# How to go from a*cosx + b*sinx TO A*cos(wt - phi)?

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1. Mar 10, 2016

### Elvis 123456789

Moved from non-homework forum section, so homework template is not present.
Express Cos(t - pi/8) + Sin(t - pi/8) in the form A*Cos(wt - phi).

I got sqrt(2)*Cos(t-3pi/4).

Not sure if thats right though

Last edited by a moderator: Mar 13, 2016
2. Mar 10, 2016

The simplest way to see this is to use $Acos(\omega*t)+Bsin(\omega*t)=sqrt(A^2+B^2)*(A/srqrt(A^2+B^2))cos(\omega*t)+(B/sqrt(A^2+B^2))sin(\omega*t)$
Let $A/sqrt(A^2+B^2)=cos(\phi)$ and $B/sqrt(A^2+B^2)=sin(\phi)$ You should recognize the expanded terms as $cos(\omega*t-\phi)$ with a sqrt(A^2+B^2) in front of it and $\phi$ is the angle that has $tan(\phi)=B/A$ This actually comes up quite a lot in different physics calculations and is something worth looking over a couple of times until you have it completely memorized. I get $3\pi/8$ though. Please check the arithmetic.

3. Mar 11, 2016

### Qwertywerty

Thanks.

4. Mar 13, 2016

### SammyS

Staff Emeritus