Problem 78 - Algebra and Trigonometry

Question

Show that \(\cos 100^{\circ}-\cos 200^{\circ}=\sin 50^{\circ}\).

Step-by-Step Solution

Answer

The expression \(\cos 100^{\circ} - \cos 200^{\circ}\) simplifies to \(\sin 50^{\circ}\).

Step 1: Identify the trigonometric identities

To solve the problem, recognize that trigonometric identities can be used to simplify the expression. We are going to use product-to-sum identities, specifically the identity: \[a) \cos A - \cos B = -2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right).\]

Step 2: Apply the identity to the given expression

Apply the identity to the expression \(\cos 100^{\circ} - \cos 200^{\circ}\):\[\cos 100^{\circ} - \cos 200^{\circ} = -2 \sin \left(\frac{100^{\circ} + 200^{\circ}}{2}\right) \sin \left(\frac{100^{\circ} - 200^{\circ}}{2}\right).\]

Step 3: Simplify the expression

Now, calculate each part of the expression:- The average of angles \(100^{\circ}\) and \(200^{\circ}\) is \(\frac{100^{\circ} + 200^{\circ}}{2} = 150^{\circ}\).- The difference of angles \(100^{\circ}\) and \(200^{\circ}\) is \(\frac{100^{\circ} - 200^{\circ}}{2} = -50^{\circ}\).Substitute these into the formula:\[\cos 100^{\circ} - \cos 200^{\circ} = -2 \sin(150^{\circ}) \sin(-50^{\circ}).\]

Step 4: Evaluate and simplify trigonometric functions

We know that \(\sin(-x) = -\sin(x)\), therefore, \(\sin(-50^{\circ}) = -\sin(50^{\circ})\).Also, \(\sin(150^{\circ}) = \sin(180^{\circ} - 30^{\circ}) = \sin(30^{\circ}) = \frac{1}{2}\).Hence, \[\cos 100^{\circ} - \cos 200^{\circ} = -2 \left( \frac{1}{2} \right) \left( - \sin(50^{\circ}) \right).\]

Step 5: Final calculation

Simplify the final expression:\[-2 \cdot \frac{1}{2} \cdot (-\sin(50^{\circ})) = \sin(50^{\circ}).\]Thus, we have shown that the original expression is equal to \(\sin(50^{\circ})\).

Key Concepts

Product-to-Sum Identities

Trigonometric identities can sometimes be daunting, but they are powerful tools for simplifying trigonometric expressions. One important group of these identities is called product-to-sum identities. The product-to-sum identities help us transform the product of trigonometric functions into a sum or difference, thereby simplifying the expression greatly.

In this exercise specifically, we apply the formula:

• \( \cos A - \cos B = -2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right) \)

This identity is beneficial because it reduces the task of dealing with trigonometric products to dealing with simpler sums or differences. Understanding and applying these identities make complex calculations far more manageable.

Trigonometric Simplification

Simplification is a key step when working with trigonometric expressions. The aim is to make the expression as simple as possible. In our exercise, after applying the product-to-sum identity, we are left with sine functions that need further simplification.

Here's how it's done:

• Calculate \( \frac{A+B}{2} \) and \( \frac{A-B}{2} \).
• Calculate the sine of these mid-values and angle differences.

For example, transforming \( \cos 100^{\circ} - \cos 200^{\circ} \) gives us \( -2 \sin(150^{\circ}) \sin(-50^{\circ}) \). Nicely, the trigonometric function \( \sin(x) \) has properties like:

• \( \sin(-x) = -\sin(x) \)

These properties and simplifications lead us to a form that is much easier to work with.

Angle Difference Identities

Another cornerstone of trigonometry is understanding and correctly applying angle difference identities. While not directly applied in this specific solution, the concept is crucial for understanding the behavior of trigonometric functions over different angles.

For example:

• The sine and cosine function changes slightly with certain shifts in angles: \( \sin(\theta) = \sin(180^{\circ} - \theta) \).
• Such transformations allow us to look at angles on the unit circle from a different perspective.

In this problem, through simplifications like \( \sin(150^{\circ}) \), which is rewritten using the identity \( \sin(180^{\circ} - \theta) \), it becomes equal to \( \sin(30^{\circ}) \).

This step-by-step look at angles is how angle difference identities strengthen our overall grasp of trigonometric calculations and equip us to handle different trigonometric scenarios deftly.