Problem 33 - Algebra and Trigonometry

Question

Verify the identity. $$ \cot (-\alpha) \cos (-\alpha)+\sin (-\alpha)=-\csc \alpha $$

Step-by-Step Solution

Answer

The identity is verified; both sides simplify to \(-\csc(\alpha)\).

Step 1: Substitute Negative Angles

Recall the trigonometric identities for negative angles: \[ \cos(-\alpha) = \cos(\alpha) \quad \text{and} \quad \sin(-\alpha) = -\sin(\alpha). \] The cotangent function has the identity: \[ \cot(-\alpha) = -\cot(\alpha). \] Substitute these into the left side of the equation:\[ \cot(-\alpha)\cos(-\alpha) + \sin(-\alpha) = -\cot(\alpha)\cos(\alpha) - \sin(\alpha). \]

Step 2: Simplify the Expression

Using the trigonometric identity \( \cot(\alpha) = \frac{\cos(\alpha)}{\sin(\alpha)} \), replace \( \cot(\alpha) \):\[ -\left(\frac{\cos(\alpha)}{\sin(\alpha)}\right)\cos(\alpha) - \sin(\alpha). \] Simplify the expression by multiplying:\[ -\frac{\cos^2(\alpha)}{\sin(\alpha)} - \sin(\alpha). \]

Step 3: Combine the Terms

Combine the two terms into a single fraction:\[ -\frac{\cos^2(\alpha)}{\sin(\alpha)} - \frac{\sin^2(\alpha)}{\sin(\alpha)}. \] This becomes:\[ -\frac{\cos^2(\alpha) + \sin^2(\alpha)}{\sin(\alpha)}. \]

Step 4: Use the Pythagorean Identity

The Pythagorean identity states:\[ \cos^2(\alpha) + \sin^2(\alpha) = 1. \] Substitute this identity into the expression:\[ -\frac{1}{\sin(\alpha)}. \] This simplifies to:\[ -\csc(\alpha). \]

Step 5: Verify the Identity

The expression \(-\csc(\alpha)\) matches the right-hand side of the original equation. Thus, the identity \[ \cot (-\alpha) \cos (-\alpha) + \sin (-\alpha) = -\csc \alpha \] is verified.

Key Concepts

Negative Angle Identities

Negative angle identities are essential tools in trigonometry that help us manipulate and simplify expressions involving angles. These identities let us understand how trigonometric functions behave when their angles are negated.
For the basic trigonometric functions, we have:

• \( \cos(-\alpha) = \cos(\alpha) \)
• \( \sin(-\alpha) = -\sin(\alpha) \)
• \( \cot(-\alpha) = -\cot(\alpha) \)

These identities stem from the symmetry and periodic nature of trigonometric functions on the unit circle. The cosine function is even, meaning it is symmetrical about the y-axis, while the sine and cotangent functions are odd, displaying symmetry about the origin.
By using these identities, you can transform complex trigonometric expressions with negative angles into simpler forms, making further manipulations more straightforward.

Cotangent Identity

The cotangent function, represented as \( \cot(\alpha) \), is one of the more intriguing trigonometric functions. It is defined as the reciprocal of the tangent function, which implies:

• \( \cot(\alpha) = \frac{1}{\tan(\alpha)} \)
• \( \cot(\alpha) = \frac{\cos(\alpha)}{\sin(\alpha)} \)

This identity is particularly useful because it connects both cosine and sine in a ratio, offering a way to break down expressions involving \( \cot(\alpha) \).
Understanding this identity is important when simplifying expressions, as it allows transformation from cotangent to sine and cosine, thus opening up paths to utilize other trigonometric identities, such as the Pythagorean identity, when evaluating or verifying equations.

Pythagorean Identity

One of the most fundamental identities in trigonometry is the Pythagorean identity:

• \( \cos^2(\alpha) + \sin^2(\alpha) = 1 \)

This identity comes from the Pythagorean theorem applied to a right-angled triangle on the unit circle, where the hypotenuse is 1, the radius of the circle.
The Pythagorean identity helps us in several ways:

• It simplifies expressions involving both sine and cosine.
• It allows conversion between these two functions when other identities aren't directly applicable.
• It plays a crucial role in verifying identities, as seen in converting expressions to show equivalence.

Through this identity, any appearance of \( \cos^2(\alpha) + \sin^2(\alpha) \) in an equation can be replaced with 1, which often results in significant simplification.

Cosecant Function

The cosecant function, denoted \( \csc(\alpha) \), is less commonly used than sine or cosine, but serves an important role as the reciprocal of the sine function. This means:

• \( \csc(\alpha) = \frac{1}{\sin(\alpha)} \)

Because it is the reciprocal, \( \csc(\alpha) \) is undefined wherever \( \sin(\alpha) = 0 \), leading it to have vertical asymptotes at multiples of \( \pi \).
When verifying expressions, like in the given problem, expressions containing \( \csc(\alpha) \) often involve simplification steps that employ its reciprocal nature.
The theme of taking reciprocal functions also emphasizes a recurring concept in trigonometry: expressing unidentified functions in terms of more fundamental or frequently encountered functions like sine, which then enables use of known identities to simplify or verify trigonometric statements.