1. ## Trigonometric "transformation"?

Rewrite $\displaystyle 6cos(3t) - 5sin(3t)$ in the form
a. $\displaystyle A_{1}sin(3t + \phi_{1})$
b. $\displaystyle A_{2}cos(3t - \phi_{2})$

I'm a bit puzzled by this. Can anyone offer insight?

--marc

2. Hello, Marc!

This takes some Olympic-level gymnastics . . .

Rewrite $\displaystyle 6\cos3t - 5\sin3t$ in the form:

$\displaystyle (a)\;A\sin(3t + \phi)$

$\displaystyle (b)\;A\cos(3t - \phi)$
I'll do part (a) . . .

We have: .$\displaystyle 6\cos3t - 5\sin3t$ .[1]

Use the two coefficients as legs of a right triangle.

. . Find the hypotenuse: .$\displaystyle h \:=\:\sqrt{6^2 + 5^2} \:=\:\sqrt{61}$

Multiply by $\displaystyle \frac{h}{h}\!:\quad\frac{\sqrt{61}}{\sqrt{61}}\big g[6\cos3t - 5\sin3t\bigg] \;=\;\sqrt{61}\bigg[\frac{6}{\sqrt{61}}\cos3t - \frac{5}{\sqrt{61}}\sin3t\bigg]$ .[2]

Let $\displaystyle \phi$ be in a right triangle with: .$\displaystyle opp = 6,\; adj = 5,\; hyp = \sqrt{61}$
Code:
                  *
* *
__    *   *
√61  *     * 6
*       *
* φ       *
*  *  *  *  *
5
Then: .$\displaystyle \sin\phi \,=\,\frac{6}{\sqrt{61}},\;\;\cos\phi \,=\,\frac{5}{\sqrt{61}}$

Substitute into [2]: . $\displaystyle \sqrt{61}\bigg[\sin\phi\cos3t - \cos\phi\sin3t\bigg]$

Therefore, we have: . $\displaystyle \sqrt{61}\,\sin(3t - \phi)$