Deriving trigonometric identities

Let’s assume that we need a formula for \(\sin(a + b)\) in terms of \(\sin(a)\), \(\sin(b)\), \(\cos(a)\) and \(\cos(b)\), but we don’t remember it, nor do we know how to get it easily with SymPy. We will derive this formula from scratch using Taylor series expansions and a little symbolic manipulation.

Let’s start with definition of symbols and the expression in consideration:

>>> var('a,b')
(a, b)

>>> f = sin(a + b)
>>> f
sin(a + b)

Now let’s expand \(f\) as a power series with respect to \(b\) around 0:

>>> f.series(b, 0, 10)
                     2           3           4           5           6           7           8           9
                    b ⋅sin(a)   b ⋅cos(a)   b ⋅sin(a)   b ⋅cos(a)   b ⋅sin(a)   b ⋅cos(a)   b ⋅sin(a)   b ⋅cos(a)
sin(a) + b⋅cos(a) - ───────── - ───────── + ───────── + ───────── - ───────── - ───────── + ───────── + ───────── + O(b**10)
                        2           6           24         120         720         5040       40320       362880

This isn’t very readable but we can clearly see a pattern around \(\sin(a)\) and \(\cos(a)\). Let’s collect terms with respect to those two expressions:

>>> collect(_, [sin(a), cos(a)])
⎛   9       7      5    3    ⎞          ⎛   8      6    4    2    ⎞
⎜  b       b      b    b     ⎟          ⎜  b      b    b    b     ⎟
⎜────── - ──── + ─── - ── + b⎟⋅cos(a) + ⎜───── - ─── + ── - ── + 1⎟⋅sin(a) + O(b**10)
⎝362880   5040   120   6     ⎠          ⎝40320   720   24   2     ⎠

>>> _.removeO()
⎛   8      6    4    2    ⎞          ⎛   9       7      5    3    ⎞
⎜  b      b    b    b     ⎟          ⎜  b       b      b    b     ⎟
⎜───── - ─── + ── - ── + 1⎟⋅sin(a) + ⎜────── - ──── + ─── - ── + b⎟⋅cos(a)
⎝40320   720   24   2     ⎠          ⎝362880   5040   120   6     ⎠

>>> g = _

We got two subexpression that look very familiar. Let’s expand \(\sin(b)\) in \(b\) around 0 and remove the order term:

>>> sin(b).series(b, 0, 10)
     3     5     7       9
    b     b     b       b
b - ── + ─── - ──── + ────── + O(b**10)
    6    120   5040   362880

>>> _.removeO()
   9       7      5    3
  b       b      b    b
────── - ──── + ─── - ── + b
362880   5040   120   6

This is clearly the second subexpression, so let’s substitute it for \(\sin(b)\):

>>> g.subs(_, sin(b))
⎛   8      6    4    2    ⎞
⎜  b      b    b    b     ⎟
⎜───── - ─── + ── - ── + 1⎟⋅sin(a) + sin(b)⋅cos(a)
⎝40320   720   24   2     ⎠

>>> h = _

Now let’s repeat this procedure for \(\cos(b)\):

>>> cos(b).series(b, 0, 10)
     2    4     6      8
    b    b     b      b
1 - ── + ── - ─── + ───── + O(b**10)
    2    24   720   40320

>>> _.removeO()
   8      6    4    2
  b      b    b    b
───── - ─── + ── - ── + 1
40320   720   24   2

>>> h.subs(_, cos(b))
sin(a)⋅cos(b) + sin(b)⋅cos(a)

This gave us a formula for \(\sin(a + b)\):

>>> Eq(f, _)
sin(a + b) = sin(a)⋅cos(b) + sin(b)⋅cos(a)

There is, however, a much simpler way to get the same result:

>>> Eq(f, sin(a + b).expand(trig=True))
sin(a + b) = sin(a)⋅cos(b) + sin(b)⋅cos(a)

Tasks

1. Repeat this procedure but expand with respect to \(a\) in the first step.

   (solution)

2. Use this procedure to derive a formula for \(\cos(a + b)\).

   (solution)