Proving by induction

**BobRoss** · September 23rd 2012, 11:55 AM

If sinx cannot = 0, use mathematical induction to show that

$cosx \cdot cos2x \cdot cos4x... cos2^{n-1} x = \frac{sin2^n x}{2^n sinx}$ for every integer $n \geq 1$

So for the induction step what I have so far is:

Assume n=k is true, show true for n=k+1

$cosx \cdot cos2x \cdot cos4x... cos2^{k-1} x \cdot cos2^{(k+1)-1}x = \frac{sin2^{k+1} x}{2^{k+1} sinx}$

$\frac{sin2^{k}x}{2^{k}sinx} \cdot cos2^{k}x = \frac{sin2^{k+1}x}{2^{k+1}sinx}$

$\frac{(sin2^{k}x)(cos2^{k}x)}{2^{k}sinx} = \frac{sin2^{k+1}x}{2^{k+1}sinx}$

Assuming any of this is correct so far, I am having trouble showing that both sides are equal from here on.

**emakarov** · September 23rd 2012, 12:03 PM

Note that sin(2x) = 2sin(x)cos(x).

**BobRoss** · September 23rd 2012, 12:26 PM

So then I'll have:

$\frac{(sin2^{k}x)(cos2^{k}x)}{2^{k}sinx} = \frac{2sin^{k+1}x cos^{k+1}x}{2^{k+1}sinx}$

Is that what you meant I should do next? Where do I go from here then?

**emakarov** · September 23rd 2012, 12:31 PM

Originally Posted by BobRoss

So then I'll have:

$\frac{(sin2^{k}x)(cos2^{k}x)}{2^{k}sinx} = \frac{2sin^{k+1}x cos^{k+1}x}{2^{k+1}sinx}$

Is that what you meant I should do next?

No, the right-hand side is wrong.. We have $\sin(2^{k+1}x)=\sin(2\cdot2^kx)=2\sin(2^kx)\cos(2^ kx)$ .

**BobRoss** · September 23rd 2012, 12:47 PM

Originally Posted by emakarov

No, the right-hand side is wrong.. We have $\sin(2^{k+1}x)=\sin(2\cdot2^kx)=2\sin(2^kx)\cos(2^ kx)$ .

Why does $\sin(2^{k+1}x)=\sin(2\cdot2^kx)$

?

**DIOGYK** · September 23rd 2012, 12:59 PM

Originally Posted by BobRoss

Why does $\sin(2^{k+1}x)=\sin(2\cdot2^kx)$

?

For example:
$a^2*a^3=a^{2+3}=a^{5}.$ .
Similarly,
$\sin(2^1*2^kx)=\sin(2^{k+1}x)$

**BobRoss** · September 23rd 2012, 03:04 PM

Oh okay that makes sense. Then I can do the same thing with the term $2^{k+1}sinx$ to get $2 \cdot 2^{k}sinx$ And doing that makes the left side equal the right side, and that proves the equation, correct?

**emakarov** · September 23rd 2012, 10:16 PM

Originally Posted by BobRoss

Oh okay that makes sense. Then I can do the same thing with the term $2^{k+1}sinx$ to get $2 \cdot 2^{k}sinx$ And doing that makes the left side equal the right side, and that proves the equation, correct?

Yes.

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