Solution:

Given: \(\displaystyle{f{{\left({t}\right)}}}={\cos{{\left({R}{t}-{7}\right)}}}\) take R as 70

\(\displaystyle\Rightarrow{f{{\left({t}\right)}}}={\cos{{\left({70}{t}-{7}\right)}}}\)

Step 2

Note that \(\displaystyle{\cos{{\left({A}-{B}\right)}}}={\cos{{A}}}\cdot{\cos{{B}}}+{\sin{{A}}}\cdot{\sin{{B}}}\)

\(\displaystyle\Rightarrow{\cos{{\left({70}{t}-{7}\right)}}}={\cos{{\left({70}{t}\right)}}}\cdot{\cos{{7}}}+{\sin{{\left({70}{t}\right)}}}\cdot{\sin{{7}}}\)

Now , \(L\left\{f(t)\right\}\)

\(=L\left\{\cos(70t-7)\right\}\)

\(=L\left\{\cos(70t) \cdot \cos 7 + \sin (70t) \cdot \sin 7 \right\}\)

\(=\cos 7 \cdot L\left\{\cos(70t)\right\}+\sin 7 \cdot L\left\{\sin(70t)\right\}\)

\(\displaystyle{\left(\because{\sin{{7}}}\ \text{ and }\ {\cos{{7}}}\ \text{ are constants}\right)}\)

\(\displaystyle={\cos{{7}}}{\left[{\frac{{{s}}}{{{s}^{{2}}+{\left({70}\right)}^{{2}}}}}\right]}+{\sin{{7}}}{\left[{\frac{{{70}}}{{{s}^{{2}}+{\left({70}\right)}^{{2}}}}}\right]}\)

\((\because L\left\{\sin at\right\}=\frac{a}{s^2+a^2} , L\left\{\cos at\right\}=\frac{s}{s^2+a^2})\)

\(\displaystyle={\frac{{{s}\cdot{\cos{{\left({7}\right)}}}}}{{{s}^{{2}}+{4900}}}}+{\frac{{{70}\cdot{\sin{{\left({7}\right)}}}}}{{{s}^{{2}}+{4900}}}}\)

Hence, Laplace transform of \(\displaystyle{f{{\left({t}\right)}}}={\cos{{\left({70}{t}-{7}\right)}}}\ \text{ is }\ {\frac{{{s}\cdot{\cos{{\left({7}\right)}}}+{70}\cdot{\sin{{\left({7}\right)}}}}}{{{s}^{{2}}+{4900}}}}\)