Define the following four rational numbers. $$ a = \frac{4243257079864535154162785598448178442416}{41016602865234375} \\ b = -\frac{143308384621912247542172258992236503771301}{1210966757832031250} \\ c = \frac{350687399375274064088342133116344593437371021}{4109863607096484375000} \\ d = -\frac{762284492856611655417326017768244278005511063}{12085448243163671875000} $$ Let $$ p = \frac{501}{10}, \qquad m = \frac{499}{10}. $$ Compute $$ \mathrm{Result} = a \cos [p] + b \cos [m] + c \sin [p] + d \sin [m]. $$ Each term in this sum is roughly $10^{23}$. There is a curious cancellation (using 40 digits) happening amongst these four terms; the correct answer is $\mathrm{Result}=7.32 \times 10^{-18}$.
My question: Where does this cancellation come from, analytically? Can you massage the terms into a form where the cancellation is manifest, and machine precision can evaluate the answer with a semblance of accuracy?
(If you're curious, the result came from the analytic integral of a highly oscillatory function.)
Here's two ideas that might be useful. They are in no way a solution.
Idea 1:
Let $r = 50, s=\frac1{10}$. Then $p = r+s, m = r-s$ so that $\cos(p) =\cos(r+s) =\cos(r)\cos(s)-\sin(r)\sin(s), \cos(m) =\cos(r-s) =\cos(r)\cos(s)+\sin(r)\sin(s), \sin(p) =\sin(r+s) =\sin(r)\cos(s)+\cos(r)\sin(s), \sin(m) =\sin(r-s) =\sin(r)\cos(s)-\cos(r)\sin(s) $.
Therefore
$\begin{array}\\ \mathrm{Result} &= a \cos [p] + b \cos [m] + c \sin [p] + d \sin [m]\\ &= a (\cos(r)\cos(s)-\sin(r)\sin(s)) + b (\cos(r)\cos(s)+\sin(r)\sin(s))\\ &\quad + c (\sin(r)\cos(s)+\cos(r)\sin(s)) + d (\sin(r)\cos(s)-\cos(r)\sin(s))\\ &=\cos(r)(a\cos(s)+b\cos(s)+c\sin(s)-d\sin(s))\\ &\quad+\sin(r)(-a\sin(s)+b\sin(s)+c\cos(s)+d\cos(s))\\ &=\cos(r)((a+b)\cos(s)+(c-d)\sin(s)) +\sin(r)((-a+b)\sin(s)+(c+d)\cos(s))\\ &=\cos(r)(w\cos(s+x)) +\sin(r)(y\sin(s+z))\\ \end{array} $
where $w^2 =(a+b)^2+(c-d)^2, y^2 =(-a+b)^2+(c+d)^2, x =\arctan(\frac{c-d}{a+b}), z =\arctan(\frac{-a+b}{c+d}) $.
Computing these values might be interesting.
Idea 2:
$\begin{array}\\ \mathrm{Result} &= a \cos [p] + b \cos [m] + c \sin [p] + d \sin [m]\\ &= a \cos [p] + c \sin [p]+ b \cos [m] + d \sin [m]\\ &= u \sin [p+v] + w \sin [m+x]\\ \end{array} $
where $u^2 = a^2+c^2, w^2 =b^2+d^2, v = \arctan(\frac{a}{c}), x = \arctan(\frac{b}{d})$.
I don't know where to go from here.