Problem involving cos + sin: express 5cos(3t) + 12sin(3t) as "A cos(wt + a)"

[Bazoya]

Problem involving cos + sin: express 5cos(3t) + 12sin(3t) as "A cos(wt + a)"

\(\displaystyle \mbox{Express }\, 5\, \cos(3t)\, +\, 12\, \sin(3t),\, \mbox{ in the form}\)

\(\displaystyle A\, \cos(\omega t\, +\, \alpha),\, \alpha\, \geq\, 0\, \mbox{ and can be written}\)

\(\displaystyle \mbox{as degrees or radians.}\)

If any help as to how to solve this type of problem (not just this one in particular) I would be incredibly grateful.

Thanks

Bazoya

 

[skaa]

You should represent your expression as \(\displaystyle A(\frac{5}{A}\cos(3t)+\frac{12}{A}\sin(3t))\)
where \(\displaystyle (\frac{5}{A})^2+(\frac{12}{A})^2=1\)
\(\displaystyle A^2=169\)
\(\displaystyle A=13\)
\(\displaystyle 13(\frac{5}{13}\cos(3t)+\frac{12}{13}\sin(3t))=13(\sin(\arcsin\frac{5}{13})\cos(3t)+\cos( \arcsin\frac{5}{13} )\sin(3t))=13\sin(3t+\arcsin\frac{5}{13})\)

 

[HallsofIvy]

You need to know that cos(A+B)= cos(A)cos(B)- sin(A)sin(B). So cos(wt+ a)= cos(a)cos(wt)- sin(a)sin(wt).

To get 5 cos(3t)+ 12 sin(3t) from that formula we would have to have cos(a)= 5 and sin(a)= -12. That is, of course, impossible, because cosine and sin must be between -1 and 1. Further, \(\displaystyle sin^2(a)+ cos^2(a)\) must equal 1.

We can fix that by multiplying and dividing by \(\displaystyle \sqrt{5^2+ 12^2}= \sqrt{169}= 13\): \(\displaystyle 13\left(\frac{5}{13}cos(wt)- \frac{12}{13}sin(wt)\right)\). Now, those numbers are between -1 and 1 and \(\displaystyle \left(\frac{5}{13}\right)^2+ \left(\frac{12}{13}\right)^2= 1\). That means we must have \(\displaystyle cos(a)= \frac{5}{13}\) and \(\displaystyle sin(a)= -\frac{12}{13}\). What is a?