Problem 12 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

If \(\cos \left[\tan ^{-1}\left\\{\sin \left(\cot ^{-1} \sqrt{3}\right)\right\\}\right]=y\), then the value of \(y\) is (a) \(y=\frac{4}{5}\) (b) \(y=\frac{2}{\sqrt{5}}\) (c) \(y=-\frac{2}{\sqrt{5}}\) (d) \(y=\frac{\sqrt{3}}{2}\)

Step-by-Step Solution

Answer

None of the given options are correct. The correct value of \(y\) is \(\frac{\sqrt{2}}{2}\).

Step 1: Evaluate Inverse Cotangent Function

Firstly, let's evaluate \(\cot ^{-1}\sqrt{3}\). It's important to remember that the result will be an angle. In this case, \(\cot ^{-1}\sqrt{3} = \frac{\pi}{6}\) because cotangent of \( \frac{\pi}{6}\) is \(\sqrt{3}\).

Step 2: Evaluate Sin Function

Now, substitute \(\frac{\pi}{6}\) in \( \sin\left(\cot ^{-1} \sqrt{3}\right)\). We get \(\sin\left(\frac{\pi}{6}\right)= \frac{1}{2}\).

Step 3: Evaluate Inverse Tangent Function

Next step is to evaluate \( \tan^{-1}\left(\sin \left(\cot^{-1} \sqrt{3}\right)\right)\) which will be \( \tan^{-1}\left(\frac{1}{2}\right)\). By definition of inverse tangent function, \( \tan^{-1}\left(\frac{1}{2}\right)= \frac{\pi}{4}\) because tangent of \( \frac{\pi}{4}\) is \(\frac{1}{2}\).

Step 4: Evaluate Cos Function

Now, use \( \frac{\pi}{4}\) in \( \cos\left[\tan^{-1}\left(\sin\left(\cot^{-1} \sqrt{3}\right)\right)\right]\) to find \(y\). Thus, \(y = \cos\left(\frac{\pi}{4}\right)\). Since, cosine of \( \frac{\pi}{4}\) is \(\frac{\sqrt{2}}{2}\), then \(y = \frac{\sqrt{2}}{2}\).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are special functions that allow us to determine angles when given a specific trigonometric ratio. They are the reverse operations of the standard trigonometric functions. For instance, if we know that the sine of an angle is \sin(\theta) = x\, then the inverse sine function, \sin^{-1}(x)\, will return the angle \(\theta\). These functions are essential in solving equations where the angle is unknown, particularly when dealing with right triangles and certain trigonometric identities. They also help to evaluate expressions within a specific range.

Key inverse trigonometric functions include:

• \sin^{-1}(x)\ - Arcsine
• \cos^{-1}(x)\ - Arccosine
• \tan^{-1}(x)\ - Arctangent
• \cot^{-1}(x)\ - Arccotangent
• \csc^{-1}(x)\ - Arccosecant
• \sec^{-1}(x)\ - Arcsecant

When working with these functions, it is crucial to consider their range of values, as they are restricted to principal values to maintain the inverse relationship. For instance, \cot^{-1}( ext{some value})\ returns an angle between 0 and \pi\, which is essential to solve problems that involve relationships between different trigonometric functions, as seen in the original exercise. Understanding these concepts can simplify complex trigonometric expressions by breaking them into manageable parts that relate known quantities to unknown angles.

Trigonometric Identities

Trigonometric identities are equations that involve trigonometric functions and hold true for all values within their domains. They are vital in simplifying trigonometric expressions and solving equations.

There are several fundamental identities:

• Pythagorean Identity: \(\sin^2(\theta) + \cos^2(\theta) = 1\)
• Sum and Difference Formulas: including \(\sin(a \pm b)\) and \(\cos(a \pm b)\)
• Double Angle Formulas: such as \(\sin(2\theta) = 2\sin(\theta)\cos(\theta)\)
• Reciprocal identities: like \(\sec(\theta) = \frac{1}{\cos(\theta)}\)

Applying these identities can transform an initial expression into something much simpler or more suitable for solving a problem. For example, if we evaluate \cos[\tan^{-1}( an(\theta))]\, using trigonometric identities helps us sidestep directly calculating complicated expressions.

Proper use of trigonometric identities enables combining, simplifying, or changing the form of expressions to reach desired values or results in exercises such as the one provided. They are stepping stones to deeper understanding and problem-solving in trigonometry, particularly useful under time constraints in competitive examinations like IIT JEE.

Angles in Radians

Radians offer a natural way to describe circular angles and are the standard unit of angular measure used in mathematics. Unlike degrees, which divide a circle into 360 parts, radians consider the arc's length for defining angles. One radian is the angle created when the arc's length equals the radius of the circle.

The fundamental formulas relate degrees and radians as follows:

• \[\pi \, ext{radians} = 180^\circ\]
• \[1 \, ext{radian} = \frac{180^\circ}{\pi} \, ext{degrees}\]

Radians simplify many calculations in calculus and analytical geometry. For example, the use of radians automatically gives the derivative of trigonometric functions its simple and recognizable form, making them highly efficient for working with sinusoidal models. In trigonometry problems like the one in the original exercise, recognizing angles such as \(\frac{\pi}{6}\) or \(\frac{\pi}{4}\) in terms of radians is crucial for streamlining computations and understanding the problem's framework. Familiarity with radians aids in analyzing problems and allows for seamless transitions between geometry and calculus, making it essential knowledge for students tackling competitive tests like the IIT JEE.