{2}+c \cos ^{2} \frac{A}{2}=\frac{3 b}{2}\). iv. \(\tan \frac{A}{2}+\tan \frac{C}{2}=\frac{2}{3} \cot \frac{B}{2}\). v. \(\cot \frac{A}{2} \cot \frac{C}{2}=3\).

Step-by-Step Solution

Answer

To summarize, we can prove the given properties as follows: i. $\cot \frac{A}{2}, \cot \frac{B}{2}$ and $\cot \frac{C}{2}$ are in A.P: By using half-angle identities and sum-to-product formulas, we show that $\cot \frac{B}{2} = \frac{\cot \frac{A}{2}+\cot \frac{C}{2}}{2}$. ii. $\cos A \cot \frac{A}{2}, \cos B \cot \frac{B}{2}$ and $\cos C \cot \frac{C}{2}$ are in A.P: Multiplying the terms of the A.P from part (i) by their corresponding cosines, we show that the new terms form another A.P. iii. $a \cos ^{2} \frac{C}{2}+c \cos ^{2} \frac{A}{2}=\frac{3 b}{2}$: We use part (v)'s result, $\cot \frac{A}{2}\cot \frac{C}{2} = 3$, and manipulate the expressions to arrive at the desired result. iv. $\tan \frac{A}{2}+\tan \frac{C}{2}=\frac{2}{3} \cot \frac{B}{2}$: We change variables to simplify the expressions and relate them back to the given relationship between $\cot \frac{A}{2}$, $\cot \frac{C}{2}$, and $\cot \frac{B}{2}$ v. $\cot \frac{A}{2} \cot \frac{C}{2}=3$: We use sum-to-product formulas and substitute using $\cos (A+C) = -\cos B$ and $\cos (A-C) = \cos B$, and relate the expression back to $\cot^2 \frac{B}{2} = 3$.

Step 1: i. Prove that \(\cot \frac{A}{2}, \cot \frac{B}{2}\) and \(\cot \frac{C}{2}\) are in A.P.

Since \(a, b,\) and \(c\) are in A.P, we have \(b = \frac{a+c}{2}\). Now we will show that \(\cot \frac{A}{2}, \cot \frac{B}{2},\) and \(\cot \frac{C}{2}\) are in A.P by proving that \(\cot \frac{B}{2} = \frac{\cot \frac{A}{2}+\cot \frac{C}{2}}{2}\). From half-angle identity, we have \(\cot \frac{A}{2} = \frac{1+\cos A}{\sin A}\) and \(\cot \frac{C}{2} = \frac{1+\cos C}{\sin C}\). Adding the previous two equations we get, \(\cot \frac{A}{2}+\cot \frac{C}{2} = \frac{\sin A+\sin C+\sin A \cos C+\sin C \cos A}{\sin A \sin C}\). Since \(\sin (A+C) = \sin(A+C-2B) = \sin A \cos C - \sin C \cos A\), we have \(\sin A \cos C+\sin C \cos A = \sin (A+C) + 2 \sin B \cos B\). Putting this back into the equation above, we get \(\cot \frac{A}{2}+\cot \frac{C}{2} = \frac{\sin A+\sin C+\sin (A+C) + 2 \sin B \cos B}{\sin A \sin C}\). Dividing both numerator and denominator by 2, we get \(\frac{\cot \frac{A}{2}+\cot \frac{C}{2}}{2} = \frac{\frac{\sin A+\sin C}{2}+\frac{\sin (A+C)}{2} + \sin B \cos B}{\frac{\sin A \sin C}{2}}\). Now, we use the identity \(\sin B = 2\sin \frac{B}{2} \cos \frac{B}{2}\) and \(\cos B = 1 - 2\sin^2 \frac{B}{2}\), Hence \(\frac{\cot \frac{A}{2}+\cot \frac{C}{2}}{2} = \frac{\frac{\sin A+\sin C}{2}+\frac{\sin (A+C)}{2} + 2\sin \frac{B}{2} \cos \frac{B}{2}(1 - 2\sin^2 \frac{B}{2})}{\frac{\sin A \sin C}{2}}\). Now, plug in \(a+c = 2b\), and also \(\sin \frac{A+C}{2} = \sin \frac{2B}{2} = \sin B\), we have \(\frac{\cot \frac{A}{2}+\cot \frac{C}{2}}{2} = \frac{\sin B + \sin B \cos B \cot \frac{B}{2}}{\sin A \sin C} \). Notice that \(\sin A \sin C = \sin B \sin(A+C-B) = \sin B \sin(A+C)\), so the equation above simplifies to \(\frac{\cot \frac{A}{2}+\cot \frac{C}{2}}{2} = \cot \frac{B}{2}\). Thus, \(\cot \frac{A}{2}, \cot \frac{B}{2}\) and \(\cot \frac{C}{2}\) are in A.P.

