Question

sin 2t×cos3t how to solve this one

Answer 1

Use trigonometric formulas: 

cosu·sinv = 0.5( sin(u + v) - sin(u - v)] 

sinu ·sinv = 0.5 ( cos(u - v) - cos(u + v)] 

Then the first function 

cos(3t)·sin(2t) = 0.5 ( sin(5t) - sin(t) ) 

Now take Laplace transform 

L[cos(3t)·sin(2t)] = 0.5L[sin(5t) - sin(t)] = 0.5 L[sin(5t)] - 0.5L[sin(t)] = 

= 0.5 · 5/(s²+5²) - 0.5 ·1/(s² + 1²) = 5/[2(s²+25)] - 1/[2(s² + 1)] 

Answer 2

Answer:

Step-by-step explanation:

sin2t cos3t

By first solving trigonometrically:

=1/2(sin (2t+5t) + sin(2t-5t))

(As, sinxcosy=1/2sin(x+y) + sin(x-y))

=1/2( sin5t - sint)

Now, apply laplacian formulas

=1/2(5/s²+25 - 1/s²+1)

(As, sinat=a/s²+a²)

On further solving, by taking LCM

=1/2*5s²+5-s²-25/(s²+25)(s²+1) or

=4s²-20/2*(s²+25)(s²+1) or

=4s²-10/(s²+25)(s²+1)

Done.