Question

Group-BHigher Trigonometry9.1. State and prove De Moivre's theorem for rational index​

Answer 1

Step-by-step explanation:

Proving (cos θ + isin θ)n = cos (nθ) + isin (nθ)

Case 1 : If n is a positive number

L.H.S = (cos θ + isin θ)n

= (cos θ + isin θ) × (cos θ + isin θ) × (cos θ + isin θ) × (cos θ + isin θ)

= cos (nθ) + isin (nθ)

= R.H.S

Case 2 : If n is a negative number

Consider, n = -m

L.H.S = (cos θ + isin θ)-m

= 1 / (cos θ + isin θ)m

= 1 / (cos mθ + isin mθ)

= ( 1 / cos mθ + isin mθ ) × (cos mθ - isin mθ / cos mθ - isin mθ)

= (cos mθ - isin mθ) / (cos² mθ + sin² mθ)

= (cos mθ - isin mθ) / 1

= cos (-mθ) + isin (-mθ)

= cos (nθ) + isin (nθ)

= R.H.S

Case 3 : If n is a fractional number

Consider, n = p/q

L.H.S = (cos θ + isin θ)p/q

= (cos pθ + isin pθ)1/q

= (cos (pθ)/q + isin (pθ)/q)

= (cos (p/q)θ + isin (p/q)θ)

= cos (nθ) + isin (nθ)

= R.H.S

So, the De Moivre's Theorem is proved for all rational numbers.