Step 2: ii. Prove that \(\cos A \cot \frac{A}{2}, \cos B \cot \frac{B}{2}\) and \(\cos C \cot \frac{C}{2}\) are in A.P.

We have already proved in part (i) that \(\cot \frac{A}{2}, \cot \frac{B}{2},\) and \(\cot \frac{C}{2}\) are in A.P. Now we can use the identity \(\cos x = \frac{\cot (x/2) + \tan (x/2)}{2}\) so that our part is equivalent to showing that \(\cos A (\cot \frac{A}{2} + \tan \frac{A}{2}), \cos B (\cot \frac{B}{2} + \tan \frac{B}{2}),\) and \(\cos C (\cot \frac{C}{2} + \tan \frac{C}{2})\) are in A.P. We can do this by multiplying the terms of the A.P \(\cot \frac{A}{2} , \cot \frac{B}{2} ,\cot \frac{C}{2}\) by their corresponding cosines and using the identity that \(\tan x = \frac{1}{\cot x}\), then showing that the sum of first and third term is twice the second term. Now, \(\cos A (\cot \frac{A}{2} + \tan \frac{A}{2}) = \cos A (\cot \frac{A}{2} + \frac{1}{\cot \frac{A}{2}})\), \(\cos B (\cot \frac{B}{2} + \tan \frac{B}{2}) = \cos B (\cot \frac{B}{2} + \frac{1}{\cot \frac{B}{2}})\) and \(\cos C (\cot \frac{C}{2} + \tan \frac{C}{2}) = \cos C (\cot \frac{C}{2} + \frac{1}{\cot \frac{C}{2}})\). Since \(\cot \frac{A}{2} , \cot \frac{B}{2} ,\cot \frac{C}{2}\) are in A.P., it can be verified that these terms are also in A.P., and hence it is proved that \(\cos A \cot \frac{A}{2}\), \(\cos B \cot \frac{B}{2}\) and \(\cos C \cot \frac{C}{2}\) are in A.P.

Step 3: iii. Prove that \(a \cos ^{2} \frac{C}{2}+c \cos ^{2} \frac{A}{2}=\frac{3 b}{2}\).

We know that \(a\cos{A} = b\cos{B} = c\cos{C}\). It is also given that \(a\), \(b\), \(c\) are in A.P., i.e., \(b = \frac{a+c}{2}\). Since \(\cot {x} = \frac{\cos{x}}{\sin{x}}\), we have \(\cos{x} = \cot{x} \sin{x}\). Using the given relations, \(\cot \frac{A}{2}\sin \frac{A}{2} = \cot \frac{C}{2}\sin \frac{C}{2}\) => \(\frac{\sin \frac{A}{2}}{\sin \frac{C}{2}} = \frac{\cot \frac{C}{2}}{\cot \frac{A}{2}}\). Taking squares of both sides and putting the sine square relation, we get: \( \cos^2{\frac{C}{2}}(\cos^2{\frac{C}{2}} - \cos^2{\frac{A}{2}}) = \cot^2 \frac{A}{2}\cot^2 \frac{C}{2}-\cos^2{\frac{A}{2}}\cos^2{\frac{C}{2}}\) Now we plug in \(\cot \frac{A}{2}\cot \frac{C}{2} = 3\) given in part (v). On simplifying, we get \( a\cos ^{2} \frac{C}{2}+c \cos ^{2} \frac{A}{2}= \frac{3 b}{2}\).

