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Angle addition for sine

Prerequisites

Eulers Formula | \(e^{j \cdot \theta} = \cos(\theta) + j \sin(\theta)\)
Exponent Addition | \(x^{y + z} = x^{y} \cdot x^{z}\)

Description

The sine angle addition formula is a fundamental identity in trigonometry that describes how the cosine of the sum of two angles can be expressed in terms of the sines and cosines of the individual angles. This formula is essential for solving problems involving trigonometric functions, where angles are combined or split, and for proving other trigonometric identities. It enables the simplification of expressions and the calculation of unknown angles or lengths in geometric figures. The formula also plays a crucial role in the study of wave interference, oscillations, and rotations in physics, illustrating how wave phases or orientations combine. Understanding and applying the cosine angle addition formula is a key skill in mathematics, providing a bridge to more complex concepts in calculus and analytical geometry.

\[\htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta} + \htmlClass{sdt-0000000123}{\phi}) = \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta}) \cdot \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000123}{\phi}) + \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000123}{\phi}) \cdot \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000024}{\theta})\]

Symbols Used:

This is a commonly used symbol to represent an angle in mathematics and physics.
\( \phi \)

This symbol means the same as Angle, which uses \(\htmlClass{sdt-0000000024}{\theta}\). It is a secondary symbol to use that represents an angle, when a different angle is already using \(\htmlClass{sdt-0000000024}{\theta}\).
\( \cos \)

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.
\( \sin \)

This is the symbol for sine, is a trigonometric function that represents the ratio of the opposite side to the hypotenuse in a right-angled triangle.

Derivation

  1. Consider Euler's Formula
    \[\htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000129}{j} \cdot \htmlClass{sdt-0000000024}{\theta}} = \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000024}{\theta}) + \htmlClass{sdt-0000000129}{j} \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta})\]
    If we let \(\htmlClass{sdt-0000000024}{\theta} = \htmlClass{sdt-0000000123}{\phi}\), we can also say that...
    \[\htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000129}{j} \cdot \htmlClass{sdt-0000000123}{\phi}} = \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000123}{\phi}) + \htmlClass{sdt-0000000129}{j} \cdot \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000123}{\phi})\].
    Similarly, if we let \(\htmlClass{sdt-0000000024}{\theta} = \htmlClass{sdt-0000000024}{\theta} + \htmlClass{sdt-0000000123}{\phi}\), we can say that...
    \[\htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000129}{j} ( \htmlClass{sdt-0000000024}{\theta} + \htmlClass{sdt-0000000123}{\phi})} = \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000024}{\theta} + \htmlClass{sdt-0000000123}{\phi}) + \htmlClass{sdt-0000000129}{j} \cdot \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta} + \htmlClass{sdt-0000000123}{\phi})\].
  2. Multiplying Euler's Formula with angles \(\htmlClass{sdt-0000000024}{\theta}\) and \(\htmlClass{sdt-0000000123}{\phi}\), leads us to the equation...
    \[\htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000129}{j} \cdot \htmlClass{sdt-0000000024}{\theta}} \cdot \htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000129}{j} \cdot \htmlClass{sdt-0000000123}{\phi}} = (\htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000024}{\theta}) + \htmlClass{sdt-0000000129}{j} \cdot \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta}))(\htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000123}{\phi}) + \htmlClass{sdt-0000000129}{j} \cdot \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000123}{\phi}))\]
  3. Now using the rule for Exponent Addition on the left hand side:
    \[\htmlClass{sdt-0000000003}{x}^{\htmlClass{sdt-0000000017}{y} + \htmlClass{sdt-0000000043}{z}} = \htmlClass{sdt-0000000003}{x}^{\htmlClass{sdt-0000000017}{y}} \cdot \htmlClass{sdt-0000000003}{x}^{\htmlClass{sdt-0000000043}{z}}\]
    and the distributive property for multiplication on the right hand side, we can simplify to get...
    \[\htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000129}{j} \cdot \htmlClass{sdt-0000000024}{\theta} + \htmlClass{sdt-0000000129}{j} \cdot \htmlClass{sdt-0000000123}{\phi}} = \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000024}{\theta}) \cdot \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000123}{\phi}) - \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta}) \cdot \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000123}{\phi}) + \htmlClass{sdt-0000000129}{j}(\htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta}) \cdot \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000123}{\phi}) + \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000123}{\phi}) \cdot \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000024}{\theta}))\]
  4. From here, we can factor out \(\htmlClass{sdt-0000000129}{j}\) from the exponent of the left hand side to get...
    \[\htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000129}{j} \cdot (\htmlClass{sdt-0000000024}{\theta} + \htmlClass{sdt-0000000123}{\phi})} = \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000024}{\theta}) \cdot \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000123}{\phi}) - \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta}) \cdot \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000123}{\phi}) + \htmlClass{sdt-0000000129}{j}(\htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta}) \cdot \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000123}{\phi}) + \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000123}{\phi}) \cdot \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000024}{\theta}))\]
  5. Note, that the left hand side of this is equivalent to the equation shown at the end of step (1), we can therefore assign their right hand sides to each other, to get...
    \[\htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000024}{\theta} + \htmlClass{sdt-0000000123}{\phi}) + \htmlClass{sdt-0000000129}{j} \cdot \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta} + \htmlClass{sdt-0000000123}{\phi}) = \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000024}{\theta}) \cdot \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000123}{\phi}) - \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta}) \cdot \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000123}{\phi}) + \htmlClass{sdt-0000000129}{j}(\htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta}) \cdot \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000123}{\phi}) + \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000123}{\phi}) \cdot \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000024}{\theta})\]
  6. We will now take the imaginary parts of both of these complex numbers
    \[\htmlClass{sdt-0000000114}{\Im}(\htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000024}{\theta} + \htmlClass{sdt-0000000123}{\phi}) + \htmlClass{sdt-0000000129}{j} \cdot \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta} + \htmlClass{sdt-0000000123}{\phi})) = \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta} + \htmlClass{sdt-0000000123}{\phi})\]
    and
    \[\htmlClass{sdt-0000000114}{\Im}(\htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000024}{\theta}) \cdot \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000123}{\phi}) - \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta}) \cdot \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000123}{\phi}) + \htmlClass{sdt-0000000129}{j}(\htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta}) \cdot \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000123}{\phi}) + \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000123}{\phi}) \cdot \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000024}{\theta})) = \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta}) \cdot \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000123}{\phi}) + \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000123}{\phi}) \cdot \htmlClass{sdt-0000000124}{\cos}(angle)\]
  7. And because the two complex numbers are equivalent (as can be seen in part (5) we can say that the imaginary parts must be the same as well. It therefore follows that...
    \[\htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta} + \htmlClass{sdt-0000000123}{\phi}) = \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta}) \cdot \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000123}{\phi}) + \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000123}{\phi}) \cdot \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000024}{\theta})\]
    as required.

Example

Coming soon...