Problem 102 - Precalculus with Limits

Question

In Exercises 99 - 102, use the sum-to-product formulas to find the exact value of the expression. \( \sin \dfrac{5\pi}{4} - \sin \dfrac{3\pi}{4} \)

Step-by-Step Solution

Answer

The exact value of the expression \( \sin \dfrac{5\pi}{4} - \sin \dfrac{3\pi}{4} \) is \( -\sqrt{2} \).

Step 1: Recall The Formula for the Difference of Sine Functions

Sum-to-product formulas are a set of equalities that express the sum or difference of trigonometric functions in terms of their products. The specific formula for the difference of two sine functions is \( \sin a - \sin b = 2 \cos \dfrac{a + b}{2} \sin \dfrac{a - b}{2} \).

Step 2: Identify a and b and Apply the Formula

Identify \( a = \dfrac{5\pi}{4} \) and \( b = \dfrac{3\pi}{4} \) from the original expression and substitute into the formula: \( 2 \cos \dfrac{\dfrac{5\pi}{4} + \dfrac{3\pi}{4}}{2} \sin \dfrac{\dfrac{5\pi}{4} - \dfrac{3\pi}{4}}{2} \). Simplifying inside the cosine function gives \( \cos \pi \) and inside the sine function gives \( \sin \dfrac{\pi}{4} \).

Step 3: Simplify to Find the Final Answer

Replace \( \cos \pi \) and \( \sin \dfrac{\pi}{4} \) with their exact values, -1 and \( \dfrac{\sqrt{2}}{2} \) respectively, yielding \( 2 \cdot (-1) \cdot \dfrac{\sqrt{2}}{2} \) which simplifies to \( -\sqrt{2} \).

Key Concepts

Trigonometric Identities

Trigonometric identities are fundamental tools used to simplify expressions involving trigonometric functions. They are equations that are true for all values of the variables involved. Trigonometric identities help in transforming and simplifying expressions, especially when dealing with angles.

Some of the major categories of these identities include:

• Reciprocal identities, which involve relationships such as \( \frac{1}{\sin x} = \csc x \)
• Pythagorean identities, like \( \sin^2 x + \cos^2 x = 1 \)
• Sum-to-product and product-to-sum identities, which convert sums or differences of functions to products, and vice versa

These identities are immensely helpful in various areas of mathematics and physics, allowing us to switch between different forms of equations and solve problems more simply. In exercises like these, **sum-to-product formulas** assist in finding exact trigonometric values, thus simplifying the calculation process.

Difference of Sine Functions

When dealing with the difference of sine functions, a specific sum-to-product formula is extremely useful. This formula is:\[\sin a - \sin b = 2 \cos \frac{a + b}{2} \sin \frac{a - b}{2}\]This formula lets us express the difference of two sine values as a product of cosine and sine functions. It can make complicated differences much easier to work with, as we can transform them into simple multiplications.

For the given problem, where we have \( \sin \frac{5\pi}{4} - \sin \frac{3\pi}{4} \), using this formula allows us to convert this expression into a form that is directly computable. By identifying \( a \) and \( b \) as \( \frac{5\pi}{4} \) and \( \frac{3\pi}{4} \) respectively, the formula becomes a straightforward application. Then, it is just a matter of inserting these values to simplify and solve the expression further.

Exact Trigonometric Values

Calculating exact trigonometric values is a crucial part of solving problems involving trigonometric expressions. The use of precise values for sine and cosine at key angles, like \(0, \frac{\pi}{4}, \frac{\pi}{2}, \) and \(\pi\), makes solving these problems more manageable.

To find the exact value of \( \sin \frac{5\pi}{4} - \sin \frac{3\pi}{4} \), we evaluated each component separately:

• For the formula \( \cos \pi \), the exact value is \(-1\).
• For \( \sin \frac{\pi}{4} \), it equals \( \frac{\sqrt{2}}{2} \).

These are derived from the unit circle and the properties of triangles, which persist as the same regardless of the method used. By multiplying these values in their respective places within the identity, it results in the final exact answer \(-\sqrt{2}\), illustrating how precise trigonometric values can simplify what might initially seem a complex calculus.