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The Laplace transform of f(t) = cos² t is equal to the following expression: [tex]\mathcal {L} \{\cos^{2} t\} = \frac{1}{2\cdot s} - \frac{s}{2\cdot (s^{2}+2)}[/tex].
In this we need to rewrite the given function in terms of sines and cosines, whose Laplace transforms are well known. There is the following trigonometric formula:
[tex]\cos ^{2} t = \frac{1-\cos 2t}{2}[/tex] (1)
Now we proceed to apply the Laplace transforms:
[tex]\mathcal {L} \{f(t)\} = \frac{1}{2}\cdot \mathcal {L} \left\{1 \right\}-\frac{1}{2}\cdot \mathcal {L}\left\{\cos 2t\right\}[/tex]
[tex]\mathcal {L} \{\cos^{2} t\} = \frac{1}{2\cdot s} - \frac{s}{2\cdot (s^{2}+2)}[/tex]
The Laplace transform of f(t) = cos² t is equal to the following expression: [tex]\mathcal {L} \{\cos^{2} t\} = \frac{1}{2\cdot s} - \frac{s}{2\cdot (s^{2}+2)}[/tex]. [tex]\blacksquare[/tex]
The statement is poorly formatted, the correct form is shown below:
Find [tex]\mathcal {L}\{f(t)\}[/tex] by first using a trigonometric expression. (Write your answer as a function of s). f(t) = cos² t
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