Step 1: Given Variables & Initial Setup

We are given the value of $\sec x$ and the location of the angle $x$:

• $\sec x = \frac{13}{5}$
• $x$ lies in the fourth quadrant (IV Quadrant).

[Per the definitions of trigonometric signs, in the IV Quadrant, only $\cos x$ and $\sec x$ are positive; $\sin x, \csc x, \tan x,$ and $\cot x$ are negative.]

Step 2: Determining $\cos x$

Using the reciprocal identity: $\cos x = \frac{1}{\sec x}$ [By Reciprocal Identity]

$\cos x = \frac{1}{13/5} = \frac{5}{13}$

Step 3: Calculating $\sin x$ using Pythagorean Identity

Using the identity $\sin^2 x + \cos^2 x = 1$ [By Pythagorean Identity]:

$\sin^2 x = 1 - \cos^2 x$

$\sin^2 x = 1 - \left(\frac{5}{13}\right)^2 = 1 - \frac{25}{169}$

$\sin^2 x = \frac{169 - 25}{169} = \frac{144}{169}$

$\sin x = \pm \sqrt{\frac{144}{169}} = \pm \frac{12}{13}$

Since $x$ is in the IV Quadrant, $\sin x$ must be negative. Thus, $\sin x = -\frac{12}{13}$.

Step 4: Calculating $\csc x$, $\tan x$, and $\cot x$

Calculating $\csc x$:

$\csc x = \frac{1}{\sin x} = \frac{1}{-12/13} = -\frac{13}{12}$

Calculating $\tan x$:

$\tan x = \frac{\sin x}{\cos x} = \frac{-12/13}{5/13} = -\frac{12}{5}$

Calculating $\cot x$:

$\cot x = \frac{1}{\tan x} = \frac{1}{-12/5} = -\frac{5}{12}$

Visual Representation

Summary Table of Functions

Final Solution: The remaining trigonometric functions are $\sin x = -\frac{12}{13}$, $\cos x = \frac{5}{13}$, $\tan x = -\frac{12}{5}$, $\csc x = -\frac{13}{12}$, and $\cot x = -\frac{5}{12}$.

Step 1: Application of Sum-to-Product Trigonometric Identities

To simplify the expression, we apply the standard sum-to-product transformation formulas. Specifically, we utilize the following identities:

• Numerator identity: $\cos C - \cos D = -2 \sin\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$ [Per trigonometric transformation laws]
• Denominator identity: $\sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$ [Per trigonometric transformation laws]

Step 2: Substituting variables for the Numerator

Given the numerator $\cos 9x - \cos 5x$:

• Let $C = 9x$ and $D = 5x$.
• $\cos 9x - \cos 5x = -2 \sin\left(\frac{9x+5x}{2}\right) \sin\left(\frac{9x-5x}{2}\right)$
• $= -2 \sin\left(\frac{14x}{2}\right) \sin\left(\frac{4x}{2}\right)$
• $= -2 \sin(7x) \sin(2x)$

Step 3: Substituting variables for the Denominator

Given the denominator $\sin 17x - \sin 3x$:

• Let $C = 17x$ and $D = 3x$.
• $\sin 17x - \sin 3x = 2 \cos\left(\frac{17x+3x}{2}\right) \sin\left(\frac{17x-3x}{2}\right)$
• $= 2 \cos\left(\frac{20x}{2}\right) \sin\left(\frac{14x}{2}\right)$
• $= 2 \cos(10x) \sin(7x)$

Step 4: Algebraic Reduction of the Fraction

Now, construct the full fraction using the results from Step 2 and Step 3:

$\text{LHS} = \frac{-2 \sin(7x) \sin(2x)}{2 \cos(10x) \sin(7x)}$

Observing the common term $\sin(7x)$ in both the numerator and denominator, we perform the cancellation (assuming $\sin(7x) \neq 0$):

$\text{LHS} = \frac{-2 \cdot \cancel{\sin(7x)} \cdot \sin(2x)}{2 \cdot \cos(10x) \cdot \cancel{\sin(7x)}}$

$\text{LHS} = -\frac{2 \sin(2x)}{2 \cos(10x)}$

Step 5: Final Simplification

Dividing the constant factor $2$ from the numerator and denominator yields:

$\text{LHS} = -\frac{\sin(2x)}{\cos(10x)}$

Visual Representation of the Harmonic Relationship

Final Solution: The left-hand side simplifies to $-\frac{\sin 2x}{\cos 10x}$, which is equal to the right-hand side. Hence, the identity is proven.

