Problem 54

Question

53–58 Find the value of the product or sum. $$3 \cos 37.5^{\circ} \cos 7.5^{\circ}$$

Step-by-Step Solution

Verified
Answer
The value is \(\frac{3}{4}(\sqrt{2} + \sqrt{3})\).
1Step 1: Identify the Trigonometric Formula
We need to simplify the expression \(3 \cos 37.5^{\circ} \cos 7.5^{\circ}\). This expression is a product of cosines. We can use the product-to-sum identities for cosines which state: \( 2 \cos A \cos B = \cos(A+B) + \cos(A-B) \).
2Step 2: Apply the Product-to-Sum Identity
Using the formula from Step 1, substitute \(A = 37.5^{\circ}\) and \(B = 7.5^{\circ}\):\[ 2 \cos 37.5^{\circ} \cos 7.5^{\circ} = \cos(37.5^{\circ} + 7.5^{\circ}) + \cos(37.5^{\circ} - 7.5^{\circ}) \].Calculate the angles: \(37.5^{\circ} + 7.5^{\circ} = 45^{\circ}\) and \(37.5^{\circ} - 7.5^{\circ} = 30^{\circ}\). Thus:\[ 2 \cos 37.5^{\circ} \cos 7.5^{\circ} = \cos 45^{\circ} + \cos 30^{\circ} \].
3Step 3: Substitute Known Cosine Values
Use the known values of cosine for these specific angles: \(\cos 45^{\circ} = \frac{\sqrt{2}}{2}\) and \(\cos 30^{\circ} = \frac{\sqrt{3}}{2}\). Substitute these values into the equation:\[ 2 \cos 37.5^{\circ} \cos 7.5^{\circ} = \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \].
4Step 4: Solve for the Original Product
Now that we have simplified the equation to:\[ 2 \cos 37.5^{\circ} \cos 7.5^{\circ} = \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \].Multiply both sides by \( \frac{3}{2} \) to account for the factor of 3 originally in the problem:\[ 3 \cos 37.5^{\circ} \cos 7.5^{\circ} = \frac{3}{2} \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right) \].
5Step 5: Simplify the Result
Perform the multiplication:\[ 3 \cos 37.5^{\circ} \cos 7.5^{\circ} = \frac{3}{4}(\sqrt{2} + \sqrt{3}) \].This is the simplified value of the original problem.

Key Concepts

Product-to-Sum IdentitiesCosine ValuesSimplifying Trigonometric Expressions
Product-to-Sum Identities
The product-to-sum identities are powerful tools in trigonometry that simplify the multiplication of trigonometric functions into sums or differences. These identities help to transform products like products of cosines into a more approachable form. For example, the identity for cosines is:
  • \[ 2 \cos A \cos B = \cos(A + B) + \cos(A - B) \]
By using this identity, one can break down complex expressions into simpler, more manageable parts.
In our problem, this identity facilitated the simplification of the product \(3 \cos 37.5^\circ \cos 7.5^\circ \), allowing us to express it as a sum of individual cosine values.
This transformation helps in evaluating the expression more easily by working with known angles.
Cosine Values
In trigonometry, knowing the standard values for cosine at certain angles is crucial. These values are memorized or derived from the unit circle, which represents cosine as the x-coordinate.
Some commonly used cosine values include:
  • \(\cos 45^\circ = \frac{\sqrt{2}}{2}\)
  • \(\cos 30^\circ = \frac{\sqrt{3}}{2}\)
By substituting these known values, calculations become much simpler. In the given exercise, after applying the product-to-sum identity, we used these standards to plug in actual cosine values of \(45^\circ\) and \(30^\circ\).
This substitution step is key in transforming the algebraic expression into numerical form, leading directly to the solution.
Simplifying Trigonometric Expressions
Trigonometric expressions can often appear daunting, but the goal is to simplify them. Simplifying involves:
  • Applying appropriate identities such as product-to-sum identities.
  • Substituting known angle values.
  • Simplifying algebraically to get a concise expression or number.
For example, in this exercise, we started with a complex product \(3 \cos 37.5^\circ \cos 7.5^\circ\). By applying the product-to-sum identity and substituting known values, we calculated each part separately and combined them.
Finally, multiplying by the original factor of 3 led us to the simplified solution \( \frac{3}{4}(\sqrt{2} + \sqrt{3}) \).
Such step-by-step simplification helps in understanding and solving trigonometric problems more effectively.