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Trigonometric identities are equations that hold true for all values within their domains of the trigonometric functions involved. They are the backbone of solving complex trigonometric equations and simplifying expressions. A profound understanding of these identities enables students to transform and manipulate trigonometric expressions, thereby solving a wide range of problems.

For instance, in the given problem, the identity used is the product-to-sum formula. It allows the conversion between products of cosines into a sum of cosines. This formula is essential when dealing with such products. It’s expressed as \[\cos p \cos q = \frac{1}{2}(\cos(p + q) + \cos(p - q))\]
By applying this identity, complex trigonometric products can be decomposed into a more workable sum or difference of two cosines, which can then be evaluated or compared as needed.

A trigonometric proof demonstrates the truth of a given trigonometric identity using logical steps and known identities. The goal is to show that the two sides of an equation are equivalent. When working with proofs, like in the presented equation \(\cos 3A + \cos 5A + \cos 7A + \cos 15A = 4 \cos 4A \cos 5A \cos 6A\), we follow a sequence of logical steps, applying identities and algebraic manipulations to simplify and ultimately match the two sides.

In this problem, the proof involves multiple applications of the product-to-sum formula and the even property of the cosine function—that is, \(\cos (-x) = \cos x\). This even property is particularly helpful in recognizing that terms \(\cos (-3A)\) can be replaced with \(\cos (3A)\), simplifying the process towards proving the given identity.

Trigonometric equation solving involves finding the values of the variable that make the equation true. To solve such equations, it's often necessary to employ various trigonometric identities, transforming the original equation into a more familiar or solvable form. This approach is versatile and can be used to tackle complex equations.

In the context of the example problem, we didn't solve for a specific variable such as \(A\), but rather proved the equivalence of two expressions. However, the methodology is similar—using transformations like the product-to-sum formula to reach a desirable state. For equations where we do need to solve for a specific variable, after simplifying the equation with identities, we could isolate the variable on one side and use inverse trigonometric functions to find its values. Important to remember is that trigonometric functions can have multiple solutions within their cycles, so all possible solutions within the given domain must be considered.