Question

Use mathematical induction to show that \({\left( {cosx + isinx} \right)^n} = cosnx + isinnx\)whenever nis a positive integer.

(Here \(i\)is the square root of \( - 1\).) (Hint:Use

the identities \(cos\left( {a + b} \right) = cosa cosb - sina sinb\) and \(sin\left( {a + b} \right) = sina\,cosb + cosa\,sinb\).)

Short answer

Using mathematical induction, it is proved that \({\left( {\cos x + i\sin x} \right)^n} = \cos nx + i\sin nx\).

Step-by-step solution

Principle of Mathematical Induction

To prove that \(P\left( n \right)\)is true for all positive integers n, where\(P\left( n \right)\)is a propositional function, it completes two steps:

Basis Step:

It verifies that\(P\left( 1 \right)\)is true.

Inductive Step:

It shows that the conditional statement \(P\left( k \right) \to P\left( {k + 1} \right)\)is true for all positive integers k.

Proving the basis step

Let\(P\left( n \right)\): “\({\left( {\cos x + i\sin x} \right)^n} = \cos nx + i\sin nx\)”.

In the basis step, it needs to prove that\(P\left( 1 \right)\)is true.

For finding statement\(P\left( 1 \right)\)substituting\(1\)for\(n\)in the statement.

\(\begin{aligned}{l}{\left( {\cos x + i\sin x} \right)^1} &= \cos \left( 1 \right)x + i\sin \left( 1 \right)x\\\cos x + i\sin x &= \cos x + i\sin x\end{aligned}\)

From the above, it can see that the statement \(P\left( 1 \right)\) is true this is also known as the basis step of the proof.

Proving the Inductive step

In the inductive step, it needs to prove that, if\(P\left( k \right)\)is true, then\(P\left( {k + 1} \right)\)is also true.

That is,\(P\left( k \right) \to P\left( {k + 1} \right)\)is true for all positive integers k.

In the inductive hypothesis, it assumes that\(P\left( k \right)\)is true for any arbitrary positive integer\(k\).

That is, it is written as:

\({\left( {\cos x + i\sin x} \right)^k} = \cos kx + i\sin kx\)

Now it must have to show that\(P\left( {k + 1} \right)\)is also true.

\({\left( {\cos x + i\sin x} \right)^{k + 1}} = {\left( {\cos x + i\sin x} \right)^k}\left( {\cos x + i\sin x} \right)\)

From the inductive hypothesis, it knows\({\left( {\cos x + i\sin x} \right)^k} = \cos kx + i\sin kx\).

Therefore, it is written as:

\(\begin{aligned}{c}{\left( {\cos x + i\sin x} \right)^{k + 1}} &= \left( {\cos kx + i\sin kx} \right)\left( {\cos x + i\sin x} \right)\\ &= \cos kx\,\,\cos x + i\sin kx\,\,\cos x + i\cos kx\,\,\sin x - \sin kx\,\,\sin x\\ &= \cos kx\,\cos x - \sin kx\sin x + i\left( {\sin kx\,\cos x + \cos kx\,\sin x} \right)\end{aligned}\)

Now applying the formula:

\(\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b\)

\(\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b\)

From the above, it can see that\(P\left( {k + 1} \right)\)is also true.

Hence,\(P\left( {k + 1} \right)\)is true under the assumption that\(P\left( k \right)\)is true. This completes the inductive step.

Hence using mathematical induction, it is proved that \({\left( {\cos x + i\sin x} \right)^n} = \cos nx + i\sin nx\).