# Commentary on "Simplify trig expressions with the double angle formula"

A bit of reflexion leads me to propose a "brute force" answer to a question wher the answer could be divined with a bit of "intuition".

The original question was :

I am trying to simplify the following expression in sage:

(sqrt(3)/3cos(x)+1/3sin(x))

the resulting expression should be:

2/3*cos(pi/6-x) .simplify_full(), trig_reduce(), or simplify_trig() cannot produce this simplification.

Is sage currently capable of doing this?

See below for the steps of the solution. Comments, criticisms and even lazzi welcome...

edit retag close merge delete

Sort by » oldest newest most voted
var("x,y,k")
f=(sqrt(3)/3*cos(x)+1/3*sin(x))


The original expression is of the form $a\cos x + b\sin x$. if it turns out that $a^2+b^2=1$, we can rewrite this as either $\sin(x+y)$ or $\cos(x+y)$, which will give us a solution. So, let's factor out a constant leaving the sum of squares of the coeffs of trig_function(x) at 1 :

k=sum([f.coefficient(g(x),1)^2 for g in [sin,cos]]).sqrt()
print k


2/3

Now that this is out of the way, let's search systematically :

fr=f/k
h1=sin(x+y).trig_expand()
L1=[fr.coefficient(g(x),1)==h.coefficient(g(x),1) for g in [sin,cos]]
print L1


[(1/2) == cos(y), 1/2*sqrt(3) == sin(y)]

print solve(L1,y)


[[y == 1/3pi + 2pi*z388]]

Check that solution :

bool(fr.subs(solve(L1,y))*k==f)


True

One notes with interest that this check does not require the specification of the free integer constant z248.

Similarly:

h2=cos(x+y).trig_expand()
L2=[fr.coefficient(g(x),1)==h.coefficient(g(x),1) for g in [sin,cos]]
print solve(L2,y)


[[y == -1/6pi + 2pi*z402]]

bool(fr.subs(solve(L2,y))*k==f)


True

This second expression is the "original" sought solution.

more