1.

Solve the 4th one using mathematical induction ​

Answer»

We have,

\displaystyle\longrightarrow\sf{P(<klux>N</klux>):(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(n\theta)\quad \forall n\in\mathbb{N}}

CONSIDER P(1).

\displaystyle\longrightarrow\sf{(\cos\theta+i\sin\theta)^1=\cos(1\cdot\theta)+i\sin(1\cdot\theta)}

\displaystyle\longrightarrow\sf{\cos\theta+i\sin\theta=\cos\theta+i\sin\theta}

We see P(1) is TRUE. Hence ASSUME that P(k) is true.

\displaystyle\longrightarrow\sf{P(k):(\cos\theta+i\sin\theta)^k=\cos(k\theta)+i\sin(k\theta)\quad \forall k\in\mathbb{N}}

Consider P(k + 1).

\displaystyle\longrightarrow\sf{P(k+1):(\cos\theta+i\sin\theta)^{k+1}=\cos((k+1)\theta)+i\sin((k+1)\theta)\quad \forall k\in\mathbb{N}}

Let's check whether P(k + 1) is true.

\displaystyle\longrightarrow\sf{LHS}

\displaystyle\longrightarrow\sf{(\cos\theta+i\sin\theta)^{k+1}}

As \sf{a^{m+n}=a^m\cdot a^n,}

\displaystyle\longrightarrow\sf{(\cos\theta+i\sin\theta)^k\cdot(\cos\theta+i\sin\theta)}

By P(k),

\displaystyle\longrightarrow\sf{(\cos(k\theta)+i\sin(k\theta))\,(\cos\theta+i\sin\theta)}

\displaystyle\longrightarrow\sf{\cos(k\theta)\cos\theta+i\sin\theta\cos(k\theta)+i\sin(k\theta)\cos\theta-\sin(k\theta)\sin\theta}

\displaystyle\longrightarrow\sf{[\cos(k\theta)\cos\theta-\sin(k\theta)\sin\theta]+i[\,\sin\theta\cos(k\theta)+\cos\theta\sin(k\theta)\,]}

As \sf{\cos A\cos B-\sin A\sin B=\cos(A+B)} and \sf{\sin A\cos B+\cos A\sin B=\sin(A+B),}

\displaystyle\longrightarrow\sf{\cos(k\theta+\theta)+i\sin(k\theta+\theta)}

\displaystyle\longrightarrow\sf{\cos((k+1)\theta)+i\sin((k+1)\theta)}

\displaystyle\longrightarrow\sf{RHS}

Hence P(k + 1) is true whenever P(k) is true.

THEREFORE P(n) is true \sf{\forall n\in\mathbb{N}.}

Hence Proved!



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