Sinful sum

First, we rewrite the second equation:

$0 = \sin a + \sin b + \sin c + \sin d \\ \ \ = 2 \sin(\frac{a+b}2) \cos(\frac{a-b}2) + 2 \sin(\frac{c+d}2) \cos(\frac{c-d}2)$

From the first equation, it follows that $a+b = -(c+d)$ , hence we get:

$0 = 2 \sin(\frac{a+b}2) (\cos(\frac{a-b}2) - \cos(\frac{c-d}2))$

For this to hold, one (or more) of three things must be true:

1. $\sin{\frac{a+b}2} = 0 \iff a+b=0 \text{ and } c+d=0$
2. $\frac{a-b}2 = \frac{c-d}2 \iff a+d=0 \text{ and } b+c=0$
3. $\frac{a-b}2 = -\frac{c-d}2 \iff a+c=0 \text{ and } b+d=0$

In each case, the relation $a+b+c+d=0$ was used. (For example, in the second case, we have $a-b-c+d = 0$ and $a+b+c+d=0$ . Adding and subtracting these two relations gives the two relations given (times two).)

We now use the inclusion-exclusion principle to count all quadruples that meet at least one of these criteria.

First of all, there are $21^2$ quadruples that meet the first criterion: $a$ and $c$ can take on any value, after which $b$ and $d$ are fixed. The same is true for the second and third criterion, giving a total of $3 \cdot 21^2$ .

However, all quadruples that meet both the first and the second criterion were counted twice. We should thus subtract them. These quadruples meet the relation is $a = -b = c = -d$ , of which there are trivially $21$ . We have the same number of quadruples that meet both the first and the third, and the second and the third criterion. In total, we subtract $3 \cdot 21$ .

That leaves the single quadruple $(0,0,0,0)$ that satisfies all three criteria. We added it three times, then subtracted it three times, so it needs to be added again.

The net total is thus $3 \cdot 21^2 - 3 \cdot 21 + 1 = \boxed{1261}$

We consider three cases:

• * Case I *: four 0's

• * Case II *: two 0's

• * Case III *: no 0

Note that $|a|,|b|,|c|,|d|$ are positive integers and not a multiple of $\pi$ , sum of sines of them can never be $0$ .

Let $A'$ be the addictive inverse of $A$ . e.g: if $A=6$ , $A' = -6$

For Case I : $a=b=c=d=0$ , denote $0000$ as a solution. 1 solution.

For Case II : $00AA'$ with $A = 1,2,3, \ldots ,10$ . So the total arrangements is $\frac {4!}{2!} \cdot 10 = 120$

For Case III : $AA'BB'$ and $AAA'A'$ are solutions with $A \ne B$ .

• For the subcase $AA'BB'$ , total arrangements is $4! \cdot {10 \choose 2} = 1080$ .

• For the subcase $AAA'A'$ , total arrangements is $\frac {4!}{2! \space 2!} \cdot {10 \choose 1} = 60$

Hence, $N = 1 + 120 + 1080 + 60 = 1261 \Rightarrow$ answer is $\boxed{261}$

Since $a+b+c+d=0$ , we have $a+b=-(c+d)$ ; hence $\sin(a)+\sin(b)+\sin(c)+\sin(d)=2\sin((a+b)/2)\cos((a-b)/2)+2\sin((c+d)/2)\cos((c-d)/2)=2\sin((a+b)/2)[\cos((a-b)/2)-\cos((c-d)/2)]=2\sin((a+b)/2)*2\sin((a-b+c-d)/4)\sin((a-b-c+d)/4)$ ;

Noting that $a,b,c$ ,and $d$ are integers, we obtain that $a+b=c+d=0$ , or $a+c=b+d=0$ , or $a+d=b+c=0$ . And the total number of solutions is 1261;

Before you're tempted to disdain a computer solution notice that it exposes a lovely pattern. Generate all possible tuples then look for those where values 'cancel'. No need to calculate values of the sine function.

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count=0
for a,b,c,d in ((a,b,c,d) for a in range(-10,11) \
    for b in range(-10,11) for c in range(-10,11) \
    for d in range(-10,11) if a+b+c+d==0):

    m,n,p,q=sorted([a,b,c,d])
    count+=m==-q and n==-p

print (count)