Step 1: Expansion of the Trigonometric Expression

Given the expression: $LHS = (\cos x + \cos y)^2 + (\sin x - \sin y)^2$

Expanding the squared binomials using the algebraic identity $(a \pm b)^2 = a^2 \pm 2ab + b^2$:

$LHS = (\cos^2 x + 2\cos x \cos y + \cos^2 y) + (\sin^2 x - 2\sin x \sin y + \sin^2 y)$ [Expansion of Binomials]

Step 2: Grouping and Applying Pythagorean Identity

Rearrange the terms to group the sine and cosine functions of the same angle together:

$LHS = (\cos^2 x + \sin^2 x) + (\cos^2 y + \sin^2 y) + 2\cos x \cos y - 2\sin x \sin y$

Applying the fundamental Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$ [Fundamental Trigonometric Identity]:

$LHS = 1 + 1 + 2(\cos x \cos y - \sin x \sin y)$

$LHS = 2 + 2(\cos x \cos y - \sin x \sin y)$

Step 3: Application of Addition Formula

Utilize the cosine addition formula: $\cos(A + B) = \cos A \cos B - \sin A \sin B$ [Compound Angle Formula]:

$LHS = 2 + 2\cos(x + y)$

Step 4: Algebraic Factorization and Half-Angle Identity

Factor out the constant $2$:

$LHS = 2[1 + \cos(x + y)]$

Utilize the double-angle identity for cosine: $\cos(2\theta) = 2\cos^2 \theta - 1$. Rearranging this gives $1 + \cos(2\theta) = 2\cos^2 \theta$.

Let $2\theta = x + y$, therefore $\theta = \frac{x + y}{2}$:

$LHS = 2 \cdot [2\cos^2(\frac{x + y}{2})]$ [Substitution via Double Angle Identity]

$LHS = 4\cos^2(\frac{x + y}{2})$

Visual Representation of the Identity

Final Solution: $(\cos x + \cos y)^2 + (\sin x - \sin y)^2 = 4 \cos^2 \frac{x+y}{2}$

Step 1: Expanding the Left-Hand Side (LHS)

We begin by expressing $\cos 4x$ as a function of $2x$ to utilize the double-angle identity: $\cos 2\theta = 1 - 2\sin^2 \theta$ [Per the Cosine Double Angle Formula].

$\cos 4x = \cos(2 \cdot 2x)$

Applying the identity where $\theta = 2x$:

$\cos 4x = 1 - 2\sin^2(2x)$

Step 2: Applying the Sine Double Angle Identity

We know that $\sin 2x = 2\sin x \cos x$ [Per the Sine Double Angle Identity]. Substituting this into our expression from Step 1:

$\cos 4x = 1 - 2(\sin 2x)^2$

$\cos 4x = 1 - 2(2\sin x \cos x)^2$

Step 3: Algebraic Expansion and Simplification

We now apply the power of a product rule $(ab)^n = a^n b^n$ to the squared term:

$\cos 4x = 1 - 2(2^2 \cdot \sin^2 x \cdot \cos^2 x)$

$\cos 4x = 1 - 2(4 \sin^2 x \cos^2 x)$

Performing the final scalar multiplication:

$\cos 4x = 1 - 8\sin^2 x \cos^2 x$

Step 4: Geometrical Visualization of the Identity

Summary of Logic

Final Solution: $\cos 4x = 1 - 8\sin^2 x \cos^2 x$ [Q.E.D.]

Step 1: Decomposition of the Angle

To find the exact value of $\tan 15^{\circ}$, we represent $15^{\circ}$ as the difference of two standard angles whose trigonometric values are known from the unit circle. Specifically, we use $45^{\circ}$ and $30^{\circ}$.

$15^{\circ} = 45^{\circ} - 30^{\circ}$

Step 2: Applying the Trigonometric Identity

We utilize the tangent subtraction formula [Per the Trigonometric Addition/Subtraction Identities]:

$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$

Substituting $A = 45^{\circ}$ and $B = 30^{\circ}$:

$\tan(45^{\circ} - 30^{\circ}) = \frac{\tan 45^{\circ} - \tan 30^{\circ}}{1 + \tan 45^{\circ} \tan 30^{\circ}}$

Step 3: Substitution of Known Values

Using the standard values from the trigonometric table:

• $\tan 45^{\circ} = 1$
• $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$

Substitute these values into the expression:

$\tan 15^{\circ} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + (1)(\frac{1}{\sqrt{3}})}$

Step 4: Algebraic Simplification

Simplify the numerator and the denominator by finding a common denominator of $\sqrt{3}$:

$\tan 15^{\circ} = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

Step 5: Rationalization of the Denominator

To eliminate the radical from the denominator, multiply the numerator and the denominator by the conjugate $(\sqrt{3} - 1)$ [By the Property of Conjugates]:

$\tan 15^{\circ} = \frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)}$

Applying the algebraic identities $(a-b)^2 = a^2 - 2ab + b^2$ and $(a+b)(a-b) = a^2 - b^2$:

$\tan 15^{\circ} = \frac{3 - 2\sqrt{3} + 1}{3 - 1}$

$\tan 15^{\circ} = \frac{4 - 2\sqrt{3}}{2}$

$\tan 15^{\circ} = 2 - \sqrt{3}$

Final Solution: \tan 15^{\circ} = 2 - \sqrt{3}

Step 1: Analyzing the Quadrant and Range of $x/2$

Given $\cos x = -\frac{1}{3}$ and $x$ lies in the third quadrant (Quadrant III), we define the interval for $x$ as: $\pi < x < \frac{3\pi}{2}$ [Per the definition of Cartesian quadrants].

