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binarymimic
Group Title
prove if possible
(cos x)^n <= cos (nx) if 0 <= x <= pi/2 and 0 < n < 1
 one year ago
 one year ago
binarymimic Group Title
prove if possible (cos x)^n <= cos (nx) if 0 <= x <= pi/2 and 0 < n < 1
 one year ago
 one year ago

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binarymimic Group TitleBest ResponseYou've already chosen the best response.1
\[\cos^{n} x \le \cos(nx)\]
 one year ago

IsTim Group TitleBest ResponseYou've already chosen the best response.0
Not sure if I got this right, but have you used trial numbers?
 one year ago

binarymimic Group TitleBest ResponseYou've already chosen the best response.1
to test the inequality ?
 one year ago

IsTim Group TitleBest ResponseYou've already chosen the best response.0
I have no clue.
 one year ago

IsTim Group TitleBest ResponseYou've already chosen the best response.0
Grade 12 Advanced Functions?
 one year ago

binarymimic Group TitleBest ResponseYou've already chosen the best response.1
f'(x)  g'(x) > 0 implies f'(x) > g'(x) right? i think i found a method to figure it out
 one year ago

binarymimic Group TitleBest ResponseYou've already chosen the best response.1
f(x) = cos (nx) g(x) = cos^n(x) f(x)  g(x) = cos (nx)  cos^n(x) if x = 0 then f(x)  g(x) = 0  0 =0 this is our starting point, the point at which both functions are equal maybe if we show f'(x) > g'(x) on the interval then that is sufficient to prove the original inequality
 one year ago

IsTim Group TitleBest ResponseYou've already chosen the best response.0
I have no clue.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
showing (cos(nx))' .>= (cos^n(x))' is sufficient after showing they begin at some point.
 one year ago

binarymimic Group TitleBest ResponseYou've already chosen the best response.1
Ok so f(x) = cos (nx) f'(x) = n sin(nx) g(x) = cos^n(x) g'(x) = n cos^(n1)(x) * sin x f'(x) > g'(x) implies f'(x)  g'(x) = 0 so . . . if we can show n sin(nx)  (n cos^(n1)(x) * sin x) = \[n \sin(nx) + n \cos^{n1}(x) * \sin x > 0\] is true on the interval.. hm
 one year ago

binarymimic Group TitleBest ResponseYou've already chosen the best response.1
ok .. this becomes \[n \left( \sin(nx) + \cos^{p  1}(x) * \sin x \right)\]
 one year ago

binarymimic Group TitleBest ResponseYou've already chosen the best response.1
\[n \left( \sin (nx) + \frac{ \sin x }{ \cos^{1  p} (x) } \right)\] reversing the order of the exponent's difference and putting cos in the denominator
 one year ago

binarymimic Group TitleBest ResponseYou've already chosen the best response.1
ok since sin is increasing on the interval (0, pi/2), if we focus just on the two sin functions... sin(nx) + sin x we know that n is fractional, so this portion of the expression will always be positive also we know cos is positive on the interval (0, pi/2) so we can conclude with certainty that sin(nx) + sin x / cos^(1  p)(x) > 0 if we multiply this expression by positive n, we will show f'(x)  g'(x) > 0 which is what we set out to prove. seems ok to me
 one year ago

ZeHanz Group TitleBest ResponseYou've already chosen the best response.0
It looks quite ingenious to me! I think p should be n, btw... You have done a good job (yourself, that is...)!
 one year ago

binarymimic Group TitleBest ResponseYou've already chosen the best response.1
youre right, the question uses p for the exponent but i'm used to n to denote an exponent . or k
 one year ago

binarymimic Group TitleBest ResponseYou've already chosen the best response.1
wow i should seriously review what i type before i post it. lets just say p = n
 one year ago
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