Asinx + bcosy into single trig function

[RsZrg]

I know that asinx + bcosx can be put into a single trig fom, but can also the above sum be put into a single trig function?

 

[jedishrfu]

If you could convert it to this format:

A(cosy*sinx + siny*cosx)

then yes it could be converted to Asin(y+x) as an example.

However, that requires that a and b must have the following relationships:

a=Acosy and b=Asiny

which will not be true in the general case for any a and b.

 

[RsZrg]

Thanks Jedi
I am looking for a general solution for any a and b. Is this possible? I have tried the above method but currently limited by the above restrictions.

 

[jedishrfu]

I don't think there is any hope here. Usually this kind of problem might push someone to use Fourier methods but again your initial solution is probably optimal.

I am not one of the Math mentors but perhaps @Mark44 or @HallsofIvy would have a better answer to your problem.

 

[Mark44]

Not that I'm aware of...

 

[Svein]

Off the top of my head: Let A=\sqrt{a^{2}+b^{2}} and a_{1}=\frac{a}{A}, b_{1}=\frac{b}{A}. Then a_{1}^{2}+b_{1}^{2}=1, therefore you can find an angle φ such that a₁=cos(φ) and b₁=sin(φ).
∴a⋅sin(x)+b⋅cos(x) = A(cos(φ)sin(x)+sin(φ)cos(x))=A⋅sin(φ+x).

 

[pasmith]

If you could convert it to this format:

A(cosy*sinx + siny*cosx)

then yes it could be converted to Asin(y+x) as an example.

However, that requires that a and b must have the following relationships:

a=Acosy and b=Asiny

which will not be true in the general case for any a and b.

Actually it is: \tan y = b/a (there exists exactly one such y \in (-\pi/2, \pi/2) for each possible choice of a and b \neq 0) and A = \sqrt{a^2 + b^2}.

See also here.