To find the range of $\frac{x}{2}$, we divide the inequality by 2:

Since $\frac{x}{2}$ lies in the interval $(\frac{\pi}{2}, \frac{3\pi}{4})$, which corresponds to the second quadrant (Quadrant II), we establish the following signs for our target functions:

• $\sin \frac{x}{2} > 0$ (Positive)
• $\cos \frac{x}{2} < 0$ (Negative)
• $\tan \frac{x}{2} < 0$ (Negative)

Step 2: Utilizing Half-Angle Identities

We employ the standard half-angle identities derived from the double-angle formula $\cos 2\theta = 2\cos^2 \theta - 1 = 1 - 2\sin^2 \theta$ [By Trigonometric Identities]:

1. $\sin \frac{x}{2} = \sqrt{\frac{1 - \cos x}{2}}$

2. $\cos \frac{x}{2} = -\sqrt{\frac{1 + \cos x}{2}}$ (Negative due to Quadrant II)

Step 3: Calculating $\sin \frac{x}{2}$

Substitute $\cos x = -\frac{1}{3}$ into the identity:

Rationalizing the denominator: $\sin \frac{x}{2} = \frac{\sqrt{6}}{3}$

Step 4: Calculating $\cos \frac{x}{2}$

Substitute $\cos x = -\frac{1}{3}$ into the identity:

Rationalizing the denominator: $\cos \frac{x}{2} = -\frac{\sqrt{3}}{3}$

Step 5: Calculating $\tan \frac{x}{2}$

By the quotient identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$ [Per Trigonometric Ratio Definitions]:

Visual Representation of the Half-Angle Location

Final Solution: $\sin \frac{x}{2} = \frac{\sqrt{6}}{3}$, $\cos \frac{x}{2} = -\frac{\sqrt{3}}{3}$, $\tan \frac{x}{2} = -\sqrt{2}$

Step 1: Evaluation of Trigonometric Ratios

To prove the identity $\sin^2 \frac{\pi}{6} + \cos^2 \frac{\pi}{3} - \tan^2 \frac{\pi}{4} = -\frac{1}{2}$, we first determine the exact values of the individual trigonometric functions based on standard unit circle coordinates [Per the definitions of trigonometric ratios in the unit circle].

• $\sin\left(\frac{\pi}{6}\right) = \sin(30^\circ) = \frac{1}{2}$
• $\cos\left(\frac{\pi}{3}\right) = \cos(60^\circ) = \frac{1}{2}$
• $\tan\left(\frac{\pi}{4}\right) = \tan(45^\circ) = 1$

Step 2: Substitution into the Left-Hand Side (LHS)

Substitute these values into the given expression:

$LHS = \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 - (1)^2$

Step 3: Algebraic Simplification

Perform the exponentiation and arithmetic operations following the order of operations [By the Order of Operations/PEMDAS]:

$LHS = \frac{1}{4} + \frac{1}{4} - 1$

$LHS = \frac{1+1}{4} - 1$

$LHS = \frac{2}{4} - 1$

$LHS = \frac{1}{2} - 1$

$LHS = -\frac{1}{2}$

Step 4: Geometric Representation of Angles

The following SVG illustrates the unit circle positions for the angles $\frac{\pi}{6}$, $\frac{\pi}{4}$, and $\frac{\pi}{3}$ utilized in the evaluation:

Step 5: Final Conclusion

Since the simplified value of the LHS results in $-\frac{1}{2}$, which is equivalent to the Right-Hand Side (RHS), the identity is verified.

Final Solution: LHS = RHS = -\frac{1}{2}

Step 1: Theoretical Basis for Conversion

To convert a given measure from radians to degrees, we utilize the fundamental relationship between the two units of angular measurement: $1 \text{ radian} = \left( \frac{180}{\pi} \right)^\circ$ [Per the definition of circular measure]. Therefore, to convert an angle in radians ($x$) to degrees ($D$), we apply the transformation formula: $D = x \times \frac{180^\circ}{\pi}$.

Step 2: Substitution of Given Variables

We are given the radian measure: $x = \frac{5\pi}{3}$

Substituting this value into the conversion formula:

$D = \left( \frac{5\pi}{3} \right) \times \left( \frac{180^\circ}{\pi} \right)$

Step 3: Algebraic Simplification

Proceeding with the elimination of the constant $\pi$ from the numerator and denominator:

$D = \frac{5}{3} \times 180^\circ$ [By the property of multiplicative inverses: $\frac{\pi}{\pi} = 1$]

$D = 5 \times \left( \frac{180^\circ}{3} \right)$

$D = 5 \times 60^\circ$

$D = 300^\circ$

Step 4: Geometric Representation

Final Solution: The degree measure of $\frac{5\pi}{3}$ radians is $300^\circ$.

