QuestionHere is a little more review concerning trig functions. Using the formula for $\sin ($ and $\cos \theta$ of the sum of two angles. $$\begin{array}{l} 3 \cos (5 x-2)=3 \cos (2) \\ 3 \sin (2 x-2)=-3 \sin (2)-\square \cos (5 x)-\square \sin (5 x) \\ \cos (2 x)+3 \cos (2) \end{array}$$
Now reverse this formula and given the expanded version find the version with just one term. This involves solving a pair of equations -in order to get $A \cos (x)+B \sin (x)=R \sin (x+b)=R \sin (b) \cos (x)+R \cos (b) \sin (x)$ what values must you choose for $R$ and $b$ ? (Match coefficients.) By convention we'll assume that the amplitude (the first coefficient on the left hand side) is positive. $$\begin{array}{l} \cos (5 x+\square)=4 \cos (5 x)+-2 \sin (5 x) \\ \sin (2 x+\arctan (3) \quad)=6 \cos (2 x)+2 \sin (2 x) \end{array}$$
The upshot of this exercise is that we can always rewrite the sum of multiples of $\sin ()$ and $\cos 0$ as a single $\sin 0$ function with a given amplitude and phase shift. We could also write it as a single $\cos ($ ), but it would have a different phase in that case. We'll use this many times in interpreting results.