Step 1: Establishing the Theoretical Framework

To find the principal value of $y = \cot^{-1}(\sqrt{3})$, we must identify an angle $y$ such that:

• $\cot(y) = \sqrt{3}$
• $y$ falls within the principal value branch of the cotangent function.

[Per the definition of Inverse Trigonometric Functions, the range of the principal value branch of $\cot^{-1}(x)$ is $(0, \pi)$].

Step 2: Conversion to Reciprocal Trigonometric Identity

Since $\cot(y) = \frac{1}{\tan(y)}$, we can rewrite the equation as follows:

$\cot(y) = \sqrt{3} \implies \frac{1}{\tan(y)} = \sqrt{3}$ [By the Reciprocal Identity of Trigonometric Functions]

$\tan(y) = \frac{1}{\sqrt{3}}$

Step 3: Determining the Angle via Trigonometric Table Values

We know from the standard values of trigonometric functions that:

$\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$

Since $\frac{\pi}{6}$ lies within the interval $(0, \pi)$, it satisfies the principal value requirement.

Step 4: Verification of the Principal Value

The value $\frac{\pi}{6}$ corresponds to $30^\circ$. We confirm:

$\cot\left(\frac{\pi}{6}\right) = \frac{\cos(\pi/6)}{\sin(\pi/6)} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$

Since $0 < \frac{\pi}{6} < \pi$, this value is the principal value.

Visual Representation of the Principal Angle

Step 5: Final Evaluation

Given $\cot(y) = \sqrt{3}$, we set $y = \cot^{-1}(\sqrt{3})$.

$y = \frac{\pi}{6}$

Final Solution: The principal value of $\cot^{-1}(\sqrt{3})$ is $\frac{\pi}{6}$.

Step 1: Identifying the Definition and Domain

The principal value branch of the function $f(x) = \sin^{-1}(x)$ is defined as the interval $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$. We seek a value $y$ such that:

$\sin(y) = -\frac{1}{2}$, where $y \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$ [Per the definition of principal value branches of inverse trigonometric functions].

Step 2: Evaluating the Trigonometric Ratio

We know from standard trigonometric table values that $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$.

Utilizing the trigonometric identity for negative angles, specifically $\sin(-\theta) = -\sin(\theta)$ [By the Odd Function property of the Sine function], we apply the substitution:

$-\sin\left(\frac{\pi}{6}\right) = \sin\left(-\frac{\pi}{6}\right)$

Therefore, $\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}$.

Step 3: Verification of the Principal Range

We must confirm that the identified angle $y = -\frac{\pi}{6}$ satisfies the condition $y \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.

Since $-\frac{\pi}{2} \leq -\frac{\pi}{6} \leq \frac{\pi}{2}$ is a true mathematical statement, $-\frac{\pi}{6}$ lies within the principal value branch.

Step 4: Visualizing the Angle on the Unit Circle

Step 5: Final Derivation

Given $\sin(y) = -\frac{1}{2}$, then:

$y = \sin^{-1}\left(-\frac{1}{2}\right)$

$y = -\frac{\pi}{6}$

Final Solution: The principal value of $\sin^{-1} (-\frac{1}{2})$ is $-\frac{\pi}{6}$.

Step 1: Theoretical Foundation of Principal Value Branch

The inverse secant function, denoted as $\sec^{-1}(x)$, is defined such that its principal value branch lies within the interval $[0, \pi] \setminus \{\frac{\pi}{2}\}$. For a given value $x$, the principal value $y$ must satisfy the condition:

$\sec(y) = x$, where $y \in [0, \pi], y \neq \frac{\pi}{2}$ [Per the definition of the range of the inverse secant function].

Step 2: Mathematical Setup and Transformation

Given the equation: $\sec^{-1} \left( \frac{2}{\sqrt{3}} \right) = y$

By the definition of the inverse trigonometric function, we convert the expression into its equivalent trigonometric form:

$\sec(y) = \frac{2}{\sqrt{3}}$

Since $\sec(y) = \frac{1}{\cos(y)}$, we apply the reciprocal identity:

$\cos(y) = \frac{1}{\sec(y)} = \frac{1}{\frac{2}{\sqrt{3}}} = \frac{\sqrt{3}}{2}$

Step 3: Determining the Principal Value

To solve $\cos(y) = \frac{\sqrt{3}}{2}$ for $y$ within the principal interval $[0, \pi]$:

• We know from standard trigonometric table values that $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$. [By the unit circle definition of cosine].
• Since $\frac{\pi}{6}$ lies within the interval $[0, \pi]$, it satisfies the condition of the principal value branch.

