## The Trigonometric Transformation

The sine trig identities

`\sin(x+y) = \sin(x)\cos(y)+\cos(x)\sin(y)`

`\sin(x-y) = \sin(x)\cos(y)-\cos(x)\sin(y)`

hence

`\sin(x+y)+\sin(x-y) = 2\sin(x)\cos(y)`

If we take `\sin(x)`

as the function `f(x)`

, and `y`

as a constant `\Delta\theta`

, we have

`f(x+\Delta\theta)+f(x-\Delta\theta)=2\cos(\Delta\theta)\cdot f(x)`

or

`f(x+\Delta\theta)=2\cos(\Delta\theta)\cdot f(x)-f(x-\Delta\theta)`

Thus, there’s an fixed relationship between `f(x-\Delta\theta)`

, `f(x)`

and `f(x+\Delta\theta)`

.

## Discrete Sine Wave Sequence

### General Form

By defining `a_n = f(n\cdot\Delta\theta)`

, we have a discrete recursive formula of sine wave

`a_n = 2\cos(\Delta\theta)a_{n-1}-a_{n-2}`

Let `\Delta\theta=2\pi/N`

, there’s `N`

of `a_i`

in the range `[0,\ 2\pi)`

.

Moreover, if we write `a_n`

as

`a_n=k\ a_{n-1}-a_{n-2}`

, for any `k\in (-1,\ 0) \bigcap (0,\ 1)`

and `| a_{n-1} | + | a_{n-2} | \neq 0`

, `a_n`

is always a sine sequence. This is for the reason that, for any valid `k`

, there’s always a (or two) `\Delta\theta`

will satisfy `2\cos(\Delta\theta)=k`

.

### Another General Form

We can define another sequence `b_n`

from `a_n`

as

`b_n = \cfrac{k_\mathrm{L}\ a_{n} + k_\mathrm{R}\ a_{n+M}}{k_\mathrm{L}+k_\mathrm{R}}`

For any `| k_\mathrm{L} | + | k_\mathrm{R} | \neq 0`

and `k_\mathrm{L}+k_\mathrm{R}\neq 0`

. The `b_n`

is also a discrete sequence of sine wave. That’s because the following recursive formula is satisfied.

`b_n=k\ b_{n-1}-b_{n-2}`

## Graphical Illustration

By letting `N`

, `M`

, `k_\mathrm{L}`

and `k_R`

have different values, we can have different patterns of `a_n`

and `b_n`

. As shown in the figures below, `a_n`

is the dots in black, `b_n`

is the dots in color.

Note that: The straight line in the Cartesian coordinate system (left) is not corresponding to the straight line in the Polar coordinate system (right).

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