Problem 31 - Precalculus with Calculus Previews

Question

In Problems \(31-36,\) use a sum-to-product-formula in Theorem 4.7.2 to find the exact value of the expression. Do not use a calculator. $$ \sqrt{2} \sin \frac{13 \pi}{12}+\sqrt{2} \sin \frac{5 \pi}{12} $$

Step-by-Step Solution

Answer

The exact value is \( \sqrt{2} \).

Step 1: Recognize the Expression

The given expression is \( \sqrt{2} \sin \frac{13 \pi}{12} + \sqrt{2} \sin \frac{5 \pi}{12} \). Notice that both terms have the common factor \( \sqrt{2} \), so factor it out: \( \sqrt{2} ( \sin \frac{13 \pi}{12} + \sin \frac{5 \pi}{12} ) \).

Step 2: Apply the Sum-to-Product Formula

To simplify \( \sin A + \sin B \), we use the sum-to-product identity: \( \sin A + \sin B = 2 \sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) \). Here, \( A = \frac{13 \pi}{12} \) and \( B = \frac{5 \pi}{12} \).

Step 3: Calculate the Midpoint and Difference

Calculate \( \frac{A + B}{2} = \frac{13 \pi}{12} + \frac{5 \pi}{12} = \frac{18 \pi}{12} = \frac{3 \pi}{2} \). Calculate \( \frac{A - B}{2} = \frac{13 \pi}{12} - \frac{5 \pi}{12} = \frac{8 \pi}{12} = \frac{2 \pi}{3} \).

Step 4: Substitute Midpoint and Difference into Formula

Substitute these values into the sum-to-product identity: \( 2 \sin \left( \frac{3 \pi}{2} \right) \cos \left( \frac{2 \pi}{3} \right) \).

Step 5: Evaluate the Trigonometric Functions

Evaluate \( \sin \frac{3 \pi}{2} \), which is \(-1\). Evaluate \( \cos \frac{2 \pi}{3} \), which is \(-\frac{1}{2}\).

Step 6: Simplify the Expression

Simplify \( 2(-1) \left( -\frac{1}{2} \right) = 2 \times 1 \times \frac{1}{2} = 1 \).

Step 7: Final Simplification

Multiply the simplified result by the common factor we initially factored out: \( \sqrt{2} \times 1 = \sqrt{2} \).

Key Concepts

Trigonometry

Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles. In this particular problem, we are working with two trigonometric functions: sine and cosine. The sine function, denoted as \(\sin\), is associated with the y-coordinate of a point on the unit circle, whereas the cosine function, denoted as \(\cos\), is related to the x-coordinate. Both functions are periodic, meaning they repeat their values in a regular cycle.
For trigonometric problems, understanding the unit circle is essential. The unit circle allows us to find sine and cosine values for specific angles, which is crucial for evaluating expressions without a calculator. In the given exercise, angles are expressed in terms of radians, where \(\pi\) radians equate to 180 degrees. Knowing this conversion helps in identifying the precise position of the angle on the unit circle and subsequently, the sine and cosine values.

Exact Values

Exact values in trigonometry refer to the specific values of trigonometric functions that are commonly known and do not need approximation. These include values for key angles such as \(0\), \(\frac{\pi}{6}\), \(\frac{\pi}{4}\), \(\frac{\pi}{3}\), \(\frac{\pi}{2}\), and so on.
In the solution, we found the exact values for \(\sin \frac{3 \pi}{2}\) and \(\cos \frac{2 \pi}{3}\). It's crucial to remember these commonly used trigonometric values:

• \(\sin \frac{3\pi}{2} = -1\)
• \(\cos \frac{2\pi}{3} = -\frac{1}{2}\)

These values can typically be derived directly from the unit circle, enabling us to simplify expressions more accurately and efficiently. Learning these exact values helps to efficiently solve trigonometric expressions without reliance on technology.

Factoring

Factoring is a mathematical process where we express an expression as a product of simpler expressions. In the given exercise, factoring simplifies the expression by identifying and extracting a common factor.
Initially, the expression \(\sqrt{2} \sin \frac{13 \pi}{12} + \sqrt{2} \sin \frac{5 \pi}{12}\) had a common factor of \(\sqrt{2}\). By factoring this out, we simplified the problem to \(\sqrt{2}(\sin \frac{13 \pi}{12} + \sin \frac{5 \pi}{12})\).
Factoring is an invaluable first step because it often makes subsequent operations more manageable. It simplifies the work required in applying identities, such as sum-to-product identities, by reducing the expression into a form that's easier to manipulate and evaluate. This strategy is useful across various mathematical problems, offering a straightforward path towards simplification and solution.