Expert: Scott A Wilson - 4/24/2010

Question
QUESTION: If T(n) = (sin x)^n +  (cos x)^n , prove that
1. (T(3) - T(5))/T(1) = (T(5) - T(7))/T(3)
2. 2T(6) - 3T(4) + 1 = 0
3. 6T(10) - 15T(8) + 10T(6) - 1 = 0

ANSWER: 1) T(3) = sin^3(x) + cos^3(x).
T(5) = sin^5(x) + cos^5(x).  That is the same as sin³x(1-cos²x) + cos³x(1-sin²x).

When we look at T(3) - T(5), we get sin³cos²x + cos³xsin²x = (sinx + cosx)(sin²x * cos²x).
When this is divided by T(1), we get sin²x * cos²x.

Looking at T(5) - T(7) give sin^5(x) + cos^5(x) - sin^7(x) - cos^7(x).
In a similar fashion, we get cos²(x)sin^5(x) + sin²(x)cos^5(x).
This factors (sin^3(x) + cos^3(x))(sin²x * cos²x).

Note that the first term is T(3).  This is divided by T(3), so all we get is sin² * cos²x.

That sounds like what we got for the other side, so we're done with (1).

I'll keep thinking about 2&3 tonight.

It seems like using sin²x + cos²x = 1 as a substitution for sin²x = 1 - cos²x or
cos²x = 1 - sin²x would be good to do.


---------- FOLLOW-UP ----------

QUESTION: Few days ago I asked u question -"If T(n) = (sin x)^n +  (cos x)^n , prove that
1. (T(3) - T(5))/T(1) = (T(5) - T(7))/T(3)
2. 2T(6) - 3T(4) + 1 = 0
3. 6T(10) - 15T(8) + 10T(6) - 1 = 0" . You gave me the solution to the first question. You told that you will keep thinking about 2nd and 3rd question . Please give me the solution to the 2nd & 3rd question if you have solved it .

Answer
I've answered [1] and [2] already, and now for [3].

If we take A = sin²x, the cos²x = 1-A.
Note that (1-A)^3 = 1 - 3A + 3A^2 - A^3, (1-A)^4 = 1 - 4A + 6A^2 - 4A^3 + a^4, and
(1-A)^5 = 1 - 5A + 10A^2 - 10A^3 + 5A^4 - A^5.

We can make the left side of 6T(10) - 15T(8) + 10T(6) - 1 = 0 turn into
6A^5 + 6(1-A)^5 - 15A^4 - 15(1-A)^4 + 10A^3 + 10(1-A)^3 - 1.

The first two terms become 6A^5 + 6 - 30A + 60A^2 - 60A^3 + 30A^4 - 6A^5 =
6 - 30A + 60A^2 - 60A^3 + 30A^4 since the A^5 terms cancel.

The next two terms become -15A^4 + 15(1-A)^4 = -15A^4 - 15 + 60A - 90A^2 + 60A^3 - 15A^4 =
- 15 + 60A - 90A^2 + 60A^3 - 30A^4 since there are two -15A^4.

The last two terms before the -1 are 10A^3 + 10(1-A)^3 = 10A^3 + 10 - 30A + 30A^2 - 10A^3 =
10 - 30A + 30A^2, since the A^3 terms cancel.

That last term is -1.

In summary, we have
6 - 30A + 60A^2 - 60A^3 + 30A^4 - 15 + 60A - 90A^2 + 60A^3 - 30A^4 + 10 - 30A + 30A^2 - 1.

Now for the leading constants, we have 6 - 15 + 10 - 1 = 16 - 15 = 1, and that's what we want.
For terms with an A in them, we have -30 + 60 - 30 = 0.
For terms with an A^2 in them, we have 60 - 90 + 30 = 0.
For terms with an A^3 in them, we have -60 + 60 = 0.
For terms with an A^4 in them, we have 30 - 30 = 0.
For terms with an A^5 in them .... we don't have any.  In fact, that's it.

All that, just for 1.

See, if this had been done with sin² and cos², each A would have added 3 characters and
each (1-A) would have added 3 characters. From this, my final equation would have probably
taken up 3 lines or so.