Step 1: Identifying Quadrant and Sign Conventions

Given $\tan x = -\frac{4}{3}$ and $x$ lies in the second quadrant ($\frac{\pi}{2} < x < \pi$).

In the second quadrant, $\cos x$ is negative and $\sin x$ is positive. Since $x$ is in the second quadrant, the half-angle $\frac{x}{2}$ lies in the first quadrant ($\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$). Therefore, all trigonometric functions ($\sin \frac{x}{2}$, $\cos \frac{x}{2}$, $\tan \frac{x}{2}$) must be positive.

Step 2: Calculating $\cos x$

Using the fundamental identity $\sec^2 x = 1 + \tan^2 x$ [Pythagorean Identity]:

$\sec^2 x = 1 + \left(-\frac{4}{3}\right)^2 = 1 + \frac{16}{9} = \frac{25}{9}$

$\cos^2 x = \frac{1}{\sec^2 x} = \frac{9}{25}$

Since $x$ is in the second quadrant, $\cos x$ is negative:

$\cos x = -\frac{3}{5}$

Step 3: Applying Half-Angle Formulas

Using the half-angle identities for $\sin \frac{x}{2}$ and $\cos \frac{x}{2}$ [Trigonometric Half-Angle Formulas]:

1. $\sin \frac{x}{2} = \sqrt{\frac{1 - \cos x}{2}}$

$\sin \frac{x}{2} = \sqrt{\frac{1 - (-\frac{3}{5})}{2}} = \sqrt{\frac{1 + \frac{3}{5}}{2}} = \sqrt{\frac{8/5}{2}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$

2. $\cos \frac{x}{2} = \sqrt{\frac{1 + \cos x}{2}}$

$\cos \frac{x}{2} = \sqrt{\frac{1 + (-\frac{3}{5})}{2}} = \sqrt{\frac{2/5}{2}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}$

3. $\tan \frac{x}{2} = \frac{\sin(x/2)}{\cos(x/2)}$

$\tan \frac{x}{2} = \frac{2/\sqrt{5}}{1/\sqrt{5}} = 2$

Step 4: Geometric Visualization

Final Solution:
$\sin \frac{x}{2} = \frac{2\sqrt{5}}{5}$
$\cos \frac{x}{2} = \frac{\sqrt{5}}{5}$
$\tan \frac{x}{2} = 2$

Step 1: Application of Trigonometric Identity

To prove the identity $\sin^2 6x - \sin^2 4x = \sin 2x \sin 10x$, we utilize the fundamental algebraic identity for the difference of two squares: $a^2 - b^2 = (a - b)(a + b)$. Let $a = \sin 6x$ and $b = \sin 4x$.

Applying the identity [Difference of Squares]:

$\sin^2 6x - \sin^2 4x = (\sin 6x - \sin 4x)(\sin 6x + \sin 4x)$

Step 2: Utilization of Sum-to-Product Formulas

We employ the standard trigonometric Sum-to-Product formulas [Per the Sum-to-Product Identities]:

• $\sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$
• $\sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right)$

For the first term $(\sin 6x - \sin 4x)$, let $C=6x$ and $D=4x$:

$\sin 6x - \sin 4x = 2 \cos\left(\frac{6x+4x}{2}\right) \sin\left(\frac{6x-4x}{2}\right) = 2 \cos 5x \sin x$

For the second term $(\sin 6x + \sin 4x)$, let $C=6x$ and $D=4x$:

$\sin 6x + \sin 4x = 2 \sin\left(\frac{6x+4x}{2}\right) \cos\left(\frac{6x-4x}{2}\right) = 2 \sin 5x \cos x$

Step 3: Algebraic Rearrangement

Substitute these results back into the expression from Step 1:

LHS $= (2 \cos 5x \sin x) \cdot (2 \sin 5x \cos x)$

Rearranging the terms to group sine and cosine components [Per the Commutative Property of Multiplication]:

LHS $= (2 \sin 5x \cos 5x) \cdot (2 \sin x \cos x)$

Step 4: Application of Double Angle Identities

Recall the double angle formula for sine [Per the Double Angle Identity $\sin 2\theta = 2 \sin \theta \cos \theta$]:

• $2 \sin 5x \cos 5x = \sin(2 \cdot 5x) = \sin 10x$
• $2 \sin x \cos x = \sin(2 \cdot x) = \sin 2x$

Substituting these into our equation:

LHS $= \sin 10x \cdot \sin 2x$

Step 5: Geometric Visualization of Frequency

Final Solution: $\sin^2 6x - \sin^2 4x = \sin 10x \sin 2x = \text{RHS}$ (Hence Proved).

