## 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?
 Recognitions: Gold Member Homework Help Science Advisor 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?

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

 Quote by hkBattousai 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?

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 Quote by hkBattousai 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"... :)

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 Quote by hkBattousai 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.
 http://www.electro-tech-online.com/m...tml#post895294 Final expression for vL(t) greatly simplified by this trigonometric identity.
 Recognitions: Science Advisor 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.

 Quote by 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.
Wow, that's super simple, thanks!

 Tags cosine, identity, sine, trigonometric, trigonometry