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Chapter 20: Problem 30

Prove the given identities. $$\sin y+\sin y \cot ^{2} y=\csc y$$

Short Answer

Expert verified
The identity \( \sin y + \sin y \cot^2 y = \csc y \) is proven by simplifying the left side to the right side.

Step by step solution

01

Simplify the Left Side

Write the left side of the equation as a single function in terms of sine and cosine: \[ \sin y + \sin y \cot^2 y = \sin y (1 + \cot^2 y) \]Recall that \( \cot y = \frac{\cos y}{\sin y} \), so \( \cot^2 y = \frac{\cos^2 y}{\sin^2 y} \). Substitute this into the expression:\[ \sin y \left( 1 + \frac{\cos^2 y}{\sin^2 y} \right) = \sin y \left( \frac{\sin^2 y + \cos^2 y}{\sin^2 y} \right) \]
02

Use Pythagorean Identity

Use the Pythagorean identity, \( \sin^2 y + \cos^2 y = 1 \), to simplify the expression:\[ \sin y \left( \frac{1}{\sin^2 y} \right) = \frac{\sin y}{\sin^2 y} \]
03

Simplify the Fraction

Continue simplifying the fraction:\[ \frac{\sin y}{\sin^2 y} = \frac{1}{\sin y} \]
04

Recognize the Cosecant Function

Notice that \( \frac{1}{\sin y} \) is the definition of the cosecant function, \( \csc y \). Therefore, the left side simplifies to:\[ \csc y \]
05

Compare Both Sides of the Equation

The left side, which was simplified to \( \csc y \), now matches the given right side of the equation. Therefore, the identity is proven.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pythagorean Identity
In trigonometry, the Pythagorean identity is a crucial concept. It states that for any angle \( y \), the sum of the square of the sine and cosine of that angle is always equal to one. Mathematically, it is expressed as:

\[ \sin^2 y + \cos^2 y = 1 \]

This identity is derived from the Pythagorean theorem applied within the unit circle - a circle with a radius of 1 centered at the origin of the coordinate system. When working with trigonometric identities, knowing this base identity helps in simplifying many expressions.
  • It relates directly to sine and cosine functions.
  • Useful for transforming and simplifying complex trigonometric expressions.
  • Can be rearranged to find one value if the other is known. For example, \( \sin^2 y = 1 - \cos^2 y \).
Understanding and using this identity correctly allows us to break down and transform expressions involving trigonometric functions seamlessly.
Cosecant Function
The cosecant function is one of the six fundamental trigonometric functions. It is denoted as \( \csc y \). The cosecant of an angle \( y \) is defined as the reciprocal of the sine function. Therefore,

\[ \csc y = \frac{1}{\sin y} \]

This relationship tells us that when the sine of the angle is close to zero, the cosecant value will become very large, potentially leading to undefined behavior if \( \sin y \) equals zero itself.
  • It is primarily used in situations where dealing with reciprocals is more convenient.
  • Commonly used in proving trigonometric identities or solving equations.
  • In the unit circle framework, it represents the hypotenuse over the opposite side.
Remember, knowing these reciprocal relationships allows us to interchange functions in equations and often makes them easier to solve.
Cotangent
The cotangent function is another trigonometric function that complements the tangent function. Represented as \( \cot y \), it is defined as the reciprocal of the tangent function, or as:

\[ \cot y = \frac{\cos y}{\sin y} \]

This means that cotangent is the ratio of the adjacent side to the opposite side in a right triangle. Let's elucidate some key points about cotangent:
  • It becomes undefined when \( \sin y = 0 \), meaning at specific angle values such as \( y = 0, \pi, 2\pi, \ldots \)
  • Commonly appears in identities and formulas, such as when deriving others using Pythagorean relationships.
  • Using the identity \( \cot^2 y = \frac{\cos^2 y}{\sin^2 y} \), helps in proving more complex trigonometric identities as seen in the exercise.
Mastering the cotangent function enables students to understand its interplay with other functions and simplify trigonometric expressions effectively.

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