First let’s look at some trig identities.
sin(A+B)=sinAcosB+cosAsinB, sin(A-B)=sinAcosB-cosAsinB,
Therefore sin(A+B)+sin(A-B)=2sinAcosB.
cos(A+B)=cosAcosB-sinAsinB, cos(A-B)=cosAcosB+sinAsinB,
Therefore cos(A-B)-cos(A+B)=2sinAsinB.
sin(x)+sin(2x)+sin(3x)+sin(4x) is the same as:
(sin(x)+sin(3x))+(sin(2x)+sin(4x)).
If A-B=x and A+B=3x, then 2A=4x, A=2x and B=x, and:
sin(3x)+sin(x)=2sin(2x)cos(x).
If A-B=2x and A+B=4x, then 2A=6x, A=3x and B=x, and:
sin(4x)+sin(2x)=2sin(3x)cos(x).
So, putting these together:
sin(x)+sin(2x)+sin(3x)+sin(4x)=2cos(x)(sin(2x)+sin(3x)).
Similarly, we can write the denominator as:
(cos(x)-cos(3x))+(cos(2x)-cos(4x)).
Let A-B=x and A+B=3x (as before) then:
cos(x)-cos(3x)=2sin(2x)sin(x).
And let A-B=2x and A+B=4x, then:
cos(2x)-cos(4x)=2sin(3x)sin(x).
Putting these together:
cos(x)+cos(2x)-cos(3x)-cos(4x)=2sin(x)(sin(2x)+sin(3x)).
When we divide we are left with cos(x)/sin(x)=cot(x) QED, because the term 2(sin(2x)+sin(3x)) cancels out.