Prove that: (cos px+i sin px)(cos qx +isin qx)= cos (p+q)x + i sin (p+q)x

Question

use trigonometric identities

Answer 1

Question: Prove that: (cos px+i sin px)(cos qx +isin qx)= cos (p+q)x + i sin (p+q)x .

Use De Moivre's Theorem.

e^(iθ) = cos(θ) + i.sin(θ)

So, cos(px) + i.sin(px) = e^(px)

And, cos(qx) + i.sin(qx) = e^(qx)

Then (cos(px) + i.sin(px))*(cos(qx) + i.sin(qx)) = e^(px)*e^(qx) = e^(px + qx) = e^[(p+q)x]

And e^[(p+q)x] = cos[(p+q)x] + i.sin[(p+q)x]

Hence,

(cos(px) + i.sin(px))*(cos(qx) + i.sin(qx)) = cos[(p+q)x] + i.sin[(p+q)x]