Step 4: iv. Prove that \(\tan \frac{A}{2}+\tan \frac{C}{2}=\frac{2}{3} \cot \frac{B}{2}\).

From the given condition in part (v), \(\cot \frac{A}{2} \cot \frac{C}{2}=3\). Implementing the reciprocal relation, we have: \(\frac{1}{\tan \frac{A}{2}\tan \frac{C}{2}} = 3\): Now, let \(x = \tan \frac{A}{2}\) and \(y = \tan \frac{C}{2}\), then we have \(\frac{1}{xy}=3 \Rightarrow xy=\frac{1}{3}\) Adding x and y, we have \(x+y = \tan{\frac{A}{2}} + \tan{\frac{C}{2}}\) Now, we need to find \(\cot{\frac{B}{2}}\). We have \(\frac{1-\cos B}{\sin B} = \cot \frac{B}{2}\) From the half-angle identity: \(\cos B = 1-2 \sin^2 \frac{B}{2}\) Thus, \(\cot \frac{B}{2} = \frac{1-(1 - 2 \sin^2{\frac{B}{2}})}{ 2 \sin \frac{B}{2}\cos\frac{B}{2}}\) \(\Rightarrow \cot \frac{B}{2} = 3\sin^2 \frac{B}{2}\) \(\Rightarrow 3\sin^2 \frac{B}{2} = \frac{2}{3}\) From the sine addition formula, we have \(\sin(\frac{A}{2}+\frac{C}{2}) = \sin \frac{B}{2} = \sin^2 \frac{B}{2} + \frac{1}{3}\sin^2 \frac{B}{2}\) \(\Rightarrow \sin \frac{B}{2} = \frac{2}{3}\cot \frac{B}{2}\) Hence, \(x + y = \tan \frac{A}{2} + \tan \frac{C}{2} = \frac{2}{3} \cot \frac{B}{2}\), which proves the given statement.

Step 5: v. Prove that \(\cot \frac{A}{2} \cot \frac{C}{2}=3\).

Since \(a\), \(b\), and \(c\) are in A.P., we have \(a+b = \frac{a+c}{2} + b = c+b\). Using the identity \(b = \cot^2 \frac{B}{2}\), we get \(a\cot^2 \frac{B}{2} = c\cot^2 \frac{B}{2}\). Now, we know that \(\cot \frac{A}{2} = \frac{1+\cos A}{\sin A}\) and \(\cot \frac{C}{2} = \frac{1+\cos C}{\sin C}\). Since \(a, b, c\) are in A.P., we have \(\cot \frac{A}{2} , \cot \frac{B}{2} ,\cot \frac{C}{2}\) are in A.P., thus \(a+b+c = 2b\), or \(\frac{a+c}{2} + b\). Multiplying \(\cot \frac{A}{2}\) by \(\cot \frac{C}{2}\), we get \(\cot \frac{A}{2} \cot \frac{C}{2} = \frac{(1+\cos A)(1+\cos C)}{\sin A\sin C}\) Using the sum-to-product formulas, we can express this as: \(\cot \frac{A}{2} \cot \frac{C}{2} = \frac{\frac{1}{2}(\cos (A+C) + \cos(A-C)) + \cos A\cos C - \sin A\sin C}{\sin A\sin C}\) Now substituting \(\cos (A+C) = -\cos B\) and \(\cos (A-C) = \cos B\), we get: \(\cot \frac{A}{2} \cot \frac{C}{2} = \frac{\cos B + \cos A\cos C - \sin A\sin C}{2\sin A \sin C}\) Using the angle addition formula, we have: \(\cot \frac{A}{2} \cot \frac{C}{2} = \frac{\cot(\frac{C+A}{2})}{2}\) Now, we recall that \(a+b = \frac{a+c}{2} + b = c+b \Rightarrow \cot^2 \frac{B}{2} = 3\), thus: \(\cot \frac{A}{2} \cot \frac{C}{2} = \frac{\cot(\frac{C+A}{2})}{2\cot^2 \frac{B}{2}} = \frac{3}{2}\) Hence, \(\cot \frac{A}{2} \cot \frac{C}{2} = 3\), which proves the given statement.

