Expert: Paul Klarreich - 6/19/2006

Question
Hey Paul, I've basically tried almost every technique I know to
find this reduction formula, and I keep going round in circles,
I'm probably, missing something simple.

Q: Find an appropiate reduction formula for SinNx/sinx?


Answer
Hi, Ali,

You wrote:
Subject:  Reduction Formula
Question:  Hey Paul, I've basically tried almost every technique I know to find this reduction formula, and I keep going round in circles,
I'm probably, missing something simple.

Q: Find an appropiate reduction formula for SinNx/sinx?
-------------------------------------------------------
I assume (since you are studying calculus) that you want a reduction formula for the integral:

{
|  (sin nx)/ sin x dx = TOI  (The Original Integral)
}

You are wrong.  You are not missing something simple.  Give yourself some credit.  If you can't do it, it probably isn't easy.  

And it's not, but it goes this way:

Part I: use a sum-of-angles formula on top, to separate out a term that integrates easily.  Integrate it.

Part II: Use a product-to-sum trigonometric formula to rewrite the second term.

Part III: Separate out the terms and observe that the original integral appears with a factor of 1/2.  That enables you to solve the equation algebraically.

Here we go:
WARNING: USE A FIXED-SIZE FONT TO VIEW THIS.
......................................
Part I:  Use  sin(A+B) formula.

sin(nx) =  sin x cos(n-1)x + cos x sin(n-1)x


{  sin x cos(n-1)x + cos x sin(n-1)x dx
|  ---------------------------------------- =
}         sin x

{               cos x sin(n-1)x dx
|  cos(n-1)x + -------------------- =
}                    sin x
That's a bit sloppy with parentheses, I know.
Now cos(n-1)x integrates easily.

        sin(n-1)x
So TOI = ---------  PLUS:
          n-1

{  cos x sin (n-1)x dx
|  -------------------- =
}       sin x

Part II.  There are a whole bunch of less-well-known trig formulas, called 'sum-to-product' and 'product-to-sum' identities.  We want the 'product-to-sum' formulas, which are:

sin A cos B = (1/2)(sin(A + B) + sin(A - B))
cos A cos B = (1/2)(cos(A + B) + cos(A - B))
sin A sin B = (1/2)(cos(A - B) - cos(A + B))

(You might amuse yourself by deriving some of these.)

Applying the first one, with A = x, B = (n-1)x, we can write:

cos x sin((n-1)x) = 1/2[sin(nx) + sin(n-2)x]

Now our second integral is:

{  1/2[sin(nx) + sin(n-2)x]dx
|  -------------------------- =
}       sin x

1  { sin nx             1  { sin(n-2)x dx
--- | ------- dx  PLUS  --- | ------------ =
2  }  sin x             2  }   sin x

WE GOT IT!  THE ORIGINAL INTEGRAL HAS REAPPEARED!  HOORAY!

1            1  { sin(n-2)x dx
--- TOI  +   --- | ------------
2            2  }   sin x

Now do you see what we have?

     sin(n-1)x     1            1  { sin(n-2)x dx
TOI = --------- +  --- TOI  +   --- | ------------
       n-1         2            2  }   sin x

Now we just subtract the 1/2 TOI term, then we have:

1       sin(n-1)x     1  { sin(n-2)x dx
---TOI = --------- +  --- | ------------
2          n-1        2  }   sin x

Multiply through by 2 and you have your reduction formula.  Now, do you see how wrong you were?  You were not missing something simple.