Laplace transformation of L[ cos^2 3t]
Answers
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)
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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