Step 1: Theoretical Foundation for Unit Conversion

To convert an angle from radian measure to degree measure, we utilize the fundamental relationship between the two units of angular measurement: $\pi \text{ radians} = 180^\circ$ [Per the definition of circular measure].

The conversion formula is established as:

$\text{Degree measure} = \text{Radian measure} \times \left( \frac{180^\circ}{\pi} \right)$

Step 2: Substitution of Given Variables

Given the radian measure $\theta = \frac{7\pi}{6}$. Substituting this into the conversion formula:

$\text{Degree measure} = \left( \frac{7\pi}{6} \right) \times \left( \frac{180^\circ}{\pi} \right)$

Step 3: Algebraic Simplification

Perform the cancellation of the variable $\pi$ from the numerator and denominator:

$\text{Degree measure} = \left( \frac{7}{6} \right) \times 180^\circ$

[By the Multiplicative Property of Equality]

$\text{Degree measure} = 7 \times \left( \frac{180^\circ}{6} \right)$

$\text{Degree measure} = 7 \times 30^\circ$

$\text{Degree measure} = 210^\circ$

Visual Representation of the Angle

Final Verification

Since $\pi$ radians is equivalent to $180^\circ$, then $\frac{\pi}{6}$ radians is equivalent to $\frac{180^\circ}{6} = 30^\circ$. Multiplying this by the scalar coefficient of $7$ yields $7 \times 30^\circ = 210^\circ$. The constraint to use $\pi = \frac{22}{7}$ is irrelevant in this specific instance because the variable $\pi$ cancels out algebraically before numerical evaluation is required.

Final Solution: $210^\circ$

Step 1: Analyzing the Input Expression

The expression to evaluate is $\cot\left(-\frac{15\pi}{4}\right)$. We utilize the parity property of the cotangent function: $\cot(-\theta) = -\cot(\theta)$ [By the Odd/Even identity for trigonometric functions].

Therefore, $\cot\left(-\frac{15\pi}{4}\right) = -\cot\left(\frac{15\pi}{4}\right)$.

Step 2: Decomposition of the Angle

To evaluate $\cot\left(\frac{15\pi}{4}\right)$, we decompose the angle into a multiple of $\pi$ plus a principal angle. We seek the nearest multiple of $2\pi$ or $\pi$:

$\frac{15\pi}{4} = \frac{16\pi - \pi}{4} = 4\pi - \frac{\pi}{4}$

Applying the periodicity property of the cotangent function, where $\cot(n\pi - \theta) = -\cot(\theta)$ for any integer $n$ [By the periodicity and reduction formulas for trigonometric functions]:

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

Step 3: Calculating the Final Value

Substitute the reduction back into the expression from Step 1:

$-\cot\left(\frac{15\pi}{4}\right) = -(-\cot\left(\frac{\pi}{4}\right))$

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

Based on standard trigonometric ratios for specific angles, $\cot\left(\frac{\pi}{4}\right) = 1$ [By the unit circle definition where $\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$ and $\sin(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$].

Step 4: Geometric Visualization

Final Solution: 1

Given Variables & Initial Setup

To determine the angular displacement in radians per second, we define the following variables based on the problem statement:

• Total revolutions per minute ($R_{min}$) = $360$
• Time interval ($t$) = $1$ minute = $60$ seconds
• Full rotation in radians ($\theta_{rev}$) = $2\pi$ radians [Per the definition of circular measure of an angle]

Step 1: Calculating Revolutions per Second

Since the wheel completes $360$ revolutions in $60$ seconds, we calculate the frequency of rotation per second ($R_{sec}$):

$R_{sec} = \frac{R_{min}}{60}$

$R_{sec} = \frac{360}{60} = 6$ revolutions/second

Step 2: Conversion to Radians

Given that one complete revolution is equivalent to $2\pi$ radians, the total angle turned in one second ($\theta_{sec}$) is the product of the number of revolutions per second and the radians per revolution:

$\theta_{sec} = R_{sec} \times 2\pi$

$\theta_{sec} = 6 \times 2\pi$ [Substituting values from Step 1]

$\theta_{sec} = 12\pi$ radians

Visual Representation of a Single Revolution

Analytical Summary

Final Solution: The wheel turns through $12\pi$ radians in one second.

Step 1: Determining the value of $\cos x$

Given $\sin x = \frac{1}{4}$ and $x$ lies in the second quadrant (Quadrant II). In Quadrant II, $\cos x$ is negative [Per the ASTC rule].

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$:

$\cos^2 x = 1 - \sin^2 x = 1 - \left(\frac{1}{4}\right)^2 = 1 - \frac{1}{16} = \frac{15}{16}$

Since $x \in$ Quadrant II, $\cos x = -\sqrt{\frac{15}{16}} = -\frac{\sqrt{15}}{4}$ [By the square root property of trigonometric identities].

