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I'd like to address a more complex trigonometric function, which can be represented as a linear combination of sine and cosine functions: f(x) = a·cos(x)+b·sin(x) In its original form the behavior of this function is not exactly clear. But, after a proper transformation, you will see that the behavior is absolutely similar to that of the behavior of sin(x) or cos(x). Obviously, the function we are analyzing has at least one coefficient, a or b, not equal to zero. Otherwise its behavior is trivial as it represents a constant 0. Now consider a point A on a coordinate plane with coordinates A={a/√(a²+b²), b/√(a²+b²)}, that is a point with abscissa equal to a/√(a²+b²) and ordinate equal to b/√(a²+b²). Since a/√(a²+b²) + b/√(a²+b²) = 1, this point A lies on a unit circle. Let φ be the central angle inside that unit circle that corresponds to this point A. Then from the definitions of sine (ordinate of the corresponding point) and cosine (its abscissa), we can say that cos(φ)=a/√(a²+b²) and sin(φ)=b/√(a²+b²). Let's use this in the analysis of our trigonometric function. Multiplying and dividing this function by √(a²+b²), we obtain: f(x) = = √(a²+b²) · · [a·cos(x)/√(a²+b²) + b·sin(x)/√(a²+b²)] Replacing a/√(a²+b²) with cos(φ) and b/√(a²+b²) with sin(φ) where the angle φ was defined above, we get: f(x) = √(a²+b²) · · [cos(φ)·cos(x)+sin(φ)·sin(x)]. The expression in square brackets is, of course, cos(x−φ). Therefore, the transformed expression for our function is f(x) = √(a²+b²) · cos(x−φ) Now it's clear that the behavior of our rather complex trigonometric function that is represented as a linear combination of sine and cosine is the same as that of basic trigonometric functions sin or cosine, that is it's waving and periodic. The only difference is a horizontal shift by some angle φ (it's value depends on coefficients a and b, as explained above using the coordinate plane) and vertical stretch by a factor √(a²+b²). Such functions are usually called sinusoidal.