Step 4: Geometric Visualization

Step 5: Verification of Constraints

Check: Does $y = \frac{\pi}{6}$ satisfy $y \in [0, \pi] \setminus \{\frac{\pi}{2}\}$?

Since $0 \leq \frac{\pi}{6} \leq \pi$ and $\frac{\pi}{6} \neq \frac{\pi}{2}$, the value is valid. [Per the domain and range constraints of $\sec^{-1}(x)$].

Final Solution: \sec^{-1} \left( \frac{2}{\sqrt{3}} \right) = \frac{\pi}{6}

Step 1: Evaluation of the First Term $\tan^{-1}(\sqrt{3})$

Let $y_1 = \tan^{-1}(\sqrt{3})$. By the definition of the inverse trigonometric function, this is equivalent to $\tan(y_1) = \sqrt{3}$, where the principal value branch of $\tan^{-1}(x)$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

Since $\tan(\frac{\pi}{3}) = \sqrt{3}$ [Trigonometric Standard Values], and $\frac{\pi}{3} \in (-\frac{\pi}{2}, \frac{\pi}{2})$, we conclude:

$y_1 = \frac{\pi}{3}$

Step 2: Evaluation of the Second Term $\sec^{-1}(-2)$

Let $y_2 = \sec^{-1}(-2)$. By definition, $\sec(y_2) = -2$, where the principal value branch of $\sec^{-1}(x)$ is $[0, \pi] - \{\frac{\pi}{2}\}$.

Recall the identity $\sec(y) = \frac{1}{\cos(y)}$. Thus, $\cos(y_2) = -\frac{1}{2}$ [Reciprocal Identity].

We know that $\cos(\frac{\pi}{3}) = \frac{1}{2}$. Since the cosine function is negative in the second quadrant, we use the identity $\cos(\pi - \theta) = -\cos(\theta)$ [Allied Angle Formula]:

$\cos(\pi - \frac{\pi}{3}) = \cos(\frac{2\pi}{3}) = -\frac{1}{2}$

Since $\frac{2\pi}{3} \in [0, \pi]$, we conclude:

$y_2 = \frac{2\pi}{3}$

Step 3: Geometrical Representation of Principal Values

Step 4: Final Calculation

Substitute the evaluated principal values into the original expression:

$\text{Result} = y_1 - y_2$

$\text{Result} = \frac{\pi}{3} - \frac{2\pi}{3}$

$\text{Result} = -\frac{\pi}{3}$

Final Solution: -\frac{\pi}{3}

Step 1: Theoretical Foundation of Principal Value

To find the principal value of $\csc^{-1}(x)$, we must identify the angle $\theta$ such that $\csc(\theta) = x$, where $\theta$ lies within the defined principal value branch for the inverse cosecant function. The domain of $\csc^{-1}(x)$ is $(-\infty, -1] \cup [1, \infty)$ and its range is $[-\frac{\pi}{2}, \frac{\pi}{2}] \setminus \{0\}$.

Let $y = \csc^{-1}(2)$. This implies:
$\csc(y) = 2$ [By definition of inverse trigonometric functions]

Step 2: Relation to Primary Trigonometric Ratios

Since $\csc(y) = \frac{1}{\sin(y)}$, we can convert the equation into terms of sine:
$\frac{1}{\sin(y)} = 2$
$\sin(y) = \frac{1}{2}$

Step 3: Solving for the Principal Angle

We seek the angle $y \in [-\frac{\pi}{2}, \frac{\pi}{2}] \setminus \{0\}$ such that $\sin(y) = \frac{1}{2}$.
From the standard trigonometric unit circle values, we know that $\sin(\frac{\pi}{6}) = \frac{1}{2}$.

Since $\frac{\pi}{6}$ falls within the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and $\frac{\pi}{6} \neq 0$, it is the principal value.

