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

[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 that's right though

 

[Charles Link]

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.

 

[Qwertywerty]

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 that's right though

Please post your working, the next time you ask a question. It saves everybody's time.
Thanks.

 

[SammyS]

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 that's right though

You can check your answer as follows:

−3π/4 = −π/8 − 5π/8

Therefore, use the angle addition identity for cosine on:

cos( (t−π/8) − 5π/8 )