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Chapter 5: Problem 84

Use the product-to-sum formulas to write the product as a sum or difference. $$6 \sin 45^{\circ} \cos 15^{\circ}$$

Short Answer

Expert verified
The product can be written as a sum or difference as follows: \( \frac{(\sqrt{3} + 1)}{2} \)

Step by step solution

01

Identify the Trigonometric Product in the Expression

The trigonometric product in the given expression is \(6 \sin 45^{\circ} \cos 15^{\circ}\).
02

Apply the Product to Sum Identity

The expression can be written as \(2(3 \sin 45^{\circ} \cos 15^{\circ})\), which fits the given formula. Now apply the product-to-sum identity to convert this product into sum, \(2 \sin \varphi \cos \psi = \sin (\varphi + \psi) + \sin (\varphi - \psi) \). Substituting the values, the expression becomes \(2(3 (\sin 45^{\circ} \cos 15^{\circ})) = \sin (45^{\circ} + 15^{\circ}) + \sin (45^{\circ} - 15^{\circ})\).
03

Simplify the Expression

Now, simplify the degrees in the expression, that gives us \sin 60^{\circ} + \sin 30^{\circ} . Is known that \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \) and \( \sin 30^{\circ} = \frac{1}{2} \). Substituting these values simplifies the expression to \( \frac{\sqrt{3}}{2} + \frac{1}{2} \) which equals \( \frac{(\sqrt{3} + 1)}{2} \) after consolidating the fractions.

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