Step 4: Geometric Visualization

Step 5: Logical Verification

Expression	Value
$\csc(y) = 2$	Confirmed
$y \in [-\frac{\pi}{2}, \frac{\pi}{2}] \setminus \{0\}$	$\frac{\pi}{6} \approx 0.523$ radians; Condition satisfied.

Final Solution: The principal value of $\csc^{-1}(2)$ is $\frac{\pi}{6}$.

Step 1: Definition of the Inverse Sine Function

By definition, the inverse sine function, denoted as $\sin^{-1} x$ or $\arcsin x$, is the inverse function of the restricted sine function. For a function $f(x) = \sin x$ to be invertible, its domain must be restricted to the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ to ensure it is bijective (both injective and surjective) over the range $[-1, 1]$.

[Per the definition of Inverse Trigonometric Functions, the domain of $\sin x$ is restricted to $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ to map the range $[-1, 1]$ uniquely.]

Step 2: Establishing the Range for $y$

Given the equation $y = \sin^{-1} x$, the variable $y$ represents the principal value of the angle whose sine is $x$. According to the standard principal value branches of inverse trigonometric functions:

• The domain of $\sin^{-1} x$ is $x \in [-1, 1]$.
• The range (principal value branch) of $\sin^{-1} x$ is $y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Step 3: Visual Representation of the Principal Value Branch

Step 4: Formal Conclusion of the Inequality

Since $y$ is defined as the output of the principal branch of the inverse sine function, $y$ must be greater than or equal to the lower bound and less than or equal to the upper bound of the restricted interval.

Mathematically, this is expressed as:

$-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$

[By the properties of the principal value branch for the inverse sine function as established in standard trigonometric identity frameworks.]

Final Solution: $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$

Step 1: Identifying the Principle Value Branch

The function $f(x) = \tan^{-1}(x)$ has a defined principal value branch restricted to the open interval $(-\frac{\pi}{2}, \frac{\pi}{2})$. [Per the definition of the inverse tangent function].

We are tasked with finding the principal value $y$ such that:

$$y = \tan^{-1}(-\sqrt{3})$$

By the definition of inverse functions, this is equivalent to:

$$\tan(y) = -\sqrt{3}, \quad y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$

Step 2: Applying Trigonometric Identities

We observe that $\tan(\frac{\pi}{3}) = \sqrt{3}$. Using the odd function property of the tangent function, specifically $\tan(-\theta) = -\tan(\theta)$ [Per the property of odd trigonometric functions], we can write:

$$-\tan\left(\frac{\pi}{3}\right) = \tan\left(-\frac{\pi}{3}\right)$$

Therefore:

$$\tan(y) = \tan\left(-\frac{\pi}{3}\right)$$

Step 3: Verification of the Domain

We must confirm that the value $-\frac{\pi}{3}$ lies within the principal value branch $(-\frac{\pi}{2}, \frac{\pi}{2})$.

Since $-\frac{\pi}{2} < -\frac{\pi}{3} < \frac{\pi}{2}$, the value satisfies the criteria for the principal value branch [By the definition of the range of the inverse tangent function].

Step 4: Conclusion

Given $\tan(y) = \tan(-\frac{\pi}{3})$ and $y \in (-\frac{\pi}{2}, \frac{\pi}{2})$, it follows that $y = -\frac{\pi}{3}$.

Final Solution: The principal value of $\tan^{-1}(-\sqrt{3})$ is $-\frac{\pi}{3}$.

Step 1: Establishing the Theoretical Range

The inverse cosine function $\cos^{-1}(x)$ is defined such that its principal value branch lies within the interval $[0, \pi]$. Let $y$ be the principal value of $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)$. Thus, we define the relation:

$y = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right)$

This is equivalent to the trigonometric equation:

$\cos(y) = \frac{\sqrt{3}}{2}$, where $y \in [0, \pi]$ [Per the definition of the Principal Value Branch of Inverse Cosine].

Step 2: Analytical Identification of the Angle

We evaluate the specific value of the cosine function within the unit circle geometry. Recalling the standard trigonometric table values for acute angles:

• $\cos(0) = 1$
• $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$
• $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$
• $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$
• $\cos\left(\frac{\pi}{2}\right) = 0$

By comparison, $\cos(y) = \frac{\sqrt{3}}{2}$ matches the identity $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$.

