Problem 191 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { If } \cos A=\frac{3}{5}, \cos B=\frac{5}{13}, \text { prove that } \sin ^{2} \frac{A-B}{2}=\frac{1}{65}, \cos ^{2} \frac{A-B}{2}=\frac{64}{65} $$

Step-by-Step Solution

Answer

Upon calculations, \( \sin^2 \frac{A-B}{2}=\frac{1}{65} \) and \( \cos^2 \frac{A-B}{2}=\frac{64}{65} \), thus proving the identity.

Step 1: Apply the formula for cosine of difference of two angles

Calculate \( \cos (A-B) \) using the given values of \( \cos A \) and \( \cos B \). The formula for the cosine of the difference of two angles is \[ \cos(A-B)=\cos A \cos B+\sin A \sin B \]. Now we have the value of cosine of \(A\) and \(B\), but we need the value of sine of these angles. We know that \(\sin A = \sqrt{1-\cos ^2 A}\) and \( \sin B = \sqrt{1-\cos ^2 B}\). Thus, \( \sin A = \sqrt{1-3/5^2}=\frac{4}{5} \), \( \sin B = \sqrt{1-5/13^2}=\frac{12}{13} \). Substitute these values into the equation: \[ \cos(A-B)=3/5*5/13+4/5*12/13= \frac{51}{65}\].

Step 2: Calculate the sine and cosine of half angles

We will use the half angle formulas to find \( \sin^2 \frac{A-B}{2} \) and \( \cos^2 \frac{A-B}{2} \). The formulas are \( \sin^2 \frac{\alpha}{2}=\frac{1-\cos \alpha}{2} \) and \( \cos^2 \frac{\alpha}{2}=\frac{1+\cos \alpha}{2} \). Here, \( \alpha = A-B \). Substituting from step 1 gives \( \sin^2 \frac{A-B}{2}= \frac{1-\frac{51}{65}}{2}=\frac{1}{65} \) and \( \cos^2 \frac{A-B}{2}=\frac{1+\frac{51}{65}}{2}=\frac{64}{65} \).

Step 3: Verifying the identities

Upon calculation in Step 2, we find that \( \sin^2 \frac{A-B}{2}=\frac{1}{65} \) and \( \cos^2 \frac{A-B}{2}=\frac{64}{65} \). These are the identities we were asked to prove, which verifies the result.

Key Concepts

Cosine of Difference Formula

The cosine of difference formula allows us to find the cosine of the difference between two angles. This formula is handy in trigonometry to simplify expressions involving angles. To compute \( \cos(A-B) \), the formula is:

• \(\cos(A-B) = \cos A \cdot \cos B + \sin A \cdot \sin B\)

Given \( \cos A = \frac{3}{5} \) and \( \cos B = \frac{5}{13} \), our main challenge is to find the sines of these angles.
We use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \) to determine \( \sin A \) and \( \sin B \). Substituting the values:

• \(\sin A = \sqrt{1-\left(\frac{3}{5}\right)^2} = \frac{4}{5}\)
• \(\sin B = \sqrt{1-\left(\frac{5}{13}\right)^2} = \frac{12}{13}\)

Now, substitute back into the cosine of difference formula to get:

• \( \cos(A-B) = \frac{3}{5}\cdot\frac{5}{13} + \frac{4}{5}\cdot\frac{12}{13} = \frac{51}{65} \)

Half Angle Formulas

Half angle formulas are crucial for finding the sine and cosine of half angles. They work by expressing these values in terms of the full angle. This simplifies proofs and calculations in trigonometry. These formulas are:

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

For the angle \( \alpha = A-B \) with \( \cos(A-B) = \frac{51}{65} \) from our earlier calculation:

• \(\sin^2 \frac{A-B}{2} = \frac{1 - \frac{51}{65}}{2} = \frac{1}{65}\)
• \(\cos^2 \frac{A-B}{2} = \frac{1 + \frac{51}{65}}{2} = \frac{64}{65}\)

These calculations show us how half angle formulas relate angles and their halves, making them extremely useful in proving identities.

Trigonometric Functions

Trigonometric functions are the backbone of trigonometry. They relate the angles of a triangle to the lengths of its sides. The primary functions are sine, cosine, and tangent, each with their own specific roles.
Cosine and sine are especially important when dealing with identities and angle calculations:

• Sine (\(\sin\)): Measures the vertical component, calculated as opposite side/hypotenuse in a right triangle.
• Cosine (\(\cos\)): Measures the horizontal component, calculated as adjacent side/hypotenuse.
• The checks between sine and cosine, such as \(\sin^2 \theta + \cos^2 \theta = 1\), keep calculations consistent and prove essential results like the Pythagorean identity.

When dealing with expressions like those in our problem, understanding these functions allows us to easily navigate through calculations. By breaking down into simpler known forms, these functions help simplify complex trigonometric expressions.