Question

Using principle of mathematical induction, prove that
$\cos \alpha \cos 2\alpha \cos 4\alpha .....\cos \left(2^{n-1}\alpha \right)=\frac{\sin 2^n\alpha }{2^n\sin \alpha }$ for all $n\in N$.

Answer 1

Let $P\left(n\right)$ be the statement given by

$P\left(n\right):\cos \alpha \cos 2\alpha \cos 4\alpha ......\cos \left(2^{n-1}\alpha \right)=\frac{\sin \left(2^n\alpha \right)}{2^n\sin \alpha }$

Setp I: $P\left(1\right):\cos \alpha =\frac{\sin \left(2^1\alpha \right)}{2^1\sin \alpha }$

$\because \frac{\sin \left(2^1\alpha \right)}{2^1\sin \alpha }=\frac{\sin 2\alpha }{2\sin \alpha }=\frac{2\sin \alpha \cos \alpha }{2\sin \alpha }=\cos \alpha$

$\therefore P\left(1\right)$ is true.

Step II: Let $p\left(m\right)$ be true. Then,

$\cos \alpha \cos 2\alpha \cos 4\alpha ......\cos \left(2^{m-1}\alpha \right)=\frac{\sin \left(2^m\alpha \right)}{2^m\sin \alpha }$

We shall now show that $P(m+1)$ is true. For this we have to show that

$\cos \alpha \cos 2\alpha \cos 2^2\alpha ....\cos \left(2^{m-1}\alpha \right)\cos \left(2^m\alpha \right)=\frac{\sin \left(2^{m+1}\alpha \right)}{2^{m+1}\sin \alpha }$

We have,

$\cos \alpha \cos 2\alpha \cos 2^2\alpha ....\cos \left(2^{m-1}\alpha \right)\cos \left(2^m\alpha \right)$

$=\{\cos \alpha \cos 2\alpha \cos 2^2\alpha ....\cos (2^{m-1}\alpha \left)\right\}\cos \left(2^m\alpha \right)$

$=\frac{\sin \left(2^m\alpha \right)}{2^m\sin \alpha }\times \cos \left(2^m\alpha \right)$

$=\frac{2\sin \left(2^m\alpha \right)\cos \left(2^m\alpha \right)}{2^{m+1}\sin \alpha }=\frac{\sin \left({2.2}^m\alpha \right)}{2^{m+1}\sin \alpha }=\frac{\sin \left(2^{m+1}\alpha \right)}{2^{m+1}\sin \alpha }$

$\therefore P(m+1)$ is true.

Thus, $P\left(m\right)$ is true $\Rightarrow P(m+1)$ is true.

Hence, by the principle of mathematical induction $P\left(n\right)$ is true for all $n\in N$.