Step 2: Determining the quadrant of $\frac{x}{2}$

Since $\frac{\pi}{2} < x < \pi$, dividing the inequality by 2 gives:

$\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$

Therefore, $\frac{x}{2}$ lies in Quadrant I. In Quadrant I, all trigonometric functions ($\sin, \cos, \tan$) must be positive [By the definition of the unit circle in the first quadrant].

Step 3: Calculating $\sin \frac{x}{2}$, $\cos \frac{x}{2}$, and $\tan \frac{x}{2}$

We utilize the half-angle formulas derived from the double-angle identity $\cos 2\theta = 1 - 2\sin^2 \theta$ and $\cos 2\theta = 2\cos^2 \theta - 1$.

For $\sin \frac{x}{2}$:

$\sin \frac{x}{2} = \sqrt{\frac{1 - \cos x}{2}} = \sqrt{\frac{1 - (-\frac{\sqrt{15}}{4})}{2}} = \sqrt{\frac{\frac{4 + \sqrt{15}}{4}}{2}} = \sqrt{\frac{4 + \sqrt{15}}{8}}$

Multiply numerator and denominator by 2 to rationalize: $\sqrt{\frac{8 + 2\sqrt{15}}{16}} = \frac{\sqrt{8 + 2\sqrt{15}}}{4} = \frac{\sqrt{(\sqrt{5} + \sqrt{3})^2}}{4} = \frac{\sqrt{5} + \sqrt{3}}{4}$

For $\cos \frac{x}{2}$:

$\cos \frac{x}{2} = \sqrt{\frac{1 + \cos x}{2}} = \sqrt{\frac{1 + (-\frac{\sqrt{15}}{4})}{2}} = \sqrt{\frac{\frac{4 - \sqrt{15}}{4}}{2}} = \sqrt{\frac{4 - \sqrt{15}}{8}}$

Multiply numerator and denominator by 2: $\sqrt{\frac{8 - 2\sqrt{15}}{16}} = \frac{\sqrt{8 - 2\sqrt{15}}}{4} = \frac{\sqrt{(\sqrt{5} - \sqrt{3})^2}}{4} = \frac{\sqrt{5} - \sqrt{3}}{4}$

For $\tan \frac{x}{2}$:

$\tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}$

Rationalizing the denominator: $\frac{(\sqrt{5} + \sqrt{3})(\sqrt{5} + \sqrt{3})}{(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})} = \frac{5 + 3 + 2\sqrt{15}}{5 - 3} = \frac{8 + 2\sqrt{15}}{2} = 4 + \sqrt{15}$

Visual Representation

Final Solution: $\sin \frac{x}{2} = \frac{\sqrt{5} + \sqrt{3}}{4}$, $\cos \frac{x}{2} = \frac{\sqrt{5} - \sqrt{3}}{4}$, $\tan \frac{x}{2} = 4 + \sqrt{15}$

Step 1: Identifying the Trigonometric Relationship

To prove the identity $\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1$, we begin by utilizing the addition property of the arguments. Observe that $3x = 2x + x$. Applying the cotangent function to both sides, we use the identity for the cotangent of a sum:

$\cot(A + B) = \frac{\cot A \cot B - 1}{\cot A + \cot B}$ [Per the trigonometric identity for $\cot(A+B)$]

Let $A = 2x$ and $B = x$. Thus:

$\cot(2x + x) = \frac{\cot 2x \cot x - 1}{\cot 2x + \cot x}$

$\cot 3x = \frac{\cot 2x \cot x - 1}{\cot 2x + \cot x}$

Step 2: Algebraic Manipulation

Rearrange the equation by cross-multiplying the denominator to isolate the product terms:

$\cot 3x (\cot 2x + \cot x) = \cot 2x \cot x - 1$

Expanding the left side of the equation:

$\cot 3x \cot 2x + \cot 3x \cot x = \cot 2x \cot x - 1$

Step 3: Transposition to Reach the Required Identity

We rearrange the terms to match the form specified in the question ($\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1$).

Subtracting $\cot 3x \cot 2x$ and $\cot 3x \cot x$ from both sides, and adding $1$ to both sides, we obtain:

$1 = \cot 2x \cot x - \cot 3x \cot 2x - \cot 3x \cot x$

Step 4: Verification of Visual Representation

Summary of Logical Steps

Final Solution: The identity $\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1$ is proven true by the expansion of the cotangent addition formula $\cot(2x+x)$.

Step 1: Analyzing the Input Expression

The expression to evaluate is $\sin \left( -\frac{11\pi}{3} \right)$.

We apply the property of odd functions for the sine trigonometric ratio [Per the identity $\sin(-\theta) = -\sin(\theta)$]:

$\sin \left( -\frac{11\pi}{3} \right) = -\sin \left( \frac{11\pi}{3} \right)$

Step 2: Decomposition of the Angle

To evaluate $\sin \left( \frac{11\pi}{3} \right)$, we express the angle in terms of multiples of $2\pi$ to utilize the periodicity of the sine function [Per the identity $\sin(2n\pi + \theta) = \sin(\theta)$, where $n \in \mathbb{Z}$].

