Problem 49 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(x=\sin \left(2 \tan ^{-1} 2\right), y=\sin \left(\frac{1}{2} \tan ^{-1} \frac{4}{3}\right)\), then (A) \(x=1-y\) (B) \(x^{2}=1-y\) (C) \(x^{2}=1+y\) (D) \(y^{2}=1-x\)

Step-by-Step Solution

Answer

(C) \(x^2 = 1 + y\) is correct.

Step 1: Find the value of x

Start by finding \(x = \sin \left(2 \tan^{-1} 2\right)\). We can use the double angle formula for sine: \(\sin(2\theta) = \frac{2 \tan \theta}{1 + \tan^2 \theta}\).Set \(\theta = \tan^{-1}2\), so \(\tan\theta = 2\). Substituting into the formula gives:\[ x = \frac{2 \cdot 2}{1 + 2^2} = \frac{4}{1 + 4} = \frac{4}{5} \]Thus, \(x = \frac{4}{5}\).

Step 2: Find the value of y

Next, find \(y = \sin \left(\frac{1}{2} \tan^{-1} \frac{4}{3}\right)\). We use the half-angle formula: \(\sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos \theta}{2}}\).First find \(\cos \theta\) knowing \(\tan \theta = \frac{4}{3}\):\[ \cos \theta = \frac{1}{\sqrt{1 + \left(\frac{4}{3}\right)^2}} = \frac{1}{\sqrt{1 + \frac{16}{9}}} = \frac{3}{5} \]So the half angle gives:\[ y = \sqrt{\frac{1 - \frac{3}{5}}{2}} = \sqrt{\frac{2}{10}} = \frac{1}{\sqrt{5}} \]

Step 3: Check the equation options

Substitute the values of \(x\) and \(y\) into the options to determine which holds:- For (A) \(x = 1-y\), \(\frac{4}{5} eq 1 - \frac{1}{\sqrt{5}}\).- For (B) \(x^2 = 1-y\), \(\left(\frac{4}{5}\right)^2 = 1 - \frac{1}{\sqrt{5}}\). Calculating: \[ \left(\frac{4}{5}\right)^2 = \frac{16}{25} \] \[ 1 - \frac{1}{\sqrt{5}} = 1 - \frac{1}{2.24} \approx 0.552 \] \[ \text{\(x^2\) equals approximately 0.64, so (B) is incorrect.} \]- For (C) \(x^2 = 1+y\), \[ \left(\frac{4}{5}\right)^2 = 1 + \frac{1}{\sqrt{5}} \] \[ \text{\(1+y\) equals approximately 1.447, which matches \(x^2\)} \]- The last option (D) doesn't match either. Thus \(x^2 = 1 + y\).

Key Concepts

Double Angle Formula

The Double Angle Formula is a valuable tool in trigonometry. It helps us find the sine, cosine, or tangent of twice an angle. For sine, the formula is \(\sin(2\theta) = \frac{2 \tan \theta}{1 + \tan^2 \theta}\). This formula is particularly handy when you know the tangent of half the angle but need the sine of the full angle.

In our problem, we used \(\theta = \tan^{-1}2\). This simplifies our task because we already know \(\tan \theta = 2\). Plugging \(\theta\) into the formula, we found \(x = \sin(2\theta) = \frac{2 \cdot 2}{1 + 2^2} = \frac{4}{5}\).

Using these formulas accurately helps streamline complex calculations involving trigonometric expressions. It's essential to remember each formula's specific conditions, such as needing the tangent of \(\theta\) to apply the sine Double Angle Formula.

Half Angle Formula

The Half Angle Formula is a key component when you need to find the sine or cosine of an angle divided by two. For sine, the formula is \(\sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos \theta}{2}}\). This enables calculations when full angle trigonometric values are known but you need the half-angle value.

Consider our scenario where \(\theta = \tan^{-1}\frac{4}{3}\). First, we identify \(\cos \theta\), which is necessary for the half-angle calculation. Using the known tangent, \(\cos \theta\) was computed as \(\frac{3}{5}\). Now substituting into the half-angle formula gives \(y = \sqrt{\frac{1 - \frac{3}{5}}{2}} = \frac{1}{\sqrt{5}}\).

Such formulas are excellent tools in simplifying our work with angles, particularly in complex trigonometric problems or expression simplifications.

Inverse Trigonometric Functions

Inverse Trigonometric Functions are used to determine the angle when a trigonometric ratio is known. It inversely relates the ratios to specific angles, like tangents to angles in our particular exercise.

For example, \(\tan^{-1}2\) implies finding the angle whose tangent is 2. Similarly, \(\tan^{-1}\frac{4}{3}\) asks the angle with that specific tangent. These functions are crucial for reverse-calculating original angle measures from known ratios.

In practice, inverse trigonometric functions facilitate determining more complex measurements and aid in deriving other trigonometric values using known ratios. Understanding both the direction and measure provided by these functions is important for solving a variety of trigonometric problems.