a machine is subject to two vibrations.
one vibration has the form : 2 cos wt and the other has the form 3cos(wt + pie/4)
Determine resulting vibration and express it in the general form of n cos (wt+-a)
$y=2\cos(\omega t)+3[\cos(\omega t)\cos(\pi/4)-\sin(\omega t)\sin(\pi/4)]$
$y=2\cos(\omega t) + \dfrac{3\sqrt{2}}{2}[\cos(\omega t)-\sin(\omega t)]$
$y=\dfrac{4+3\sqrt{2}}{2}\cos(\omega t) - \dfrac{3\sqrt{2}}{2}\sin(\omega t)$
note ... $a\cos(x)+b\sin(x)=R\cos(x-\alpha)$
Let $a=\dfrac{4+3\sqrt{2}}{2}$ and $b=-\dfrac{3\sqrt{2}}{2}$
$R=\sqrt{a^2+b^2} \implies R \approx 4.635$
$\alpha = \arctan(b/a) \approx -0.475$
$y=4.635\cos(\omega t + 0.475)$
I prefer key lime pie.
Big hint:
$\displaystyle 2 ~ cos( \omega t) + 3 ~ cos \left ( \omega t + \frac{\pi}{4} \right )$
Use the identity $\displaystyle cos(a + b) = cos(a)~cos(b) - sin(a)~sin(b)$ to rewrite this in the form $\displaystyle a~cos(\omega t) + b~sin(\omega t + \phi)$
where $\displaystyle c ~ \sqrt{a^2 + b^2}$ and $\displaystyle \phi = atan2 \left ( \frac{b}{a} \right )$. This is a standard, but little used function. It is defined in terms of the usual $\displaystyle tan^{-1}(x)$ function.
Can you fill in the details?
-Dan
Edit: skeeter got to it first. I'll have to take him to an empty field and do some "skeet" shooting.