Problem 27 - Precalculus Mathematics for Calculus

Question

Prove the identity. $$\sin (x-\pi)=-\sin x$$

Step-by-Step Solution

Answer

\( \sin(x - \pi) = -\sin x \) after applying the sine difference identity and simplification.

Step 1: Rewrite using the Sine Difference Identity

Identify the identity to apply. The sine difference identity is given by \( \sin(a-b) = \sin a \cos b - \cos a \sin b \). In this case, set \( a = x \) and \( b = \pi \).

Step 2: Apply the Sine Difference Identity

Substitute \( a = x \) and \( b = \pi \) into the identity: \[ \sin(x - \pi) = \sin x \cos \pi - \cos x \sin \pi \].

Step 3: Evaluate Trigonometric Values at \( \pi \)

Recall the trigonometric values: \( \cos \pi = -1 \) and \( \sin \pi = 0 \). Substitute these values into the equation obtained in Step 2, yielding: \[ \sin(x - \pi) = \sin x (-1) - \cos x (0) \].

Step 4: Simplify the Expression

Simplify the expression \( \sin(x - \pi) = -\sin x \) since the term \( \cos x (0) \) equals zero. Thus, \( \sin(x - \pi) = -\sin x \).

Step 5: Conclude the Proof

The original identity \( \sin(x - \pi) = -\sin x \) has been proven to be true as the derived expression matches the given identity.

Key Concepts

Sine Difference Identity

The sine difference identity is an equation that helps us simplify the calculation of sine for the difference between two angles. It is expressed as \[ \sin(a-b) = \sin a \cos b - \cos a \sin b \]. This identity allows us to break down the sine of a difference into separate known trigonometric values.
In our exercise, we needed to replace the expression \( \sin(x-\pi) \) using this identity. By acknowledging this fundamental trigonometric concept, we set \( a = x \) and \( b = \pi \). By substituting these values into the sine difference identity, we revealed a formula that could easily be manipulated using known values like \( \cos \pi \) and \( \sin \pi \). By mastering identities like sine difference, simplifying complex expressions becomes much more manageable.

Trigonometric Values

Understanding trigonometric values is essential in applying identities effectively. Trigonometric values such as \( \cos \pi = -1 \) and \( \sin \pi = 0 \) are foundational in simplifying trigonometric expressions.
With these values on hand, we can easily substitute them into the identities to further simplify expressions. In our exercise, substituting \( \cos \pi \) and \( \sin \pi \) into the equation made simplification straightforward. These trigonometric values are often remembered from the unit circle, a critical tool in trigonometry that visualizes angles and their corresponding function values.
In the unit circle, \( \pi \) radian corresponds to 180 degrees, a point where the cosine value is -1 and sine is 0. This knowledge makes it quicker and easier to prove identities like \( \sin(x-\pi) = -\sin x \). Familiarity with these values allows one to navigate through problems with increased confidence.

Proof Techniques

Proof techniques in trigonometry involve a step-by-step approach to validate mathematical statements. It's a method of logical reasoning that begins with known identities and steps through an orderly chain of substitutions and simplifications.
Our proof validated the given identity \( \sin(x-\pi) = -\sin x \) by employing a systematic approach. Firstly, identifying the correct identity to use is crucial. After applying the sine difference identity, we carefully substituted known trigonometric values to reflect common truths. Each step was verified, ensuring the result consistently matched the expected outcome.

• **Selecting the Right Identity:** First identify which trigonometric identity fits the problem.
• **Substitute Known Values:** Use known trigonometric values to simplify expressions.
• **Logical Conclusion:** Ensure each step is logically sound and transitions smoothly into the next, ensuring the final expression matches the proposed identity.

These methods not only prove a statement but also deepen understanding of trigonometric relationships, reinforcing the learner's ability to tackle more intricate problems. Mastery of such techniques is vital in the study of trigonometry.