# Representing sum of cosine and sine as a single cosine expression

a.cos(wt) + b.sin(wt) = M.cos(wt + ϕ)

Can you give me M and ϕ in terms of a and b?

arildno
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cos(u+v)=cos(u)cos(v)-sin(u)sin(v)

That 's all you need.

The final representation was something like
M = sqrt(a^2 + b^2)
and
ϕ = arctan(-b/a)
but I'm no sure.

Can anyone confirm it for me?

arildno
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The final representation was something like
M = sqrt(a^2 + b^2)
and
ϕ = arctan(-b/a)
but I'm no sure.

Can anyone confirm it for me?
No.

Try to do it for yourself, and we can correct whatever mistakes you make.

Code:
M.cos(wt + ϕ) = a.cos(wt) + b.sin(wt)

cos(wt + ϕ) = (a/M).cos(wt) + (b/M).sin(wt)........(I)

cos(wt + ϕ) = cos(wt).cos(ϕ) - sin(wt).sin(ϕ)........(II)

From (I) and (II),
cos(ϕ) = (a/M)
sin(ϕ) = -(b/M)

cos^2(ϕ) + sin^2(ϕ) = (a^2 + b^2)/(M^2) = 1

We assume that M is always positive and we keep any negativity in the phase angle ϕ,
M = sqrt(a^2 + b^2)

sin(ϕ)/cos(ϕ) = tan(ϕ) = -(b/M)/(a/M) = -b/a

tan(ϕ) = -b/a   ==>   ϕ = arctan(-b/a)
Is there anything wrong in my derivation?

arildno
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Code:
M.cos(wt + ϕ) = a.cos(wt) + b.sin(wt)

cos(wt + ϕ) = (a/M).cos(wt) + (b/M).sin(wt)........(I)

cos(wt + ϕ) = cos(wt).cos(ϕ) - sin(wt).sin(ϕ)........(II)

From (I) and (II),
cos(ϕ) = (a/M)
sin(ϕ) = -(b/M)

cos^2(ϕ) + sin^2(ϕ) = (a^2 + b^2)/(M^2) = 1

We assume that M is always positive and we keep any negativity in the phase angle ϕ,
M = sqrt(a^2 + b^2)

sin(ϕ)/cos(ϕ) = tan(ϕ) = -(b/M)/(a/M) = -b/a

tan(ϕ) = -b/a   ==>   ϕ = arctan(-b/a)
Is there anything wrong in my derivation?
No.

I liked your way of "you must do it yourself if you want to success"... :)

arildno
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I liked your way of "you must do it yourself if you want to success"... :)
It's true, isn't it?

Yeah, either I don't like the ones who ask a big problem and wait for others to solve it for him.
In my question, this is a simple trigonometric identity, I expected a mathematician to write it for me since most of math guys have memorized and actively use these kind of identities.

Landau
You could also do the following:

t=0 gives a=M cos(ϕ),
t=pi/(2w) gives b=-M sin(ϕ).

Hence

a^2+b^2=M^2,
tan(ϕ)=-b/a.

You could also do the following:

t=0 gives a=M cos(ϕ),
t=pi/(2w) gives b=-M sin(ϕ).

Hence

a^2+b^2=M^2,
tan(ϕ)=-b/a.
Wow, that's super simple, thanks!