Question

Find the Laplace transform of \(f(t)=\sin 2t \sin 3t\).

• A. \(\dfrac{12s}{(s^2+4)(s^2+9)}\)
• B. \(\dfrac{12s}{(s^2+1)(s^2+9)}\)
• C. \(\dfrac{12s}{(s^2+1)(s^2+25)}\)
• D. \(\dfrac{12s}{(s^2+4)(s^2+25)}\)

Hint:

For products like \(\sin A\sin B\), first convert them using trigonometric identities, then apply the standard Laplace transform formula.

Answer 1

The Correct Option is C

Concept:
To find the Laplace transform of a product of two sine functions, we first use the trigonometric identity: \[ \sin A \sin B = \frac{1}{2}\left[\cos(A-B)-\cos(A+B)\right] \]

Step 1: Apply the identity. \[ \sin 2t \sin 3t = \frac{1}{2}\left[\cos(2t-3t)-\cos(2t+3t)\right] \] \[ \sin 2t \sin 3t = \frac{1}{2}\left[\cos(-t)-\cos 5t\right] \] Since \(\cos(-t)=\cos t\), \[ \sin 2t \sin 3t = \frac{1}{2}\left[\cos t-\cos 5t\right] \]

Step 2: Take Laplace transform. \[ L\{\sin 2t\sin 3t\} = \frac{1}{2}\left[L\{\cos t\}-L\{\cos 5t\}\right] \]

Step 3: Use the standard formula. \[ L\{\cos at\}=\frac{s}{s^2+a^2} \] So, \[ L\{\cos t\}=\frac{s}{s^2+1} \] and \[ L\{\cos 5t\}=\frac{s}{s^2+25} \]

Step 4: Substitute and simplify. \[ L\{\sin 2t\sin 3t\} = \frac{1}{2}\left[\frac{s}{s^2+1}-\frac{s}{s^2+25}\right] \] \[ = \frac{s}{2}\left[\frac{(s^2+25)-(s^2+1)}{(s^2+1)(s^2+25)}\right] \] \[ = \frac{s}{2}\left[\frac{24}{(s^2+1)(s^2+25)}\right] \] \[ = \frac{12s}{(s^2+1)(s^2+25)} \] \[ \therefore \text{Correct Answer is (C)} \]