This work develops the mathematical apparatus of a single space, in particular, methods of working with vector trigonometric functions. This approach allows us to formalize and simplify calculations, as well as to identify additional patterns in their structure. For the correct processing of expressions containing negative radical values, it is advisable to use hyperbolic numbers, which naturally expand the domain of definition of functions.

Special attention is paid to the problem of calculating the phase shift of the vector sine. In the classical form of the vector sine expansion, only the terms corresponding to even indices of the unit vectors are nonzero, while the terms with odd indices vanish. The introduction of a phase shift breaks this symmetry and leads to the appearance of new nonzero components, which makes the analysis of functions more complete.

When transforming a scalar sine into a vector one, only even unit vectors work, while odd ones are equal to zero \[\tag{1} \Sin\v = \i\sumn1 \j_n \sqrt{ \frac12 \cos \left( {n\pi \over 2} \right) {\a^{n} \over n!} } \] where: \(\i\) - hyperbolic unit.

In the case of a phase shift, odd nonzero vectors should also appear. To test this hypothesis, we denote the phase shift angle by the symbol \(\d\) and write the transformation from scalar sine to vector as follows: \[\tag{2} \sin (\v + \d) \to \Sin (\v + \d) \] We introduce a half-angle for convenience and simplicity of further calculations \[ \v = {\a \over 2} \] This will allow us to work further with this angle as follows: \[\tag{3} \sin (\v + \d) = \sin {\a + 2\d \over 2} \] We transform this scalar function into a vector. To do this, according to the theorem, we will sequentially determine the transformation coefficients according to the established rule: \[ a_n = \left. \sqrt{ { \left [\sin^2 {\large \a + 2\d \over 2} \right]^{\large (n)} \over n!} }\, \right|_{\large \a=0} \] Then the zeroth coefficient will be found as follows \[ a_0 = \sin(\d) \] Since \[ \sin^2 {\a + 2\d \over 2} = {1 - \cos (\a + 2\d) \over 2} \] then the remaining coefficients are now found as follows: \[ a_1 = \sqrt{ \frac12 { \left [1 - \cos (\a + 2\d) \right]_x^{'} \over 1!} } = \left. \sqrt{ \frac12 { \sin (\a + 2\d) \over 1!} }\, \right|_{\a=0} = \sqrt{\sin(2\d) \over 2\cdot 1!} \] \[ a_2 = \sqrt{ \frac12 { \left [\sin (\a + 2\d) \right]_x^{'} \over 2!} } =\left. \sqrt{ \frac12 { \cos (\a + 2\d) \over 2!} }\, \right|_{\a=0} = \sqrt{\cos(2\d) \over 2\cdot 2!} \] \[ a_3 = \sqrt{ \frac12 { \left [\cos (\a + 2\d) \right]_x^{'} \over 3!} } =\left. \sqrt{ -\frac12 { \sin (\a + 2\d) \over 3!} }\, \right|_{\a=0} = \sqrt{-{\sin(2\d) \over 2\cdot 3!}} \] \[ a_4 = \ik\sqrt{ \frac12 { \left [\sin (\a + 2\d) \right]_x^{'} \over 4!} } =\left. \sqrt{ -\frac12 { \cos (\a + 2\d) \over 4!} }\, \right|_{\a=0} = \sqrt{-{\cos(2\d) \over 2\cdot 4!}} \] \[ \ldots \] Now let's write this series in general form: \[\tag{4} \Sin (\v + \d) = \j_0\,\sin(\d) + \sumn1 \j_n \sqrt{\sin(2\d) \cdot \sin {\pi n \over 2} - \cos(2\d) \cdot \cos {\pi n \over 2}} \sqrt{\a^n \over 2 \cdot n!} \]

Compare the resulting expression with (1). As we planned, in the case of a non-zero shift angle, terms at even \(n\) are added to the vector.

Let's check formula (6) on a specific example. Let's take the following phase shift angle: \[ \d = {\pi \over 4} \] Then the vector sine with a shift of a quarter of the number Pi becomes: \[\tag{7} \Sin (\v + \pi/4) = \frac{1}{\sqrt{2}} \left( \j_0 + \sumn1 \j_n \sqrt{\sin \left({\pi n \over 2} \right) {\a^n \over n!}} \right) \] Let's compare the obtained result with the transformation of the function \(\cos(\v) + \sin(\v)\) from this work. Obviously, they coincide with an accuracy of \(\sqrt{2}\), which means that their originals also coincide with the same accuracy: \[\tag{8} \cos(\v) + \sin(\v) = \sqrt{2} \sin (\v + \pi/4) \] The last expression is a classical trigonometric identity.

For such a transformation, we take the following phase shift angle \[ \d = {\pi \over 2} \] and substitute it into formula (5): \[\tag{9} \Sin (\v + \pi/2) = \j_0 + \sumn1 \j_n \sqrt{\cos \left({\pi n \over 2} \right) {\a^n \over 2 \cdot n!}} \] In this case, the hyperbolic units will cancel out, since: \( \cos(x + \pi) = -\cos(x) \). As a result, we obtained an expression for the vector cosine, which means: \[\tag{10} \Sin(\v + \pi/2) = \Cos (\v) \] Since the vector analogs coincide, the originals also coincide: \[\tag{11} \sin(\v + \pi/2) = \cos (\v) \] The resulting equality is also a classical trigonometric identity.

The paper considers the phase shift of vector trigonometric functions using the vector sine as an example. It is shown that when introducing a non-zero shift angle, the vector expansion of the sine is enriched with new terms: unlike the original case, where only the vectors with odd indices were non-zero, the phase shift leads to the appearance of non-zero terms with even indices as well.

The step-by-step derivation of the formula is based on the transformation of the scalar sine into a vector one using the theorem on expansion coefficients and the properties of hyperbolic numbers, which allows us to correctly work with negative radical expressions. As a result, we obtain a general formula for the vector sine with a phase shift, as well as its form convenient for calculations.

The example with the shift angle \(\pi/4\) and \(\pi/2\) confirmed the correctness of the obtained expressions: the vector function exactly corresponds to the well-known trigonometric identity This demonstrates the consistency of the vector approach with classical trigonometry and its extended capabilities for analyzing functions in a single space.