Problem 67 - Precalculus Mathematics for Calculus

Question

Find the value of the product or sum. $$2 \sin 52.5^{\circ} \sin 97.5^{\circ}$$

Step-by-Step Solution

Answer

\( \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2} \)

Step 1: Identify the formula to use

For the expression \(2 \sin A \sin B\), the product-to-sum identities tell us that \(2 \sin A \sin B = \cos (A - B) - \cos (A + B)\). So, we apply this formula.

Step 2: Assign values to \(A\) and \(B\)

We have \(A = 52.5^{\circ}\) and \(B = 97.5^{\circ}\). We'll substitute these values into the formula chosen.

Step 3: Calculate \(A - B\)

Subtract \(B = 97.5^{\circ}\) from \(A = 52.5^{\circ}\):\[ A - B = 52.5^{\circ} - 97.5^{\circ} = -45^{\circ} \]

Step 4: Calculate \(A + B\)

Add \(A = 52.5^{\circ}\) and \(B = 97.5^{\circ}\):\[ A + B = 52.5^{\circ} + 97.5^{\circ} = 150^{\circ} \]

Step 5: Evaluate the cosine values

Using known cosine values:- \(\cos(-45^{\circ}) = \cos(45^{\circ}) = \frac{1}{\sqrt{2}}\)- \(\cos(150^{\circ}) = -\frac{\sqrt{3}}{2}\)

Step 6: Calculate the final result

Substitute the values into the identity:\[2 \sin 52.5^{\circ} \sin 97.5^{\circ} = \cos(-45^{\circ}) - \cos(150^{\circ})\]\[= \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2}\]Combine these to get the result.

Key Concepts

Product-to-Sum Identities

Product-to-sum identities are vital tools in trigonometry that simplify products of sine and cosine into sums or differences of trigonometric functions. This transformation makes complex trigonometric expressions easier to manage, especially when calculating angles that are not readily available on standard calculators.
For the expression \(2 \sin A \sin B\), the identity states \(2 \sin A \sin B = \cos(A - B) - \cos(A + B)\). This helps convert a product of sines into a simpler form involving cosines.

• To use this identity, identify the angles \(A\) and \(B\) in your original expression.
• Then, find \(A - B\) and \(A + B\).
• Cast the trigonometric function from sines to cosines using the identity.

This transformation is especially useful in solving problems involving wave interference, sound waves, or vibrations.

Cosine Values

Understanding how to evaluate cosine values for specific angles is crucial in trigonometric calculations. Some angles like \(-45^{\circ}\) and \(150^{\circ}\) have known cosine values that make calculations simpler. Knowing these can quickly lead to a solution without relying on a calculator.
For example, \(\cos(-45^{\circ})\) is the same as \(\cos(45^{\circ})\), because cosine is an even function. The formula states: \(\cos(-x) = \cos(x)\). Therefore, \(\cos(45^{\circ}) = \frac{1}{\sqrt{2}}\).
Furthermore, the cosine value of \(150^{\circ}\) is \(-\frac{\sqrt{3}}{2}\). This comes from the unit circle and understanding that \(150^{\circ}\) lies in the second quadrant where cosine values are negative. Practicing with these values will strengthen your ability to solve trigonometric problems swiftly.

Trigonometric Calculations

Carrying out trigonometric calculations can often seem daunting due to the nature of the functions involved. However, breaking them down step-by-step, using identities and known values, makes the process more approachable.
In the example problem, once the Product-to-Sum Identity is applied, you're left with a straightforward task of evaluating cosine at specific angles. The primary work involves:

• Subtracting and adding the angles \(A\) and \(B\) to find \(A - B\) and \(A + B\).
• Substituting these into the converted cosine expression.
• Using known values for \(\cos(-45^{\circ})\) and \(\cos(150^{\circ})\) to finish the calculation.
• Finally, combining these values to achieve a simplified result.

Approaching each step with patience and a strategy, using trigonometric identities where possible, streamlines the calculation process and boosts both understanding and proficiency.