Maharashtra State BoardHSC Science (Electronics) 11th
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Prove by method of induction, for all n ∈ N: (cos θ + i sin θ)n = cos (nθ) + i sin (nθ) - Mathematics and Statistics

Sum

Prove by method of induction, for all n ∈ N:

(cos θ + i sin θ)n = cos (nθ) + i sin (nθ)

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Solution

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

  Is there an error in this question or solution?
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APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 4 Methods of Induction and Binomial Theorem
Exercise 4.1 | Q 14 | Page 74
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