Prove by induction that $\sin\left(x\right)\ldots \sin\left(2^{n}x\right)=\frac{\sin\left(2^{n+1}x\right)}{2^{n+1}\sin\left(x\right)}$

Asked Nov 12 '20 at 23:42

Active Nov 12 '20 at 23:57

Viewed 64 times

Prove the identity by using the induction theorem: $$\sin\left(x\right)\sin\left(2x\right)\sin\left(4x\right)\ldots \sin\left(2^{n}x\right)=\frac{\sin\left(2^{n+1}x\right)}{2^{n+1}\sin\left(x\right)}$$.

Basic Step: $\sin\left(2x\right)=\frac{\sin\left(2^{2}x\right)}{2^{2}\sin\left(x\right)}$

Prove Case $S(n)$: $$S\left(n\right)$$

Assuming true: $$\sin\left(x\right)sin\left(2x\right)\sin\left(4x\right)\ldots \sin\left(2^{n+1}x\right)=\frac{\sin\left(2^{n+2}x\right)}{2^{n+2}\sin\left(x\right)}$$

LHS: $$\frac{\sin\left(2^{n+1}x\right)}{2^{n+1}\sin\left(x\right)}\text{·}\sin\left(2^{n+1}x\right)=\frac{\sin\left(2^{n+1}x\right)^{2}}{2^{n+1}\sin\left(x\right)}$$

What is the correct process in order to prove it by induction?

edited Nov 12 '20 at 23:57

S. Euler

asked Nov 12 '20 at 23:42

Diego Gallo

1

It would be so much easier to try & prove \begin{eqnarray} \cos(x) \cos(2x) \cdots \cos(2^n x)=\frac{\sin(2^{n+1}x)}{2^{n+1}\sin(x)}. \end{eqnarray} – Donald Splutterwit Nov 12 '20 at 23:55
1

The assertion isn't true: Put $n=0$ and it says $\sin x = \frac {\sin 2x} {2\sin x}$. – grand_chat Nov 13 '20 at 00:04
And the one about the cosines has surely been asked and answered on this site before, so it might be good to do a search for it. Maybe https://math.stackexchange.com/questions/2066277/is-there-an-alternative-form-identity-for-prod-k-1m-cos2kx and the questions linked there. – Gerry Myerson Nov 13 '20 at 00:06
Any thoughts about the comments, Diego? – Gerry Myerson Nov 14 '20 at 10:57
I guess not. Too bad. – Gerry Myerson Nov 15 '20 at 12:10

Prove by induction that $\sin\left(x\right)\ldots \sin\left(2^{n}x\right)=\frac{\sin\left(2^{n+1}x\right)}{2^{n+1}\sin\left(x\right)}$

0 Answers0