Question

Find Laplace transform of cost cos3t cos5t​

Answer 1

Answer:

Step-by-step explanation:

Note that

cosAcosB=12(cos(A−B)+cos(A+B))(1)

So, if we apply this identity to the first two cosines (the ones with arguments t

and 2t

), we obtain:

cos(t)cos(2t)=12(cos(−t)+cos(3t))

Hence,

cos(t)cos(2t)cos(3t)=12⎛⎝⎜⎜cos(−t)cos(3t)★+cos(3t)cos(3t)♣⎞⎠⎟⎟

We’ll apply (1)

to ★

to obtain:

cos(−t)cos(3t)=12(cos(−4t)+cos(2t))

We’ll do the same for ♣

:

cos(3t)cos(3t)=12(cos(0)+cos(6t))=12(1+cos(6t))

Finally, let’s substitute everything back in and simplify:

cos(t)cos(2t)cos(3t)=12(12(cos(−4t)+cos(2t))+12(1+cos(6t)))=14