Question 1204376
<pre>

{{{S}}}{{{""=""}}}{{{cos(x + theta) + cos(2x + 3theta) + cos(3x + 5theta) + cos(4x + 7theta) + ""*""*""*"" + cos(40x + 79theta)}}}

{{{S}}}{{{""=""}}}{{{sum((k+1)x+(2k+1)theta,k=0,40-1)}}}

We use this formula from 

https://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro

{{{sum(cos(a+kd),k=0,n-1)}}}{{{""=""}}}{{{expr(sin(n*expr(d/2))/sin(d/2))}}}{{{""*""}}}{{{cos((2a+(n-1)d)/2)}}}

We rewrite 
{{{(k+1)x+(2k+1)theta}}}{{{""=""}}}{{{kx+x+2k*theta+theta}}}{{{""=""}}}{{{k+theta+kx+2k*theta)}}}{{{""=""}}}{{{(x+theta)+k(x+2theta)}}}

So we substitute {{{a=x+theta}}}, {{{d=x+2theta}}}, {{{n=40}}}

{{{sum(cos(a+kd),k=0,n-1)}}}{{{""=""}}}{{{expr(sin(n*expr(d/2))/sin(d/2))}}}{{{""*""}}}{{{cos((2a+(n-1)d)/2)}}}


{{{sum(cos((x+theta)+k(x+2theta)),k=0,40-1)}}}{{{""=""}}}{{{expr(sin(40*expr((x+2theta)/2))/sin((x+2theta)/2))}}}{{{""*""}}}{{{cos((2(x+theta)+(40-1)(x+2theta))/2)}}}{{{""=""}}}{{{expr(sin(40*expr((x+2theta)/2))/sin((x+2theta)/2))}}}{{{""*""}}}{{{cos((2x+2theta+39x+78theta)/2)}}}

{{{S}}}{{{""=""}}}{{{(sin(20(x+2theta)^"")cos(expr(41/2)x+40theta^""))/(sin(x/2+theta))}}}

Edwin</pre>