Problem 20 - Precalculus Essentials

Question

Be sure that you've familiarized yourself with the second set of formulas presented in this section by working \(5-8\) in the Concept and Vocabulary Check. Express each sum or difference as a product. If possible, find this product's exact value. $$\cos 75^{\circ}-\cos 15^{\circ}$$

Step-by-Step Solution

Answer

The exact value of \(\cos 75^{\circ}-\cos 15^{\circ}\) is \(-\frac{\sqrt{2}}{2}\).

Step 1: Application of the Sum-To-Product Formula

Apply the sum-to-product identity \(\cos a - \cos b = -2 \sin \frac{a+b}{2} \sin \frac{a-b}{2}\). By substituting \(a=75^{\circ}\) and \(b=15^{\circ}\), obtain \(-2 \sin \frac{75^{\circ}+15^{\circ}}{2} \sin \frac{75^{\circ}-15^{\circ}}{2}\).

Step 2: Simplification

Simplify the above expression. This gives \(-2 \sin 45^{\circ} \sin 30^{\circ}\).

Step 3: Calculation of the Final Value

Calculate the value of the expression by using the known values \(\sin 45^{\circ} = \frac{\sqrt{2}}{2}\) and \(\sin 30^{\circ} = \frac{1}{2}\). This results in \(-2 . \frac{\sqrt{2}}{2} . \frac{1}{2} = -\frac{\sqrt{2}}{2}\).

Key Concepts

Cosine Difference as Product

When students grapple with trigonometry, one of the hurdles can be understanding how the difference between two cosine functions can be expressed as a product. This is where the sum-to-product identities come into play. Specifically, the identity for the cosine difference is \(\cos a - \cos b = -2 \sin \frac{a+b}{2} \sin \frac{a-b}{2}\).

This trigonometric identity allows us to convert a seemingly complex expression, like the difference between two cosines, into a simpler form involving the product of sines. Here's a step-by-step breakdown of how the identity is applied in practice:

• Identify the angles \(a\) and \(b\) from the expression \(\cos a - \cos b\).
• Plug these angles into the identity.
• Reduce the resulting equation using known angle values and algebraic simplification.

Take, for instance, the problem \(\cos 75^\circ - \cos 15^\circ\). When we apply the identity with \(a = 75^\circ\) and \(b = 15^\circ\), the complex difference of cosines simplifies down to a product involving \(\sin 45^\circ\) and \(\sin 30^\circ\), which are much more manageable to calculate with.

Trigonometric Identities

Trigonometric identities are like a secret code to unlock the mysteries of trigonometry, offering alternative ways to express trigonometric functions. These identities are true for all values of the involved variables. They come in many flavors, such as reciprocal identities, Pythagorean identities, and, pertinent to our problem, sum-to-product identities.

Understanding and memorizing these identities is vital for simplifying trigonometric expressions and solving equations. They allow us to prove equivalencies and transform trigonometric terms into forms that are easier to work with. For example, by using the sum-to-product identity, we converted a cosine difference into a product of sines, making it much easier to compute for particular angle values.

Trigonometric identities are not arbitrary but are derived from the unit circle and the relationships between the angles and sides of a right-angled triangle. Regular practice with these identities, such as exercises converting sums or differences into products, helps students gain familiarity and become more adept at manipulating trigonometric expressions.

Sine Values of Special Angles

Many trigonometric problems involve angles that have 'special' values because their sine, cosine, and tangent values are known and easy to work with. We often refer to these angles as 'special angles', typically including but not limited to 0, 30, 45, 60, and 90 degrees.

The sine values for these special angles can be remembered using various mnemonics and methods, one of the most common being the right-angled triangle definitions or the unit circle. For example, in the context of our exercise, \(\sin 45^\circ = \frac{\sqrt{2}}{2}\) and \(\sin 30^\circ = \frac{1}{2}\) are derived from the unit circle or an isosceles right-angled triangle and an equilateral triangle cut in half, respectively.

Being able to recall these specific values quickly makes it easier to solve trigonometric equations and compute expressions without the need for a calculator. This knowledge is essential and is often tested in various types of math problems, ensuring students understand the fundamental properties of these special angles.