Show that ${\mathit{C}}_{\mathbf{1}}\mathit{c}\mathit{o}\mathit{s}\mathit{\omega }\mathit{t}\mathbf{+}{\mathit{C}}_{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathit{\omega }\mathit{t}$ can be written in the form $\mathit{A}\mathit{c}\mathit{o}\mathit{s}\left(\omega t-\varphi \right)$, where $\mathit{A}\mathbf{=}\sqrt{{\mathbf{C}}_{\mathbf{1}}^{\mathbf{2}}\mathbf{+}{\mathbf{C}}_{\mathbf{2}}^{\mathbf{2}}}$and $\mathit{t}\mathit{a}\mathit{n}\mathit{\varphi }\mathbf{=}{\mathit{C}}_{\mathbf{2}}\mathbf{/}{\mathit{C}}_{\mathbf{1}}$. [Hint: Use a standard trigonometric identity with ${\mathit{C}}_{\mathbf{1}}\mathbf{=}\mathit{A}\mathit{c}\mathit{o}\mathit{s}\mathit{\varphi }\mathbf{,}{\mathit{C}}_{\mathbf{2}}\mathbf{=}\mathit{A}\mathit{s}\mathit{i}\mathit{n}\mathit{\varphi }$.] Use this fact to verify the alternate representation (8) of F(t) discussed in Example 2 on page 104.

For the solution use a standard trigonometric identity with $C_1=Acos\varphi ,C_2=Asin\varphi$.

In $C_{1}cos\omega t+C_{2}sin\omega t$, we will use $C_1=Acos\varphi ,C_2=Asin\varphi$

Therefore,

$$\begin{gathered} C_{1}cos\omega t+C_{2}sin\omega t=Acos\varphi cos\omega t+Asin\varphi sin\omega t \\ =A\left(cos\varphi cos\omega t+sin\varphi sin\omega t\right) \\ =Acos\left(\omega t-\varphi \right)(U\sin gidentitycos\left(A-B\right)=cosAcosB+sinAsinB) \end{gathered}$$

Hence, role="math" localid="1664172041931" ${\mathit{C}}_{\mathbf{1}}\mathit{c}\mathit{o}\mathit{s}\mathit{\omega }\mathit{t}\mathbf{+}{\mathit{C}}_{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathit{\omega }\mathit{t}$can be written in the form role="math" localid="1664172054217" $\mathit{A}\mathit{c}\mathit{o}\mathit{s}\left(\omega t-\varphi \right)$.

We have, $F\left(\mathrm{t}\right)={[1+(\mathrm{\omega }/\mathrm{k}\left)\right]}^{-\frac{1}{2}}cos(\mathrm{\omega t}-\mathrm{\varphi })$

Comparing its R.H.S. with $Acos\left(\omega t-\varphi \right)$,

$$\begin{gathered} A={[1+(\mathrm{\omega }/\mathrm{k}\left)\right]}^{-\frac{1}{2}} \\ A=\frac{K}{\sqrt{K^2+{\omega }^2}} \end{gathered}$$

Now as given $C_1=Acos\varphi ,C_2=Asin\varphi$

Therefore,

$C_1=\frac{K}{\sqrt{K^2+{\omega }^2}}cos\varphi ,C_2=\frac{K}{\sqrt{K^2+{\omega }^2}}sin\varphi$

Substituting these values of C₁ and C₂ in $A=\sqrt{C_1^2+C_2^2}$,

$$\begin{gathered} A=\sqrt{\frac{K^2}{K^2+{\omega }^2}co{s}^2\varphi +\frac{K^2}{K^2+{\omega }^2}si{n}^2\varphi } \\ A=\sqrt{\frac{K^2}{K^2+{\omega }^2}\left(co{s}^2\varphi +si{n}^2\varphi \right)} \\ A=\sqrt{\frac{K^2}{K^2+{\omega }^2}} \\ A=\frac{K}{\sqrt{K^2+{\omega }^2}} \\ A=\frac{1}{\sqrt{1+{(\omega /k)}^2}} \end{gathered}$$

Hence, the representation role="math" localid="1664173344588" $\mathit{F}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{[}\mathbf{1}\mathbf{+}\mathbf{(}\mathbf{\omega }\mathbf{/}\mathbf{k}\mathbf{\left)}\mathbf{\right]}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathbf{\omega t}\mathbf{-}\mathbf{\varphi }\mathbf{)}$is verified according to the given conditions .