## Precalculus (6th Edition) Blitzer

The amplitude is $\pm \sqrt{13}$ and the period is $2\pi$.
The given equation is provided: $d=2\cos t+3\sin t$ Multiply and divide the given equation by $\sqrt{{{2}^{2}}+{{3}^{2}}}=\sqrt{13}$. Then we get $d=\left( \frac{2}{\sqrt{13}}\cos t+\frac{3}{\sqrt{13}}\sin t \right)\sqrt{13}$ The $\cos$ is computed by dividing the base and hypotenuse and sine is computed by dividing the perpendicular and hypotenuse. It is given that $\tan \theta =\frac{3}{2}$. Hence, assume a right triangle with the perpendicular $=3k$ and base $=2k$ for some constant $k$ and let the hypotenuse be ‘H’. By using Pythagoras theorem: \begin{align} & {{\left( 2k \right)}^{2}}+{{\left( 3k \right)}^{2}}={{H}^{2}} \\ & H=\left( \sqrt{{{2}^{2}}+{{3}^{2}}} \right)k \\ & H=\sqrt{13}k \end{align} Then, we further solve: \begin{align} & \theta =\frac{3k}{\sqrt{13}k} \\ & =\frac{3}{\sqrt{13}}\text{ } \end{align} Now, the value of cos is: \begin{align} & \cos \theta =\frac{2k}{\sqrt{13}k} \\ & =\frac{2}{\sqrt{13}} \end{align} Put the values: \begin{align} & d\left( t \right)=\left( cos\theta \cdot \cos t+\sin \theta \cdot \sin t \right)\sqrt{13} \\ & =\left( \cos t\cdot \cos \theta +\sin t\cdot \sin \theta \right)\sqrt{13} \\ & =\cos \left( t-\theta \right)\sqrt{13}\text{ } \end{align} Further solve the equation: \begin{align} & d\left( t \right)=2\cos t+3\sin t \\ & =\sqrt{13}\cos \left( t-\theta \right)\text{ } \end{align} By applying the aforementioned formula: $\left( \cos A\cos B+\sin A\sin B=\cos \left( A-B \right) \right)$ Hence, it is proved. The amplitude will be $\sqrt{13}\left( \text{maximum or minimum of cos}\left( t-\theta \right) \right)$: \begin{align} & \text{Amplitude}=\sqrt{13}\left( \pm 1 \right) \\ & =\pm \sqrt{13} \end{align} The period of the ball's motion will be $\text{cos}\left( t-\theta \right)$ $=2\pi$.