# Thread: Tricky induction problem with trig

1. ## Tricky induction problem with trig

Okay, I'm working on a really tricky induction problem where I have to show the following (assuming that $\displaystyle \sin{(x/2)}\neq0$):
$\displaystyle \sin{(x)}+2\sin{(2x)}+\ldots +n\sin{(nx)}=\frac{\sin{[(n+1)x]}}{4\sin^2{(x/2)}}-\frac{(n+1)\cos{[(2n+1)(x/2)]}}{2\sin{(x/2)}}$
for all natural numbers $\displaystyle n$.

I've already verified the base case (with the help of ebaines), so now I'm trying to show that the above equation implies this equation:

$\displaystyle \noindent{ \sin{(x)}+2\sin{(2x)}+\ldots +(n+1)\sin{[(n+1)x]}= \frac{\sin{[(n+2)x]}}{4\sin^2{(x/2)}}-\frac{(n+2)\cos{[(2n+3)(x/2)]}}{2\sin{(x/2)}}}$

I almost hesitate to ask this as to me it seems either impossible or a huge amount of work, but perhaps that's not so for those with more experience. Using the first equation we can write the second equation as

$\displaystyle \noindent{\sin{(x)}+2\sin{(2x)}+\ldots +(n+1)\sin{[(n+1)x]}=\frac{\sin{[(n+1)x]}}{4\sin^2{(x/2)}}-\frac{(n+1)\cos{[(2n+1)(x/2)]}}{2\sin{(x/2)}}+(n+1)\sin{[(n+1)x]}}$

so what we desire to show is that

$\displaystyle \noindent{\frac{\sin{[(n+1)x]}}{4\sin^2{(x/2)}}-\frac{(n+1)\cos{[(2n+1)(x/2)]}}{2\sin{(x/2)}}+(n+1)\sin{[(n+1)x]}=\frac{\sin{[(n+2)x]}}{4\sin^2{(x/2)}}-\frac{(n+2)\cos{[(2n+3)(x/2)]}}{2\sin{(x/2)}}}$

So I'm unsure where to go from here. We already have the denominators set up neatly, so I don't want to mess with that. I thought about distributing the $\displaystyle n+1$ in the third term and then adding the resulting two terms to the first two terms in some manner, but I'm not sure how the arguments will work out. Can anyone help?

2. ## Re: Tricky induction problem with trig

In order to see if there's some kind of strategy with this, I'm trying the desired identity with $\displaystyle n=3$, giving
$\displaystyle \frac{\sin{(4x)}}{4\sin^2{(x/2)}}-\frac{4\cos{(7x/2)}}{2\sin{(x/2)}}+4\sin{(4x)}=\frac{\sin{(5x)}}{4\sin^2{(x/2)}}-\frac{5\cos{(9x/2)}}{2\sin{(x/2)}}$

How would you go about showing this sort of thing?

For example, is there a way of writing $\displaystyle \sin{(5x)}$ in terms of functions with an argument of $\displaystyle 4x$?

3. ## Re: Tricky induction problem with trig

Okay before anyone wastes time doing this I just found the answer on Chegg. Thanks anyway!

4. ## Re: Tricky induction problem with trig

Hi Ragnorak

I have a solution to this problem, took me an hour or so !
What method of solution did Chegg use ?
If different, I'll post mine.

5. ## Re: Tricky induction problem with trig

Nice! I am far too lazy to type up Chegg's whole solution but the main steps were (starting from the LHS of the desired identity, last one in my first post):

Got a common denominator.
Rewrote $\displaystyle 4(n+1)\sin{[(n+1)x]}\sin^2{(x/2)}$ in the numerator as $\displaystyle 2(n+1)\sin{[(n+1)x]}2\sin^2{(x/2)}=$
$\displaystyle =2(n+1)\sin{[(n+1)x]}(1-\cos{(x)})$ (half-angle formula).
Distributed $\displaystyle 1-\cos{(x)}$.
Added $\displaystyle \sin{[(n+1)x]}$ terms.
Factored $\displaystyle -2(n+1)$ out of remaining terms.
Used product identies on $\displaystyle \sin{(x/2)}\cos{[(2n+1)(x/2)]}$ and $\displaystyle \sin{[(n+1)x]}\cos{(x)}$ terms.
Simplified.
Used a difference identity on $\displaystyle \sin{[(n+1)x]}-\sin{[(n+2)x]}$.
Simplified and divided by denominator.

6. ## Re: Tricky induction problem with trig

Okay, I used the same identities but attacked it in a different way, (having failed to find my way through the induction approach).

Start by letting

$\displaystyle S=\sin x + 2\sin 2x + 3\sin 3x + \dots + k\sin kx.$

Then, multiplying by $\displaystyle 2\cos x,$

$\displaystyle 2S\cos x = 2\sin x\cos x + 2(2\sin 2x \cos x)+3(2\sin 3x \cos x) +\dots + k(2\sin kx \cos x)$

$\displaystyle =\sin 2x + 2(\sin 3x+\sin x)+3(\sin 4x+\sin 2x) + 4(\sin 5x + \sin 3x)+\dots + k(\sin(k+1)x + \sin(k-1)x),$

$\displaystyle =2\sin x + 4\sin 2x+ 6\sin 3x +\dots + (2k-2)\sin (k-1)x +(k-1)\sin kx + k\sin (k+1)x.$

Now subtract the original series,

$\displaystyle 2S\cos x - S = \sin x + 2\sin 2x + 3\sin 3x +\dots +(k-1)\sin (k-1)x + (k-1)\sin kx + k\sin (k+1)x - k\sin kx,$

$\displaystyle = S+ (k-1)\sin kx + k\sin (k+1)x -2k\sin kx,$

$\displaystyle = S - \sin (k+1)x +(k+1)\sin (k+1)x -(k+1)\sin kx.$

So,

$\displaystyle 2S\cos x - 2S = - \sin (k+1)x +(k+1)\sin (k+1)x -(k+1)\sin kx,$

or, multiplying both sides by a negative sign,

$\displaystyle 2S(1-\cos x)= \sin (k+1)x -(k+1)\sin (k+1)x +(k+1)\sin kx,$

$\displaystyle 4S\sin^{2}(x/2)=\sin (k+1)x -(k+1)(\sin (k+1)x - \sin kx),$

$\displaystyle =\sin(k+1)x-(k+1)2\cos((2k+1)(x/2))\sin (x/2).$

So, finally

$\displaystyle S = \frac{\sin(k+1)x}{4\sin^{2}(x/2)}-(k+1)\frac{\cos((2k+1)(x/2))}{2\sin(x/2)}.$

Hope there aren't any misprints !

7. ## Re: Tricky induction problem with trig

Very interesting! I'm going to give it a week and then try this problem again, see if I remember anything.