trigonometric linear combination

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No answer. Okay, I,ll assume this is not homework (if it is, you'll loose the benefit of finding this by yourself ; on your head be it)...):

The "naive" solution" coesn't work:

sage: var("a, b, c, d")
(a, b, c, d)
sage: E1=a*sin(x)+b*cos(x)==c*sin(x+d); E1
b*cos(x) + a*sin(x) == c*sin(d + x)
sage: solve(E1,[c,d])
[[c == (b*cos(x) + a*sin(x))/sin(r1 + x), d == r1]]

This "solution" depends on x. since we are looking for a general solution this can't be accepted. In fact, our equation, which we can rewrite as :

sage: E1.trig_expand()
b*cos(x) + a*sin(x) == (cos(x)*sin(d) + cos(d)*sin(x))*c

must hold for any x. In particular,

• it must hold for $\cos(x)=0$:

This is interesting:

sage: E1.trig_expand().subs(cos(x)==0)
a*sin(x) == c*cos(d)*sin(x)

Since $\cos(x)=0\Rightarrow \sin(x)\in \left(-1,1\right)$, this gives us a"special case" equation:

sage: E2=E1.trig_expand().subs(cos(x)==0)/sin(x); E2
a == c*cos(d)

• It must hold for $\sin(x)=0$.

By a similar reasoning, we get another "special case" equation:

sage: E3=E1.trig_expand().subs(sin(x)==0)/cos(x); E3
b == c*sin(d)

E2 and E3 together gives us a solution for d:

sage: S1=(E3/E2).trig_reduce().solve(d); S1
[d == arctan(b/a)]

This solution can be substituted in our (expanded) original equation which can now be solved for c:

sage: S2=E1.trig_expand().subs(S1).solve(c);S2
[c == a*sqrt(b^2/a^2 + 1)]

And we can check that our original equation holds:

sage: E1.subs(S1).subs(S2).trig_simplify()
b*cos(x) + a*sin(x) == b*cos(x) + a*sin(x)

We end up with a tautology. Good...

Left as an exercise for the reader: We have a solution; is this the solution ?