Divide $11$ by $3$: $11 = 3 \times 4 - 1$. Thus, $\frac{11\pi}{3} = \frac{12\pi - \pi}{3} = 4\pi - \frac{\pi}{3}$.

Substituting this back into the expression:

$-\sin \left( 4\pi - \frac{\pi}{3} \right)$

Step 3: Applying Quadrant Reduction

The angle $4\pi - \frac{\pi}{3}$ lies in the Fourth Quadrant. [Per the CAST Rule/Quadrant Analysis: In the Fourth Quadrant, sine is negative].

Using the reduction formula $\sin(2n\pi - \theta) = -\sin(\theta)$:

$-\sin \left( 4\pi - \frac{\pi}{3} \right) = - \left( -\sin \frac{\pi}{3} \right) = \sin \left( \frac{\pi}{3} \right)$

Step 4: Trigonometric Evaluation

Evaluating the standard trigonometric ratio at $\frac{\pi}{3}$ radians ($60^\circ$):

$\sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2}$

Visual Representation of Unit Circle Rotation

Final Solution: $\frac{\sqrt{3}}{2}$

Step 1: Identifying the Trigonometric Transformation

To solve the expression $\cos\left(\frac{3\pi}{4} + x\right) - \cos\left(\frac{3\pi}{4} - x\right)$, we utilize the sum-to-product identity for the difference of two cosines. The standard identity is defined as follows:

$\cos C - \cos D = -2 \sin\left(\frac{C + D}{2}\right) \sin\left(\frac{C - D}{2}\right)$

[By Sum-to-Product Identity for Cosines]

Step 2: Assigning Variables

Let $C = \left(\frac{3\pi}{4} + x\right)$ and $D = \left(\frac{3\pi}{4} - x\right)$.

We calculate the sum and difference of the arguments:

• Sum: $C + D = \left(\frac{3\pi}{4} + x\right) + \left(\frac{3\pi}{4} - x\right) = \frac{6\pi}{4} = \frac{3\pi}{2}$
• Difference: $C - D = \left(\frac{3\pi}{4} + x\right) - \left(\frac{3\pi}{4} - x\right) = 2x$

Step 3: Substituting into the Identity

Substitute the calculated values into the formula:

$\cos\left(\frac{3\pi}{4} + x\right) - \cos\left(\frac{3\pi}{4} - x\right) = -2 \sin\left(\frac{3\pi/2}{2}\right) \sin\left(\frac{2x}{2}\right)$

$\cos\left(\frac{3\pi}{4} + x\right) - \cos\left(\frac{3\pi}{4} - x\right) = -2 \sin\left(\frac{3\pi}{4}\right) \sin(x)$

Step 4: Evaluating the Constant

Evaluate $\sin\left(\frac{3\pi}{4}\right)$ using the reference angle $\pi - \frac{\pi}{4}$:

$\sin\left(\frac{3\pi}{4}\right) = \sin\left(\pi - \frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$

[Per Trigonometric Ratio properties in the Second Quadrant]

Step 5: Final Simplification

Substitute the value back into the expression:

$-2 \left(\frac{1}{\sqrt{2}}\right) \sin(x) = -\left(\frac{2}{\sqrt{2}}\right) \sin(x)$

Since $2 = \sqrt{2} \times \sqrt{2}$, we have:

$-\left(\frac{\sqrt{2} \cdot \sqrt{2}}{\sqrt{2}}\right) \sin(x) = -\sqrt{2} \sin x$

Geometric Verification (Unit Circle Concept)

Final Solution: $\cos\left(\frac{3\pi}{4} + x\right) - \cos\left(\frac{3\pi}{4} - x\right) = -\sqrt{2} \sin x$ (Proved)

Step 1: Decomposition of the Argument

To prove the identity for $\tan 4x$, we utilize the multiple angle formula for the tangent function. We begin by expressing $4x$ as $2(2x)$. [By definition of multiplication].

The standard double-angle identity for tangent is:

$\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta}$ [Standard Trigonometric Identity]

Substituting $\theta = 2x$, we obtain:

$\tan 4x = \tan(2 \cdot 2x) = \frac{2\tan 2x}{1 - \tan^2 2x}$

Step 2: Expansion of the Inner Term

We now substitute the expression for $\tan 2x$ into our equation. Again, using $\tan 2x = \frac{2\tan x}{1 - \tan^2 x}$:

$\tan 4x = \frac{2 \left( \frac{2\tan x}{1 - \tan^2 x} \right)}{1 - \left( \frac{2\tan x}{1 - \tan^2 x} \right)^2}$

Step 3: Algebraic Simplification

Focusing on the numerator and the denominator separately to simplify the complex fraction:

Numerator: $2 \cdot \frac{2\tan x}{1 - \tan^2 x} = \frac{4\tan x}{1 - \tan^2 x}$

Denominator: $1 - \frac{(2\tan x)^2}{(1 - \tan^2 x)^2} = 1 - \frac{4\tan^2 x}{(1 - \tan^2 x)^2}$

