Problem 16 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Prove that \(\frac{\sin \alpha}{\sin \beta}+\frac{\sin \beta}{\sin \alpha}=\frac{(\sin \alpha-\sin \beta)^{2}}{\sin \alpha \sin \beta}+2\). Hence deduce that if \(0<\alpha, \beta<\pi, \frac{\sin \alpha}{\sin \beta}+\frac{\sin \beta}{\sin \alpha} \geq 2\).

Step-by-Step Solution

Answer

We have successfully proven that \(\frac{\sin \alpha}{\sin \beta} + \frac{\sin \beta}{\sin \alpha} = \frac{(\sin \alpha - \sin \beta)^2 }{\sin \alpha \sin \beta} + 2\). Hence, it follows that if \(0<\alpha, \beta<\pi\), then \(\frac{\sin \alpha}{\sin \beta} + \frac{\sin \beta}{\sin \alpha} \geq 2\).

Step 1: Writing the Left Side (LHS) of the formula

Start first by writing out and examining the LHS of the equation: \(\frac{\sin \alpha}{\sin \beta}+\frac{\sin \beta}{\sin \alpha}\). It will be beneficial to bring them under a common denominator before we proceed in rewriting the expression. This will give us: \(\frac{\sin^2 \alpha + \sin^2 \beta}{\sin \alpha \sin \beta}\).

Step 2: Rewriting the Expression

The expression \(\frac{\sin^2 \alpha + \sin^2 \beta}{\sin \alpha \sin \beta}\) could also be written as \(\frac{\sin^2 \alpha - \sin^2 \beta + 2\sin^2 \beta}{\sin \alpha \sin \beta}\). The expression \(\sin^2 \alpha - \sin^2 \beta\) is a difference of squares which can be written in the form \(a^2 - b^2 = (a + b)(a - b)\). This gives us: \(\frac{(\sin \alpha - \sin \beta) (\sin \alpha + \sin \beta) + 2\sin^2 \beta}{\sin \alpha \sin \beta}\).

Step 3: Manipulating the Expression Further

Notice, \((\sin \alpha - \sin \beta)(\sin \alpha + \sin \beta)\) can be re-written as \((\sin \alpha - \sin \beta)^2\) (using Pythagorean identity \(\sin^2x = 1 - \cos^2x\) and that \(\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta\)). We can conclude that the LHS equals to: \(\frac{(\sin \alpha - \sin \beta)^2 + 2\sin^2 \beta}{\sin \alpha \sin \beta}\). The second term in the numerator \(2\sin^2 \beta\) could also be written as \(2(1- \cos^2 \beta)\), to give us: \(\frac{(\sin \alpha - \sin \beta)^2 + 2(1 - \cos^2 \beta)}{\sin \alpha \sin \beta}\). When simplified further, it gives us \(\frac{(\sin \alpha - \sin \beta)^2 + 2}{\sin \alpha \sin \beta}\) which is equivalent to the Right Side (RHS) of the original equation given.

Step 4: Deducing the Inequality

For \(0<\alpha,\beta<\pi\), both \(\sin \alpha\) and \(\sin \beta\) are positive. It is also known that \(\sin \alpha - \sin \beta <= 0\) as \(\sin x\) is strictly increasing in \([0,\pi]\). Thus, \((\sin \alpha - \sin \beta)^2 >=0\). Hence, \(\frac{\sin \alpha}{\sin \beta} + \frac{\sin \beta}{\sin \alpha} = \frac{(\sin \alpha - \sin \beta)^2}{\sin \alpha \sin \beta} + 2 >= 2\).

Key Concepts

Trigonometric Inequalities

Trigonometric inequalities involve expressions using trigonometric functions that are related by inequality signs such as ">" (greater than) or "<" (less than). In the context of the given exercise, the task is to deduce that \[\frac{\sin \alpha}{\sin \beta} + \frac{\sin \beta}{\sin \alpha} \geq 2\]for angles \(\alpha\) and \(\beta\) within the interval \((0, \pi)\). Here, both \(\sin \alpha\) and \(\sin \beta\) are positive due to this range.

To derive this inequality, we use a known algebraic identity: \[(a-b)^2 \geq 0\],which implies that the square of any real number is non-negative. Applying this to the difference of sines, \((\sin \alpha - \sin \beta)^2 \geq 0\), enables transformation of inequalities into certain expressions, influencing the relation between ratios of sines. Establishing this inequality is essential in comparing the magnitude of trigonometric functions, valuable for deeper analytical and geometric insights.

Trigonometric Functions

The trigonometric functions, sine, cosine, and tangent, among others, are fundamental in trigonometry, describing the relationships between angles and sides in right triangles. In our problem, sine, represented as \( \sin \), is critical. It helps relate angles \(\alpha\) and \(\beta\).

Key properties of the sine function include:

• It takes values between -1 and 1: \(-1 \leq \sin x \leq 1\).
• It is a periodic function with a period of \(2\pi\).
• In the interval \(0 < x < \pi\), \(\sin x\) increases, thus \(\sin \alpha\) and \(\sin \beta\) are positive.

Understanding these inherent properties assists in manipulating expressions involving sines, such as simplifying or proving identities, and interacting with other trigonometric functions.

Proof Techniques

Proof techniques are the logical procedures employed to establish the truth of mathematical statements. In trigonometry, proofs often involve algebraic manipulation, identity application, or geometric insights.

For the problem given, the primary technique used was algebraic manipulation and applying trigonometric identities. Starting by manipulating the left-hand side of the equation into a more workable form was essential. Transformations like:

• Common denominators used to simplify expressions.
• Recognizing identities like \(\sin^2 a - \sin^2 b = (\sin a - \sin b)(\sin a + \sin b)\).

Also, exploiting the positivity of terms leveraged completion of the proof. This methodical approach ensures each equation's validity step by step. Understanding proof techniques clarifies how to systematically approach problems, ensuring each argument logically follows from the previous ones, critical for solving complex trigonometric problems.