Question 886357
<pre>
sin(2A) = cos(12B)

cos(90°-2A) = cos(12B)

cos(90°-2A) - cos(12B) = 0

We use the identity:  

{{{cos(x)-cos(y)}}}{{{""=""}}}{{{-2sin((x+y)/2)sin((x-y)/2)}}} with x=90°-2A and y=12B

{{{-2sin(("90°"-2A+12B)/2)sin(("90°"-2A-12B)/2)}}}{{{""=""}}}{{{"0"}}}

Divide both sides by -2 and simplify the angles

{{{sin("45°"-A+6B)sin("45°"-A-6B)}}}{{{""=""}}}{{{"0"}}}

{{{sin("45°"-(A-6B))sin("45°"-(A+6B))}}}{{{""=""}}}{{{"0"}}}

Use zero-factor property:

Since we are only interested in A+6B we set only the second
factor = 0

{{{sin("45°"-(A+6B))}}}{{{""=""}}}{{{"0"}}}

Use the identity {{{sin(x)=-sin(x)}}} to make (A+6B) positive:

{{{-sin(-"45°"+(A+6B))}}}{{{""=""}}}{{{"0"}}}

Divide both sides by -1 and reverse the terms

{{{sin((A+6B)-"45°")}}}{{{""=""}}}{{{"0"}}}

{{{(A+6B)-"45°"}}}{{{""=""}}}{{{"180°"*n}}}

{{{A+6B}}}{{{""=""}}}{{{"45°"+"180°"*n}}}

{{{tan(A+6B)}}}{{{""=""}}}{{{tan("45°"+"180°"*n)}}

{{{tan(A+6B)}}}{{{""=""}}}{{{1}}}

Edwin</pre>