1. ## Proving by induction

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

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

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

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

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

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

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

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

2. ## Re: Proving by induction

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

3. ## Re: Proving by induction

So then I'll have:

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

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

4. ## Re: Proving by induction

Originally Posted by BobRoss
So then I'll have:

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

Is that what you meant I should do next?
No, the right-hand side is wrong.. We have $\displaystyle \sin(2^{k+1}x)=\sin(2\cdot2^kx)=2\sin(2^kx)\cos(2^ kx)$.

5. ## Re: Proving by induction

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

?

6. ## Re: Proving by induction

Originally Posted by BobRoss
Why does $\displaystyle \sin(2^{k+1}x)=\sin(2\cdot2^kx)$

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

7. ## Re: Proving by induction

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

8. ## Re: Proving by induction

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