Why does A cos(ωt) + B sin(ωt) equal C cos(ωt - σ)?

[asdf1]

why can y(t) = A cos[omega(t)]+bsin[omega(t)]
equal Ccos[omega(t)-sigma], where C=(A^2 + B^2)^(1/2) and
tan(sigma)=B/A?

 

[James R]

Well

$$\cos(a-b) = \cos a \cos b + \sin a \sin b$$

Try expanding out $C\cos(\omega t - \sigma)$ using that formula, and see what you get.

 

[asdf1]

i got C[cos(wt)cos(sigma)+sin(wt)sin(sigma)]...
am i missing a step?
@@a

 

[James R]

Now define A=Ccos(sigma) and B=Csin(sigma).

What do you get if you eliminate sigma from these two equations?

 

[asdf1]

C[cos(wt)A/C + sin(wt)B/C]= the orignal equation!
thanks!

 

[James R]

Note also that $A^2 + B^2 = C^2$ and tan(sigma) = B/A, with the given definitions.

 

[asdf1]

cool! thanks again! :)