Question 1207441
<pre>
Your problem should have been stated:
Given
{{{sin(5A)/sin(A)+cos(5A)/cos(A)}}}{{{""=""}}}{{{a+b*cos(4A)}}},
Determine the values of a and b.

We look up the formulas for sin(5A), cos(5A), and cos(4A)

{{{sin^""(5A)}}}{{{""=""}}}{{{16sin^5(A)-20sin^3(A)+5sin^""(A)}}}

{{{cos^""(5A)}}}{{{""=""}}}{{{16cos^5(A)-20cos^3(A)+5cos^""(A)}}} 

{{{cos^""(4A)}}}{{{""=""}}}{{{8cos^4(A)-8cos^2(A)+1}}}

{{{(16sin^5(A)-20sin^3(A)+5sin^""(A))/sin^""(A)+(16cos^5(A)-20cos^3(A)+5cos^""(A))/cos^""(A)}}}{{{""=""}}}{{{a+b*(8cos^4(A)-8cos^2(A)+1^""^"")}}}

{{{16sin^4(A)-20sin^2(A)+5+16cos^4(A)-20cos^2(A)+5}}}{{{""=""}}}{{{8b*cos^4(A)-8b*cos^2(A)+a+b}}}

{{{16sin^4(A)-20sin^2(A)+16cos^4(A)-20cos^2(A)+10}}}{{{""=""}}}{{{8b*cos^4(A)-8b*cos^2(A)+a+b}}}

We change the sines in the first two terms to cosines.  To change the first
term:

{{{16sin^4(A)=16(sin^2(A)^"")^2=16(1-cos^2(A))^2=16(1-2cos^2(A)+cos^4(A)^"")=
16-32cos^2(A)+16cos^4(A)}}} 

To change the second term

{{{-20sin^2(A)=-20(1-cos^2(A)^"")= -20+20cos^2(A)}}}

Substituting:

{{{16-32cos^2(A)+16cos^4(A)-20+20cos^2(A)+16cos^4(A)-20cos^2(A)+10}}}{{{""=""}}}{{{8b*cos^4(A)-8b*cos^2(A)+a+b}}}

Simplifying:

{{{32cos^4(A)-32cos^2(A)+6}}}{{{""=""}}}{{{8b*cos^4(A)-8b*cos^2(A)+a+b}}}

This will be an identity if and only if the numerical coefficients on the left
are equal to the literal coefficients of corresponding like terms on the right.

{{{system(32=8b,-32=-8b,6=a+b)}}} 

Either of the first two equations give b=4, substituting in the third
6=a+4 or a=2.

Answer: a=2, b=4

Edwin</pre>