Problem 44 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-50: Verify the identity. $$ \frac{\csc (-t)-\sin (-t)}{\sin (-t)}=\cot ^{2} t $$

Step-by-Step Solution

Answer

The identity \( \frac{\csc (-t)-\sin (-t)}{\sin (-t)}=\cot ^{2} t \) is verified.

Step 1: Simplify the Left-Hand Side

Start by expressing \( \csc(-t) \) and \( \sin(-t) \) in terms of \( t \). We know that \( \csc(-t) = -\csc(t) \) and \( \sin(-t) = -\sin(t) \). So, substitute these into the expression:\[\frac{\csc(-t) - \sin(-t)}{\sin(-t)} = \frac{-\csc(t) + \sin(t)}{-\sin(t)}\]

Step 2: Simplify Further

Now we simplify the expression:\[\frac{-\csc(t) + \sin(t)}{-\sin(t)} = \frac{\csc(t) - \sin(t)}{\sin(t)}\]This is because multiplying both the numerator and the denominator by -1 does not change the expression's value.

Step 3: Express in Terms of Fundamental Identities

Recall that \( \csc(t) = \frac{1}{\sin(t)} \). Substitute \( \csc(t) \) into the expression:\[\frac{\frac{1}{\sin(t)} - \sin(t)}{\sin(t)} = \frac{\frac{1 - \sin^2(t)}{\sin(t)}}{\sin(t)}\]

Step 4: Use Pythagorean Identity

Use the Pythagorean identity \( 1 - \sin^2(t) = \cos^2(t) \):\[\frac{\cos^2(t)}{\sin(t) \cdot \sin(t)} = \frac{\cos^2(t)}{\sin^2(t)}\]

Step 5: Identify as \( \cot^2(t) \)

Now, recognize that \( \frac{\cos^2(t)}{\sin^2(t)} \) is exactly \( \cot^2(t) \):\[\frac{\cos^2(t)}{\sin^2(t)} = \cot^2(t)\]

Step 6: Conclusion: Identity Verified

Both sides of the equation are equal, so the identity is verified. The left-hand side simplifies to \( \cot^2(t) \), which matches the right-hand side of the original equation.

Key Concepts

Cosecant Function

The cosecant function, often abbreviated as \( \csc \), is the reciprocal of the sine function. It is defined as \( \csc(t) = \frac{1}{\sin(t)} \). This means wherever sine is zero, cosecant is undefined, since division by zero is not possible. Cosecant is a less commonly used trigonometric function compared to sine or cosine, but it plays a crucial role in various trigonometric identities and equations.
Understanding the cosecant function involves knowing its behavior and graph. Just like the sine function, it is periodic with a period of \( 2\pi \). However, unlike the smooth oscillation of the sine graph, the cosecant graph consists of repeating vertical asymptotes, where the sine function crosses zero. Knowing these relationships is essential for understanding how trigonometric identities involve multiple functions.
Whenever you encounter cosecant in expressions, think of it as a simpler form that relates directly back to sine, since any changes or manipulations can often be achieved more easily by temporary conversions to sine.

Sine Function

The sine function, symbolized as \( \sin(t) \), is a fundamental building block in trigonometry. It originates from a right-angled triangle, where it represents the ratio of the length of the side opposite the angle to the hypotenuse. In the unit circle, \( \sin(t) \) corresponds to the y-coordinate of a point on the circle, given an angle \( t \).

• The sine function is periodic with a period of \( 2\pi \), repeating its values in a predictable wave-like form.
• It has a range typically between -1 and 1, fluctuating symmetrically above and below the x-axis.

These properties make sine crucial for solving various trigonometric equations and for verifying identities, such as the one provided in the original exercise. By exploiting its periodic and reciprocal nature, transformations involving sine and cosecant are possible, as shown in the simplification steps. Recognizing these relationships reduces the complexity of broader trigonometric expressions.

Cotangent Function

The cotangent function, denoted as \( \cot(t) \), is another important trigonometric function. It is the reciprocal of the tangent function and is expressed as \( \cot(t) = \frac{1}{\tan(t)} \), which further breaks down into \( \cot(t) = \frac{\cos(t)}{\sin(t)} \). This relationship is particularly useful in expressions requiring the conversion between sine and cosine components.
In the original exercise, by manipulating the given expression, we reach \( \cot^2(t) = \frac{\cos^2(t)}{\sin^2(t)} \). The understanding of \( \cot(t) \) is vital when dealing with expressions that primarily contain sine and cosine components. These transformations help students visualize the completeness of trigonometric functions and identities.
Remembering the cotangent is especially useful when resolving problems involving reciprocal identities or working with complementary angles. It’s a pivotal part of trigonometry that supports deriving simplified forms of many complex expressions.

Pythagorean Identity

One of the cornerstone identities in trigonometry is the Pythagorean identity. It connects sine and cosine in a fundamental way, expressed as \( \sin^2(t) + \cos^2(t) = 1 \). This identity stems directly from the Pythagorean theorem applied to the unit circle. It implies the intrinsic relationship between \( \sin(t) \) and \( \cos(t) \).
This identity can be manipulated to solve various trigonometric problems; for instance, rearranging it to express \( \cos^2(t) \) in terms of \( \sin^2(t) \):

• \( \cos^2(t) = 1 - \sin^2(t) \)

Such transformations are often used in simplifying expressions, as seen in the step-by-step solution provided. By using the identity, \( 1 - \sin^2(t) = \cos^2(t) \), we simplified the expression down to \( \cot^2(t) \). Recognizing and applying Pythagorean identities is crucial for verifying and simplifying trigonometric equations, making them appear more intuitive and approachable for students.