Problem 38 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

If \(\tan ^{2}\left(\frac{\pi}{4}+\frac{\theta}{2}\right)=\frac{a}{b}\), then \(\sin (\theta)\) is (a) \(\left(\frac{a-b}{a+b}\right)\) (b) \(-\left(\frac{a-b}{a+b}\right)\) (c) \(\left(\frac{a+b}{a-b}\right)\) (d) \(-\left(\frac{a+b}{a-b}\right)\)

Step-by-Step Solution

Answer

Hence, the correct option is (a) \(\left(\frac{a-b}{a+b}\right)\)

Step 1: Application of Addition Formula for Tan

Determine the value of \(\tan ^{2}\left(\frac{\pi}{4}+\frac{\theta}{2}\right)\) using the addition formula: \(tan(a+b)=\frac{tan(a)+tan(b)}{1-tan(a)*tan(b)}\). Here, a = \(\frac{\pi}{4}\) and b = \(\frac{\theta}{2}\). Hence, we have \(\tan ^{2}\left(\frac{\pi}{4}+\frac{\theta}{2}\right)=\left(\frac{1 + \tan(\frac{\theta}{2})}{1 - \tan(\frac{\theta}{2})}\right)^{2}\).

Step 2: Use the Expression for Tan θ/2 in terms of Sin θ

Using the half angle formulas, we know that \(tan(\frac{\theta}{2})=\sqrt{\frac{1-\cos(\theta)}{1+\cos(\theta)}}\). Substituting this into the equation in Step 1, we get: \(\frac{a}{b}=\left(\frac{1 + \sqrt{\frac{1-\cos θ}{1+\cos θ}}}{1 - \sqrt{\frac{1-\cos θ}{1+\cos θ}}}\right)^{2}\)

Step 3: Rearrange the Equation to find sin θ

With several steps of algebraic manipulation and substitutions, using \(cos^{2} θ = 1 - sin^{2} θ\), the equation \(\frac{a}{b}=\left(\frac{1 + \sqrt{\frac{1-\cos θ}{1+\cos θ}}}{1 - \sqrt{\frac{1-\cos θ}{1+\cos θ}}}\right)^{2}\) can eventually be rearranged to find that \(sin (\theta) = \left(\frac{a-b}{a+b}\right)\)

Key Concepts

Addition Formulas

Trigonometric addition formulas allow us to find the tangent, sine, or cosine of the sum or difference of two angles. This can be particularly useful in solving complex trigonometric equations or expressions. Let's focus on the tangent addition formula:

• For addition, the formula is: \[ \tan(a+b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)} \]
• When using this formula, it's important to understand each component. Here, 'a' and 'b' are individual angles.
• In our specific problem, \(a = \frac{\pi}{4}\) and \(b = \frac{\theta}{2}\).

To apply the formula, we substitute these values into the equation. \(\tan(\frac{\pi}{4})\) simplifies to 1, simplifying the expression to:

• \[ \tan\left(\frac{\pi}{4} + \frac{\theta}{2}\right) = \frac{1 + \tan\left(\frac{\theta}{2}\right)}{1 - \tan\left(\frac{\theta}{2}\right)} \]
• By squaring the result as given in the problem, we can use it to further determine the values needed for \(\sin(\theta)\).

Half-Angle Identities

Half-angle identities are key tools in trigonometry that allow one to find the sine, cosine, or tangent of half an angle. They are especially useful when dealing with integrals or evaluating trigonometric equations at non-standard angles. For our problem, the half-angle identity for tangent is particularly useful:

• \[ \tan\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos(\theta)}{1 + \cos(\theta)}} \]
• This identity is derived from the sine and cosine half-angle identities. It offers a way to express \(\tan\) in terms of \(\cos\), which links directly to \(\sin\) due to the identity \(\cos^2\theta + \sin^2\theta = 1\).
• Using these identities, you can derive expressions involving \(\sin(\theta)\) by substituting for \(\cos\), knowing \(\cos(\theta) = \sqrt{1 - \sin^2(\theta)}\).

In the problem, once the half-angle tangent identity is substituted into the addition formula, you can simplify and ultimately solve for \(\sin(\theta)\) through algebraic rearrangement.

Tan and Sin Relationship

Understanding the relationship between \(\tan\) and \(\sin\) is crucial in trigonometry. The function \(\tan(\theta)\) can be expressed in terms of \(\sin(\theta)\) and \(\cos(\theta)\), as follows:

• \[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \]
• This identity means that knowing either the sine or cosine of an angle can help you find \(\tan\).

In the exercise, we utilized this relationship while manipulating the identity \(\tan\left(\frac{\theta}{2}\right)\) with the half-angle formula. Given \(\tan^2\), the substitution using the half-angle identity simplifies linking the problem’s expression by bridging it to \(\sin(\theta)\) through algebraic steps.With the stated formula \(\cos^2\theta = 1 - \sin^2\theta\), you can express \(\cos(\theta)\) to include \(\sin(\theta)\) terms, linking back to tangent expressions. Through these trigonometric maneuvers, we ultimately rearranged the problem to highlight that \(\sin (\theta) = \left(\frac{a-b}{a+b}\right)\), showing how intricately \(\tan\) calculations can connect to \(\sin\). By mastering these identities and relationships, solving complex trigonometric equations becomes much clearer.