Step 3: Verification of Domain and Range

Since $\frac{\pi}{6} \approx 0.5236$ radians, and the defined range is $[0, \pi] \approx [0, 3.14159]$, it is clear that $\frac{\pi}{6} \in [0, \pi]$. [By the Property of Inverse Trigonometric Functions: $\cos^{-1}(\cos \theta) = \theta$ for $\theta \in [0, \pi]$].

Step 4: Geometric Representation (Unit Circle)

Step 5: Final Conclusion

As the value $y = \frac{\pi}{6}$ satisfies both the trigonometric equation $\cos(y) = \frac{\sqrt{3}}{2}$ and the principal branch constraint $y \in [0, \pi]$, we conclude the result.

Final Solution: The principal value is $\frac{\pi}{6}$ radians.

Step 1: Definition of Principal Value Branch

The principal value branch of the inverse cosecant function, $\text{cosec}^{-1}(x)$, is defined as:
$\text{cosec}^{-1} : \mathbb{R} \setminus (-1, 1) \rightarrow \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \setminus \{0\}$
Let $y = \text{cosec}^{-1}(-\sqrt{2})$. By the definition of inverse functions, this implies:
$\text{cosec}(y) = -\sqrt{2}$, where $y \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$ and $y \neq 0$. [Per the definition of Inverse Trigonometric Functions]

Step 2: Relation to Sine Function

Since $\text{cosec}(y) = \frac{1}{\sin(y)}$, we can rewrite the equation as:
$\frac{1}{\sin(y)} = -\sqrt{2}$
$\sin(y) = -\frac{1}{\sqrt{2}}$ [By the Reciprocal Identity of Trigonometric Functions]

Step 3: Finding the Reference Angle

We know that $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$. Since $\sin(y)$ is negative, $y$ must lie in the fourth quadrant of the principal value branch [As $\sin(\theta) < 0$ for $\theta \in [-\frac{\pi}{2}, 0)$]. Using the property $\sin(-\theta) = -\sin(\theta)$:
$\sin(y) = -\sin\left(\frac{\pi}{4}\right)$
$\sin(y) = \sin\left(-\frac{\pi}{4}\right)$ [By the Odd Function Property of Sine]

Step 4: Evaluating the Principal Value

Given the domain constraint $y \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$, we compare:
$y = -\frac{\pi}{4}$
Since $-\frac{\pi}{4} \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$ and $-\frac{\pi}{4} \neq 0$, this value satisfies the condition for the principal branch.

Step 5: Verification of Range

Function	Principal Value Branch	Result
$\text{cosec}^{-1}(x)$	$[-\frac{\pi}{2}, \frac{\pi}{2}] \setminus \{0\}$	$-\frac{\pi}{4}$

Final Solution: The principal value of $\text{cosec}^{-1}(-\sqrt{2})$ is $-\frac{\pi}{4}$.

Step 1: Definition of Principal Value Branch

The function $y = \cos^{-1}(x)$ has a restricted domain of $[-1, 1]$ and a principal value branch (range) defined as $[0, \pi]$. Let the given expression be represented as:

$y = \cos^{-1} \left( -\frac{1}{\sqrt{2}} \right)$

[By the definition of Inverse Trigonometric Functions, $y$ must satisfy $0 \le y \le \pi$].

Step 2: Conversion to Trigonometric Form

Applying the cosine function to both sides, we obtain:

$\cos(y) = -\frac{1}{\sqrt{2}}$

We know that for positive values, $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$. Since $\cos(y)$ is negative, the angle $y$ must lie in the second quadrant because the cosine function is negative in the interval $(\frac{\pi}{2}, \pi]$.

Step 3: Algebraic Manipulation using Quadrant Properties

Using the allied angle formula $\cos(\pi - \theta) = -\cos(\theta)$ [Per the Unit Circle Symmetry Principle]:

$\cos(y) = -\cos\left(\frac{\pi}{4}\right)$

$\cos(y) = \cos\left(\pi - \frac{\pi}{4}\right)$

$\cos(y) = \cos\left(\frac{3\pi}{4}\right)$

Step 4: Geometric Representation

Step 5: Verifying the Range

Since $\frac{3\pi}{4}$ lies within the interval $[0, \pi]$, it satisfies the requirement for the principal value branch [By the Range Constraint of Inverse Cosine].