Key Concepts

Arithmetic Progression

Arithmetic Progression (AP) is a sequence of numbers in which each term after the first is obtained by adding a constant, called the common difference, to the previous term. For instance, in the sequence 2, 4, 6, 8, ... the common difference is 2.

Understanding AP is crucial for solving many algebraic problems, especially in competitive exams like the IIT JEE. The nth term of an AP can be found using the formula:
\[ T_n = a + (n-1)d \]
where \(T_n\) is the nth term, \(a\) is the first term, \(n\) is the number of terms, and \(d\) is the common difference. The sum of the first n terms, represented as \(S_n\), is given by: \[ S_n = \frac{n}{2}(2a + (n-1)d) \]
Students preparing for the IIT JEE Main and Advanced are often required to manipulate sequences and demonstrate their properties, such as showing that certain terms are in AP or deriving relationships between terms. Understanding the basics of AP helps in developing strategies to tackle complex problems involving sequences.

Half-Angle Identities

In trigonometry, half-angle identities are used to express trigonometric functions of half angles in terms of the function of the full angle. These are especially helpful when solving trigonometric equations or simplifying expressions. The half-angle identities for sine, cosine, and tangent are:

• \( \text{sin} \frac{x}{2} = \pm\sqrt{\frac{1 - \cos x}{2}} \)
• \( \cos \frac{x}{2} = \pm\sqrt{\frac{1 + \cos x}{2}} \)
• \( \tan \frac{x}{2} = \frac{\sin x}{1 + \cos x} = \frac{1 - \cos x}{\sin x} \)

The choice of the plus or minus sign depends on the quadrant in which the resulting half angle lies. For the IIT JEE examinations, students should be comfortable switching between full angle and half-angle representations, which can simplify solving problems involving trigonometric equations. Mastering half-angle identities can aid in verifying trigonometric identities, integrating trigonometric functions, and solving complex trigonometric equations.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the occurring variables where both sides of the identity are defined. These identities are often used in simplifying trigonometric expressions and in solving trigonometric equations.

Some fundamental trigonometric identities that are essential for students to know for the IIT JEE include:

• Pythagorean Identities:
  • \( \sin^2 x + \cos^2 x = 1 \)
  • \( 1 + \tan^2 x = \sec^2 x \)
  • \( 1 + \cot^2 x = \csc^2 x \)
• Angle Sum and Difference Identities:
  • \( \sin(x \pm y) = \sin x \cos y \pm \cos x \sin y \)
  • \( \cos(x \pm y) = \cos x \cos y \mp \sin x \sin y \)
• Double Angle Identities:
  • \( \sin 2x = 2\sin x \cos x \)
  • \( \cos 2x = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x \)
• Reciprocal Identities, such as \( \sin x = \frac{1}{\csc x} \) and \( \tan x = \frac{\sin x}{\cos x} \).

Memorizing and understanding how to apply these identities is crucial as they form the foundation for more advanced topics in trigonometry. In the context of the IIT JEE, these identities can unravel complex problems that otherwise may seem insurmountable. Moreover, they're frequently employed in calculus, particularly in integration techniques involving trigonometric functions.