Problem 117 - Precalculus Essentials

Question

Exercises \(116-118\) will help you prepare for the material covered in the next section. In each exercise, use exact values of trigonometric functions to show that the statement is true. Notice that each statement expresses the product of sines and/or cosines as a sum or a difference. $$\cos \frac{\pi}{2} \cos \frac{\pi}{3}=\frac{1}{2}\left[\cos \left(\frac{\pi}{2}-\frac{\pi}{3}\right)+\cos \left(\frac{\pi}{2}+\frac{\pi}{3}\right)\right]$$

Step-by-Step Solution

Answer

The equation is true as the left-hand side is \(0\) and the right-hand side of the equation also simplifies to \(0\). Thus, the statement is valid.

Step 1: Compute the left side of the equation

The left side of the equation is \(\cos \frac{\pi}{2} \cos \frac{\pi}{3}\). The cosine of \(\frac{\pi}{2}\) is 0 and the cosine of \(\frac{\pi}{3}\) is \(\frac{1}{2}\), hence their product will be \(0 × \frac{1}{2} = 0\).

Step 2: Compute the right side of the equation

The right-hand side of the equation is \(\frac{1}{2}[\cos (\frac{\pi}{2}-\frac{\pi}{3})+\cos(\frac{\pi}{2}+\frac{\pi}{3})]\). First, find the value inside the brackets i.e. \(cos(\frac{\pi}{2}-\frac{\pi}{3})\) which simplifies to \(\cos(\frac{\pi}{6})\) and its value is \(\frac{\sqrt{3}}{2}\). Similarly, the second part \(\cos(\frac{\pi}{2}+\frac{\pi}{3})\) simplifies to \(\cos(\frac{5\pi}{6})\) and its value is \(-\frac{\sqrt{3}}{2}\). Adding these two values inside the bracket we get \(0\). Hence the right hand side of the equation simplifies to \(\frac{1}{2} × 0 = 0\)

Step 3: Compare both sides

After computing both sides of the equation, we find that both sides equal to \(0\), hence the given equation is correct.

Key Concepts

Cosine Addition Formula

The cosine addition formula is a powerful tool in trigonometry that allows us to find the cosine of the sum or difference of two angles. This formula states: \[ \cos(\alpha \pm \beta) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta \] In simple terms, this formula can be used to separate or combine angles when you're dealing with sums or differences of angles. In the context of our exercise, the formula helps verify the truth of the statement by expressing the product of two cosines as a combination of cosine terms. By breaking down these angles into simpler components, you can calculate each part individually, simplifying complex identities for more straightforward computations.

Exact Values of Trigonometric Functions

Knowing the exact values of trigonometric functions for specific angles is invaluable when solving trigonometry problems. Some of the most common angles you'll encounter include \( 0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2} \), and so on.

• For instance, \( \cos \frac{\pi}{2} = 0 \).
• Meanwhile, \( \cos \frac{\pi}{3} = \frac{1}{2} \).
• On a related note, \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \), and \( \cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2} \).

These exact values are memorized and used in identities to simplify and solve trigonometric equations. Understanding these values allows you to tackle a wide range of problems efficiently.

Product-to-Sum Formulas

The product-to-sum formulas are particularly useful when transforming the product of sines or cosines into a sum or difference. They relate products of trig functions to simpler terms: \[ \cos A \cos B = \frac{1}{2} [\cos(A + B) + \cos(A - B)] \] These formulas reduce complex products into simpler expressions that are easier to evaluate. In our specific exercise, using the product-to-sum identity enabled the clarification of the given statement by breaking down the cosine products into known cosine sum values. These identities are essential for dealing with more challenging trigonometric expressions, allowing for a clear path to verifying identities or solving intricate problems.