trigonometry - Sine series: angle multipliers add to 1

It is known that in an sine series with angles in arithmetic progression (I refer to this question):

$\sum_{k=0}^{n-1}\sin (a+k \cdot d)=\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \sin\biggl( \frac{2 a + (n-1) \cdot d}{2}\biggr)$

What if $k$ does not go from $0$ to $n-1$, but its elements are strictly positive rational numbers,

with $0and $\sum_{i=1}^{n} k_i=1$
and $k$ is monotonically increasing

Is there a way to simplify:

$\sum_{i=1}^{n} \sin (a+k_i \cdot d)$

in a way analogous to the first formula? Maybe using the property that k adds to $1$ and using $\sin(\pi/2)=1$ ??

e.g.:
$k_i = (0.067,0.133,0.200,0.267,0.333)$ (5 increasing elements between $0$ & $1$ which sum to 1)
$a=90$
$d=40$

Sum = $(90+0.067\cdot40)+(90+0.133\cdot40)+(90+0.200\cdot40)+(90+0.267\cdot40)+(90+0.333\cdot40)$

Answer

There is no way to give this a general form. The definition of $k_i$ is too general. I have been trying to come up with a function that would give $k_i$ while also fitting your terms, but I am having difficulty. This would need to have specific values of $k_i$ and each term of the sum would need to be calculated individually. I did try to change the sum into something else (work below), but it seems that this is only more complicated. $$\sum_{i=1}^n \sin (a + k_i d)$$ We can separate the sum using the Trigonometric Addition Formula: $$\sin (\alpha + \beta) = \sin\alpha \cos\beta + \sin\beta \cos\alpha$$ $$\sum_{i=1}^n [\sin (a) \cos (k_i d) + \sin (k_i d) \cos (a)]$$ $$= \sin a \sum_{i=1}^n [\cos(k_i d)] + \cos a \sum_{i=1}^n[\sin(k_i d)]$$ Past this point, there is no general form. You can attempt to use the Multiple-Angle Formulas, but this only seems to complicate things. In conclusion, I believe your best option would be to just use the original sum and calculate it normally. Unless you have a more rigorous definition for $k_i$, there is no general form that will apply.