Problem 105 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \frac{2 \sin (A-C) \cos C-\sin (A-2 C)}{2 \sin (B-C) \cos C-\sin (B-2 C)}=\frac{\sin A}{\sin B} $$

Step-by-Step Solution

Answer

After applying various trigonometric identities and simplifying the equation, we successfully prove the given identity \(\frac{2 \sin (A-C) \cos C-\sin (A-2 C)}{2 \sin (B-C) \cos C-\sin (B-2 C)}=\frac{\sin A}{\sin B}\).

Step 1: Manipulate the Left-hand Side(LHS)

To start, let's examine the LHS of our equation. We can apply the difference identity \(\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta\) and the double angle identity \(\sin2\alpha = 2\sin\alpha\cos\alpha\). Applying these, we find: LHS = \(\frac{2 \sin A \cos C \cos C - \sin A \cos^2C - 2\sin A \sin^2C}{2 \sin B \cos C \cos C - \sin B \cos^2C - 2 \sin B \sin^2C}\)Simplify this to get:LHS = \(\frac{\sin A (\cos^2C - 2\sin^2C)}{\sin B (\cos^2C - 2\sin^2C)}\)

Step 2: Use Trigonometric Identity

Now, using the Pythagorean identity \(\sin^2x+\cos^2x=1\), replace \(\cos^2C\) with \(1 - \sin^2C\) in the above equation. After simplifying, you get:LHS = \(\frac{\sin A (1 - 2\sin^2C)}{\sin B (1 - 2\sin^2C)}\)

Step 3: Simplify the Expression

You can simplify further by canceling out the common factor \(1 - 2\sin^2C\) in the numerator and denominator. This gives:LHS = \(\frac{\sin A}{\sin B}\)

Step 4: Conclusion

The left hand side (LHS) has been simplified to \(\frac{\sin A}{\sin B}\), which equals the right hand side (RHS). Therefore, we have proven that the given trigonometric identity is true.

Key Concepts

Pythagorean Identity

The Pythagorean identity states: \( \sin^2x + \cos^2x = 1 \). This is one of the most fundamental identities in trigonometry. Think of it as the trigonometric equivalent of the Pythagorean Theorem in geometry. It relates the squares of the sine and cosine of an angle to 1. This is because the hypotenuse in a right triangle with sides \( \sin x \) and \( \cos x \) is always 1. This identity allows you to express \( \cos^2x \) in terms of \( \sin^2x \) or vice versa, which is useful in simplifying expressions. If you know one, you can find the other through rearranging the identity:

• \( \cos^2x = 1 - \sin^2x \)
• \( \sin^2x = 1 - \cos^2x \)

This identity was used in the solution to simplify \( \cos^2C \) into a form that can easily be cancelled out, showing its practical utility in solving problems.

Double Angle Identity

Another crucial identity in trigonometry is the Double Angle Identity, specifically the formula \( \sin2\alpha = 2 \sin\alpha \cos\alpha \). It provides a way to express the sine of double an angle in terms of the sine and cosine of the original angle. Think of it as generating bigger angles from smaller ones.In the context of our exercise, realizing that a doubled angle sine can break down into simpler parts is key. This was used to rewrite and simplify terms in the equation by expressing complex sine terms with angles like \( A - 2C \) into more manageable forms involving \( \sin A \) and \( \cos C \). This identity simplifies manipulation of trigonometric expressions by breaking a complex angle into familiar components.It's especially valuable for solving equations and identities where direct comparison of angles or attempting common denominators are involved.

Difference Identity

The Difference Identity for sine is described as \( \sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta \). It's essential for splitting up expressions with angles expressed as a difference. Understanding this identity helps when you must decompose expressions like \( \sin(A - C) \). It shows how you can break them into constituent parts that involve simpler trigonometric functions. This kind of breakdown is handy for simplifying expressions and solving equations.In our exercise, employing the difference identity allowed the equation's left side to express initially combined angles into separable terms. By doing so, it facilitates solving and proving the equation by aligning angles, making it easier to spot similar patterns and simplify accordingly.Using the difference identity effectively turns a complex situation into a straightforward comparison and calculation, vital for handling more intricate trigonometric expressions.