Final Solution: \frac{3\pi}{4}

Step 1: Definition of Principal Value Branch

For the function $f(x) = \cos^{-1}(x)$, the principal value branch is defined as the interval $[0, \pi]$. [Per the definition of Inverse Trigonometric Functions].

We are required to find $y$ such that:

$y = \cos^{-1}\left(-\frac{1}{2}\right)$

By the definition of the inverse function, this is equivalent to:

$\cos(y) = -\frac{1}{2}$, where $y \in [0, \pi]$

Step 2: Reference Angle Analysis

First, identify the reference angle for the positive value $\frac{1}{2}$:

$\cos(\theta) = \frac{1}{2} \implies \theta = \frac{\pi}{3}$ [Standard trigonometric table values].

Step 3: Determining the Quadrant

Since $\cos(y) = -\frac{1}{2}$ is negative, the angle $y$ must lie in the second quadrant, as the range of $\cos^{-1}(x)$ is restricted to $[0, \pi]$ and the cosine function is negative in $( \frac{\pi}{2}, \pi ]$. [Per the ASTC rule for trigonometric signs].

Step 4: Algebraic Transformation

Using the identity for supplementary angles in the second quadrant:

$\cos(\pi - \theta) = -\cos(\theta)$ [By the Trignometric Identity: $\cos(\pi - \alpha) = -\cos(\alpha)$]

Substituting $\theta = \frac{\pi}{3}$ into the identity:

$\cos\left(\pi - \frac{\pi}{3}\right) = -\cos\left(\frac{\pi}{3}\right)$ $\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$

Step 5: Visual Representation

Step 6: Verification of Range

Check if the calculated value $y = \frac{2\pi}{3}$ satisfies the principal branch condition:

• $0 \leq \frac{2\pi}{3} \leq \pi$
• $0 \leq 2.094 \leq 3.141$ (True)

Since $\frac{2\pi}{3}$ is within the interval $[0, \pi]$, it is the valid principal value.

Final Solution: \cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}

Step 1: Definition of the Principal Value Branch

To find the principal value of $y = \tan^{-1}(-1)$, we must refer to the range of the principal value branch of the inverse tangent function. The domain of the function $f(x) = \tan^{-1}(x)$ is $\mathbb{R}$, and its range is the open interval $(-\frac{\pi}{2}, \frac{\pi}{2})$.

Thus, we seek an angle $y$ such that:

$\tan(y) = -1$

where $y \in (-\frac{\pi}{2}, \frac{\pi}{2})$ [By Definition of Principal Value Branches].

Step 2: Evaluating the Trigonometric Identity

We know from the standard trigonometric table that $\tan(\frac{\pi}{4}) = 1$. Utilizing the property of odd functions for the tangent ratio, specifically $\tan(-\theta) = -\tan(\theta)$ [Per the property of odd functions in trigonometry], we can express the negative value as follows:

$\tan(-y) = -\tan(y)$

Substitute $\theta = \frac{\pi}{4}$ into the identity:

$\tan\left(-\frac{\pi}{4}\right) = -\tan\left(\frac{\pi}{4}\right)$

Since $\tan\left(\frac{\pi}{4}\right) = 1$, it follows that:

$\tan\left(-\frac{\pi}{4}\right) = -1$

Step 3: Verification of the Range Constraint

The value obtained is $y = -\frac{\pi}{4}$. We must verify if this value lies within the defined range $(-\frac{\pi}{2}, \frac{\pi}{2})$.

• Lower bound: $-\frac{\pi}{2} \approx -1.5707$ radians
• Upper bound: $\frac{\pi}{2} \approx 1.5707$ radians
• Calculated value: $-\frac{\pi}{4} \approx -0.7854$ radians

Since $-\frac{\pi}{2} < -\frac{\pi}{4} < \frac{\pi}{2}$, the value $-\frac{\pi}{4}$ is indeed the principal value [By the definition of the inverse tangent range].