Finding a common denominator for the total expression:

Denominator $= \frac{(1 - \tan^2 x)^2 - 4\tan^2 x}{(1 - \tan^2 x)^2}$

Step 4: Expansion and Final Assembly

Expanding the binomial $(1 - \tan^2 x)^2 = 1 - 2\tan^2 x + \tan^4 x$ [By Binomial Theorem $(a-b)^2 = a^2 - 2ab + b^2$]:

Denominator $= \frac{1 - 2\tan^2 x + \tan^4 x - 4\tan^2 x}{(1 - \tan^2 x)^2} = \frac{1 - 6\tan^2 x + \tan^4 x}{(1 - \tan^2 x)^2}$

Combining the numerator and denominator:

$\tan 4x = \frac{4\tan x}{1 - \tan^2 x} \cdot \frac{(1 - \tan^2 x)^2}{1 - 6\tan^2 x + \tan^4 x}$

By canceling one factor of $(1 - \tan^2 x)$ from the numerator and denominator:

$\tan 4x = \frac{4\tan x (1 - \tan^2 x)}{1 - 6\tan^2 x + \tan^4 x}$

Geometric Representation (Unit Circle Basis)

Final Solution: $\tan 4x = \frac{4\tan x (1 - \tan^2 x)}{1 - 6\tan^2 x + \tan^4 x}$ (Hence Proved)

Step 1: Given Variables & Initial Setup

To determine the angle subtended at the center of the circle, we identify the following parameters provided in the problem statement:

• Radius of the circle, $r = 100 \text{ cm}$
• Length of the arc, $l = 22 \text{ cm}$
• Constant value, $\pi = \frac{22}{7}$

Step 2: Mathematical Relationship (Radian Measure)

The relationship between the angle $\theta$ (in radians), the arc length $l$, and the radius $r$ of a circle is given by the formula:

$\theta = \frac{l}{r} \text{ radians}$ [Per the definition of radian measure]

Substituting the given values:

$\theta = \frac{22}{100} \text{ radians} = \frac{11}{50} \text{ radians}$

Step 3: Conversion to Degree Measure

To convert the angle from radians to degrees, we utilize the conversion factor: $1 \text{ radian} = \left( \frac{180}{\pi} \right)^\circ$.

$\text{Angle in degrees} = \theta \times \left( \frac{180}{\pi} \right)^\circ$

Substituting $\theta = \frac{11}{50}$ and $\pi = \frac{22}{7}$:

$\text{Angle} = \left( \frac{11}{50} \right) \times \left( \frac{180}{22/7} \right)^\circ$

$\text{Angle} = \left( \frac{11}{50} \right) \times \left( \frac{180 \times 7}{22} \right)^\circ$

Step 4: Algebraic Simplification

Simplifying the expression through arithmetic reduction:

$\text{Angle} = \frac{11}{50} \times \frac{1260}{22}$

$\text{Angle} = \frac{1}{50} \times \frac{1260}{2} = \frac{1260}{100} = 12.6^\circ$

Step 5: Representing in Degrees and Minutes

To express $0.6^\circ$ in minutes: $1^\circ = 60'$, therefore $0.6^\circ = 0.6 \times 60' = 36'$.

Final Solution: The angle subtended at the centre is $12.6^\circ$ or $12^\circ 36'$.

Step 1: Analyzing the Argument and Periodicity

The given angle is $\theta = \frac{19\pi}{3}$. To evaluate the trigonometric function $\tan(\theta)$, we utilize the periodic property of the tangent function, which is given by $\tan(\theta + n\pi) = \tan(\theta)$ for any integer $n$ [Per the Periodicity Property of Tangent].

Step 2: Reducing the Angle to the Principal Domain

We express the angle in terms of multiples of $2\pi$ to simplify the calculation:

$\frac{19\pi}{3} = \frac{18\pi + \pi}{3} = \frac{18\pi}{3} + \frac{\pi}{3} = 6\pi + \frac{\pi}{3}$

Here, $6\pi$ represents $3$ full revolutions around the unit circle [By the concept of Coterminal Angles].

Step 3: Applying Reduction Formulae

Since $\tan(6\pi + \theta) = \tan(\theta)$, we substitute $\theta = \frac{\pi}{3}$ into the expression:

$\tan\left(\frac{19\pi}{3}\right) = \tan\left(6\pi + \frac{\pi}{3}\right)$

$\tan\left(\frac{19\pi}{3}\right) = \tan\left(\frac{\pi}{3}\right)$

Step 4: Evaluating the Trigonometric Value

From the standard values of trigonometric ratios in the first quadrant, we know that $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$ [Per the Standard Trigonometric Table for Special Angles].

Step 5: Summary Table of Transformation

Final Solution: \tan\left(\frac{19\pi}{3}\right) = \sqrt{3}