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]