Question

Laplace transformation of L[ cos^2 3t]​

Answer 1

Step-by-step explanation:

cos^2 x = 1+cos2x/2

cos^2(3t) = 1+cos(2(3t))/2

= 1+cos(6t)/2

= 1/2+1/2×cos(6t)

L{cos^2(3t)} = 1/2L{1} + 1/2×L{cos(6t)} k=6

= 1/2×1/s + 1/2×(s/s^2+6^2)

= 1/2s + 1/2×(s/s^2+36)

Answer 2

L[cos²3t]= 1/2s+ s/2(s²+36)

Given:

cos²(3t)

To Find:

Laplace transform of cos²3t

Solution:

We know that,

cos2x= 2cos²x-1

From this formula, we can find the value of cos²x

2cos²x= cos2x+1

cos²x= (1+cos2x)/2

Put x=3t in above equation:

cos²3t= (1+cos2(3t))/2

cos²3t= (1+cos6t)/2

We can write above in another way i.e.

cos²3t= 1/2+cos6t/2   _(1)

Taking Laplace transform both side of equation_(1)

L[cos²3t]= l[1/2+cos6t/2]

L[cos²3t]= (1/2)L[1]+(1/2)L[cos6t]    _(2)

Now L[1]= 1/s

and L[cos at]= s/(s²+ a²)

Putting a=6

L[cos 3t]= s/(s²+6²)

Putting value of L[1]  and L[cos 3t] in equation _(2)

L[cos²3t]= 1/2s+ s/2(s²+36)

Hence, L[cos²3t]= 1/2s+ s/2(s²+36)

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