91Ó°ÊÓ

The sum-to-product identities are very useful for simplifying trigonometric equations. These identities transform the sum of two trigonometric functions into a product, which often makes it easier to solve equations. For example, the identity for the sum of sines is given by:

• \(\sin a + \sin b = 2\sin\frac{a+b}{2}\cos\frac{a-b}{2}\)

This means that the sum of \(\sin a\) and \(\sin b\) can be expressed as a product of sine and cosine functions. Similarly, the identity for the sum of cosines is:

• \(\cos a + \cos b = 2\cos\frac{a+b}{2}\cos\frac{a-b}{2}\)

These identities are particularly helpful when both sine and cosine terms are present in an equation. In the original exercise, applying these identities helps convert the equation \(\sin x + \sin 3x + \cos x + \cos 3x\) into a simpler form that can be easily factored.

Factoring is an algebraic technique used to simplify expressions and solve equations. When dealing with trigonometric expressions, this method is especially beneficial after applying identities such as the sum-to-product identities. In the given problem, after rewriting the equation using these identities, we end up with:

• \(2\sin 2x \cos x + 2\cos 2x \cos x = 0\)

The next step is to factor out the common elements. In this case, \(2\) and \(\cos x\) are common to both terms, allowing you to factor it out:

• \(2\cos x(\sin 2x + \cos 2x) = 0\)

Factoring this way leverages the property that if a product of factors is zero, at least one of the factors must be zero. This provides individual equations for each factor, which can be solved separately.

Solving trigonometric equations often requires breaking the equation into simpler parts, just as we did with factoring. Once the equation is factored, each part can be solved individually. Consider the factored equation from the exercise:

• \(\cos x = 0\)
• \(\sin 2x + \cos 2x = 0\)

The first equation \(\cos x = 0\) yields solutions such as \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\), found by considering where cosine equals zero. For the second equation, rearranging in terms of tangent gives \(\tan 2x = -1\), which yields solutions based on known tangent values, such as \(x = \frac{5\pi}{8}\) and \(x = \frac{9\pi}{8}\). Solving these equations typically requires knowledge of trigonometric function values and their periodic properties, enabling us to find solutions within specific intervals.