Challenging Mathematics 2009

When n = 1,
cosx – sinx = 1
cos(π/4)cosx – sin(π/4)sinx = cos(π/4) since sin(π/4) = cos(π/4) = √2/2
cos(x+π/4) = cos(π/4)
x+π/4 = 2nπ +/- π/4
x = 2nπ or 2nπ – π/2
x = 0 or 3π/2
For the sake of presentation simplicity, only angle x where 0<=x<2π will be considered in the analysis, but it should be clear that 2kπ + relevant angles also apply.
When n = 2,
cos2x – sin2x = 1
cos2x = 1
2x = 2nπ
x = 0 or π
For n>2, let's define f(x) = cosnx – sinnx
f'(x) = ncosn-1x(-sinx) – nsinn-1xcosx … (1)
f'(x) = -nsinxcox(cosn-2x + sinn-2x) … (2)
Differentiate (1) gives, after some simplification,
f"(x) = -n2(cosnx-sinnx) + n(n-1)(cosn-2x-sinn-2x) … (3)
Equate (2) f'(x) = 0;
either case I: sinxcosx=0
or case II: cosn-2x + sinn-2x=0
Case I : sin2x = 0 => 2x = nπ => x = nπ/2
Or x = 0, π/2, π and 3π/2
Using (3);
f"(0) = -n2 + n(n-1) = -n < 0 => max
f(0) = 1
f"(π/2) = n2 - n(n-1) = n > 0 => min
f(π/2) = -1
f"(π) = -n2(-1)n + n(n-1)(-1)n-2
= (-1)n+1n
is negative (max) when n is even and f(π) = 1
is positive (min) when n is odd and f(π) = -1
f"(3π/2) = -n2(-1)n + n(n-1)(-1)n-2
= (-1)nn
is negative (max) when n is odd and f(3π/2) = 1
is positive (min) when n is even and f(3π/2) = -1
Case II: cosn-2x + sinn-2x=0
when n is even, there is no real solution.
when n is odd, tann-2x = -1 => x = 3π/4 or 7π/4
f"(3π/4) = -n2[(-√2/2)n-(√2/2)n] + n(n-1)[(-√2/2)n-2-(√2/2)n-2]
=(√2/2)n[2n2 – 2n(n-1)(2)]
=(√2/2)n2n(2-n) < 0 (max)
f(3π/4) = (-√2/2)n-(√2/2)n = -2(2-n)/2 > -1
f"(7π/4) = -(√2/2)n2n(2-n) > 0 (min)
f(7π/4) = (√2/2)n-(-√2/2)n = 2(2-n)/2 < 1
Since cosnx-sinnx is a periodic function, the above indicate f(x) is bounded within -1 and +1, for n>2.
It attains the maximum value of 1 when
x = 0 or x = π when n is even; OR
x = 0 or x = 3π/2 when n is odd.
This is consistent with the cases when n=1 and 2. Of course 2kπ + angle all apply.

2009-07-18 18:06:49 補充：
Graph for f(x) when n = 12 and 13
http://img249.imageshack.us/img249/7696/cossin.jpg

Yes , but I think the ans. should be in terms of n

好明顯對任何 n , x = 0 為其解之一。