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Cosine quarter cycle zeros identity

Prerequisites

Angle addition for cosine | \(\cos(\theta + \phi) = \cos(\theta) \cdot \cos(\phi) - \sin(\theta) \cdot \sin(\phi)\)

Description

The "Cosine Quarter Cycle Zeros Identity" highlights a fundamental property of the cosine function, indicating that it equals zero at specific intervals, namely at \(\htmlClass{sdt-0000000126}{\pi} k + \frac{\htmlClass{sdt-0000000126}{\pi}}{2}\)​ for any integer \(k\). This identity pinpoints the instances where the cosine wave crosses the horizontal axis, occurring precisely at the quarter points of its periodic cycle. By setting \(k\) to any integer value, we can systematically identify all points along the cosine function where its value transitions from positive to negative or vice versa, without directly calculating the cosine of those angles. These zero points are crucial for understanding wave behavior, signal processing, and the symmetrical properties of trigonometric functions. The "Cosine Quarter Cycle Zeros Identity" serves as a bridge between abstract mathematical concepts and practical applications, offering a clear and predictable pattern for locating the zeros of the cosine function across its infinite domain.

\[\htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000126}{\pi} k + \frac{\htmlClass{sdt-0000000126}{\pi}}{2}) = 0, k \in \htmlClass{sdt-0000000122}{\mathbb{Z}}\]

Symbols Used:

This symbol represents the set of all integers, which includes all positive and negative whole numbers, as well as zero.
\( \cos \)

This is the symbol for cosine, a trigonometric function that calculates the ratio of the adjacent side to the hypotenuse of a right-angled triangle.
\( \pi \)

This is the symbol for pi, mathematical constant representing the ratio of a circle's circumference(\( \htmlClass{sdt-0000000128}{c} \)) to its diameter(\( \htmlClass{sdt-0000000034}{d} \)).

Derivation

  1. Consider:
    \[\htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000024}{\theta} + \htmlClass{sdt-0000000123}{\phi}) = \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000024}{\theta}) \cdot \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000123}{\phi}) - \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta}) \cdot \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000123}{\phi})\]
  2. Now we apply this with \(a = \htmlClass{sdt-0000000126}{\pi} k\) and \(b = \frac{\htmlClass{sdt-0000000126}{\pi}}{2}\):
    \[\htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000126}{\pi} k + \frac{\htmlClass{sdt-0000000126}{\pi}}{2}) = \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000126}{\pi} k) \htmlClass{sdt-0000000124}{\cos}(\frac{\htmlClass{sdt-0000000126}{\pi}}{2}) - sine(\htmlClass{sdt-0000000126}{\pi} k ) \htmlClass{sdt-0000000127}{\sin}(\frac{\htmlClass{sdt-0000000126}{\pi}}{2})\]
  3. Evaluate \(\htmlClass{sdt-0000000124}{\cos}(\frac{\htmlClass{sdt-0000000126}{\pi}}{2})\) and \(\htmlClass{sdt-0000000127}{\sin}(\frac{\htmlClass{sdt-0000000126}{\pi}}{2})\):
    \[\htmlClass{sdt-0000000124}{\cos}(\frac{\htmlClass{sdt-0000000126}{\pi}}{2}) = 0\]
    \[\htmlClass{sdt-0000000127}{\sin}(\frac{\htmlClass{sdt-0000000126}{\pi}}{2}) = 1\]
  4. Since:
    \[\htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000126}{\pi} k) = 0, k \in \htmlClass{sdt-0000000122}{\mathbb{Z}} (make page for this later)\]
  5. We can now substitute these values into the equation from step 2 to get:
    \[ \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000126}{\pi} k + \frac{\htmlClass{sdt-0000000126}{\pi}}{2}) = \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000126}{\pi} k) \cdot 0 - 0 \cdot 1 = 0\]
  6. We can therefore conclude that:
    \[\htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000126}{\pi} k + \frac{\htmlClass{sdt-0000000126}{\pi}}{2}) = 0, k \in \htmlClass{sdt-0000000122}{\mathbb{Z}}\]