Problem 44 - Precalculus: Functions and Graphs

Question

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 Trigonometric Functions

First, we note the identity for the cosecant and sine of negative angles: \( \csc(-t) = \frac{1}{\sin(-t)} \) and \( \sin(-t) = -\sin(t) \). Thus, the given expression becomes \( \frac{\frac{1}{-\sin(t)} + \sin(t)}{-\sin(t)} \). Simplifying this gives us \( \frac{-\frac{1}{\sin(t)} - \sin(t)}{-\sin(t)} \).

Step 2: Simplify the Fraction

Let's simplify the expression \( \frac{-\frac{1}{\sin(t)} - \sin(t)}{-\sin(t)} \). Combine the terms in the numerator to obtain a single fraction: \( \frac{-1 - \sin^2(t)}{-\sin(t)} \).

Step 3: Simplify Negative Signs

Factor out the negative sign from the numerator: \( \frac{-1 - \sin^2(t)}{-\sin(t)} = \frac{-(1 + \sin^2(t))}{-\sin(t)} \). Cancelling the negative signs, we get \( \frac{1 + \sin^2(t)}{\sin(t)} \).

Step 4: Use Pythagorean Identity

Using the Pythagorean identity, \( \sin^2(t) + \cos^2(t) = 1 \), we can substitute for 1 to get \( \cos^2(t) + \sin^2(t) = 1 + \sin^2(t) \). Substituting into our fraction: \( \frac{\cos^2(t)}{\sin(t)} \).

Step 5: Express Result in Terms of Cotangent

Note that \( \cot(t) = \frac{\cos(t)}{\sin(t)} \). Thus, \( \cot^2(t) = \left(\frac{\cos(t)}{\sin(t)}\right)^2 = \frac{\cos^2(t)}{\sin^2(t)} \). Thus, \( \frac{\cos^2(t)}{\sin(t)} = \frac{\cos^2(t)\cdot \sin(t)}{\sin^2(t)} = \cot^2(t) \), verifying the identity.

Key Concepts

Cosecant Function

The Cosecant function, denoted as \( \csc(t) \), is a trigonometric function that is the reciprocal of the sine function. In simple terms, this means it is the inverse of sine. Therefore, the relationship can be expressed as:

• \( \csc(t) = \frac{1}{\sin(t)} \)

It is important to note that, unlike sine, the cosecant function is undefined at points where the sine function equals zero. These are the points where you cannot divide by zero. For negative angles, the cosecant function has the identity:

• \( \csc(-t) = \frac{1}{\sin(-t)} \)

This identity shows that the cosecant of a negative angle is the same as the negative of the cosecant of the original angle. This property helps simplify expressions involving trigonometric identities, just like in the problem we are solving here.

Sine Function

The Sine function, represented as \( \sin(t) \), is one of the fundamental trigonometric functions that relates an angle of a right-angled triangle to the ratio of the length of the side opposite the angle to the hypotenuse. For any angle \( t \) in a unit circle, sine can also be interpreted as the y-coordinate of a point on the circle. For negative angles, the sine function holds the identity:

• \( \sin(-t) = -\sin(t) \)

This property states that sine is an odd function, meaning it flips sign when the angle becomes negative. In our exercise this identity helps in rewriting the terms involving \( \sin(-t) \) to simplify the expressions. Understanding these core properties of the sine function assists in solving various trigonometric equations and deepens comprehension of trigonometric behaviors.

Pythagorean Identity

The Pythagorean Identity is a vital tool in trigonometry, based on the Pythagorean Theorem. It states:

• \( \sin^2(t) + \cos^2(t) = 1 \)

This identity connects sine and cosine functions, emphasizing the relation that on any point on the unit circle, the sum of the squares of the sine and cosine of an angle equals one. Practically, this identity is used anytime an equation involving sine squared or cosine squared needs simplification. In the exercise, we used this identity to substitute \( 1 \) with \( \cos^2(t) + \sin^2(t) \) when simplifying fractions, which streamlined verifying the given trigonometric identity. Thoroughly grasping the Pythagorean Identity and its implications enables efficient manipulation and transformation in trigonometric equations.