Answer 1

Let P(n) ≡ (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ, for all n  ∈ N.

Step 1:

For n = 1, L.H.S. = cos θ + i sin θ

R.H.S. = cos θ + i sin θ

∴ L.H.S. = R.H.S. 

∴ P(1) is true.

Step 2:

Let us assume that for some k ∈ N, P(k) is true,

i.e., (cos θ + i sin θ)^k = cos kθ + i sin kθ    ...(1)

Step 3: To prove that P(k + 1) is true,

i.e., to prove that

(cos θ + i sin θ)^k+1 = cos (k + 1)(θ + i sin (k + 1) θ

L.H.S. = (cos θ + i sin θ)^k+1

= (cos θ + i sin θ)^k · (cos θ + i sin θ)

= (cos kθ + i sin kθ) · (cos θ + i sin θ)    ...[By (1)]

= cos kθ · cos θ + i cos kθ sin θ + i sin kθ cos θ + i² sin kθ sin θ

= (cos kθ cos θ – sin kθ sin θ) + i(sin kθ cos θ + cos kθ sin θ)  ......[as i² = – 1]

= cos(kθ + 0) + i sin (kθ + θ)

= cos (k + 1)θ + i sin(k + 1)θ

= R.H.S.

∴ P(k + 1) is true.

Step 4:

From all the above steps and by the principle of mathematical induction, P(n) is true for all n ∈ N,

i.e., (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ. for all n ∈ N