Step 4: Geometric Representation of the Inverse Tangent

Final Solution: The principal value of $\tan^{-1}(-1)$ is $-\frac{\pi}{4}$.

Step 1: Evaluation of the Principal Value of $\tan^{-1}(1)$

Let $x = \tan^{-1}(1)$. By definition of the inverse tangent function, $\tan(x) = 1$. The principal value branch of $\tan^{-1}(x)$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$. [Per the definition of Inverse Trigonometric Functions].

Since $\tan(\frac{\pi}{4}) = 1$, and $\frac{\pi}{4} \in (-\frac{\pi}{2}, \frac{\pi}{2})$:

$\tan^{-1}(1) = \frac{\pi}{4}$

Step 2: Evaluation of the Principal Value of $\cos^{-1}(-\frac{1}{2})$

Let $y = \cos^{-1}(-\frac{1}{2})$. By definition, $\cos(y) = -\frac{1}{2}$. The principal value branch of $\cos^{-1}(x)$ is $[0, \pi]$.

We know that $\cos(\frac{\pi}{3}) = \frac{1}{2}$. Since the cosine function is negative in the second quadrant, we use the identity $\cos(\pi - \theta) = -\cos(\theta)$ [Per the Trigonometric Quadrant Rules]:

$\cos(\pi - \frac{\pi}{3}) = \cos(\frac{2\pi}{3}) = -\frac{1}{2}$

Since $\frac{2\pi}{3} \in [0, \pi]$:

$\cos^{-1}(-\frac{1}{2}) = \frac{2\pi}{3}$

Step 3: Evaluation of the Principal Value of $\sin^{-1}(-\frac{1}{2})$

Let $z = \sin^{-1}(-\frac{1}{2})$. By definition, $\sin(z) = -\frac{1}{2}$. The principal value branch of $\sin^{-1}(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

We know that $\sin(\frac{\pi}{6}) = \frac{1}{2}$. Using the property $\sin(-\theta) = -\sin(\theta)$ [Per the Odd Function property of Sine]:

$\sin(-\frac{\pi}{6}) = -\frac{1}{2}$

Since $-\frac{\pi}{6} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$:

$\sin^{-1}(-\frac{1}{2}) = -\frac{\pi}{6}$

Step 4: Summation of the Evaluated Values

Substitute the obtained values into the original expression:

Expression $= \frac{\pi}{4} + \frac{2\pi}{3} + (-\frac{\pi}{6})$

To perform the addition, find the Least Common Multiple (LCM) of the denominators $4, 3,$ and $6$, which is $12$:

$= \frac{3\pi}{12} + \frac{8\pi}{12} - \frac{2\pi}{12}$

$= \frac{3\pi + 8\pi - 2\pi}{12}$

$= \frac{9\pi}{12}$

Reducing the fraction by dividing the numerator and denominator by $3$:

$= \frac{3\pi}{4}$

Final Solution: \frac{3\pi}{4}

Step 1: Evaluation of $\cos^{-1}\left(\frac{1}{2}\right)$

To evaluate $\cos^{-1}\left(\frac{1}{2}\right)$, we seek an angle $\theta_1$ such that $\cos(\theta_1) = \frac{1}{2}$, where $\theta_1 \in [0, \pi]$ [Per the principal value branch of the cosine inverse function].

• Since $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$ and $\frac{\pi}{3}$ lies within the interval $[0, \pi]$, we have:
• $\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}$

Step 2: Evaluation of $\sin^{-1}\left(\frac{1}{2}\right)$

To evaluate $\sin^{-1}\left(\frac{1}{2}\right)$, we seek an angle $\theta_2$ such that $\sin(\theta_2) = \frac{1}{2}$, where $\theta_2 \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ [Per the principal value branch of the sine inverse function].

• Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ and $\frac{\pi}{6}$ lies within the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, we have:
• $\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$

Step 3: Substitution and Calculation

Substitute the derived principal values into the original expression:

$E = \cos^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right)$

$E = \frac{\pi}{3} + 2\left(\frac{\pi}{6}\right)$

$E = \frac{\pi}{3} + \frac{\pi}{3}$

$E = \frac{2\pi}{3}$

Visual Representation of Unit Circle Positions

Final Solution: \frac{2\pi}{3}