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anonymous
 3 years ago
prove if possible
(cos x)^n <= cos (nx) if 0 <= x <= pi/2 and 0 < n < 1
anonymous
 3 years ago
prove if possible (cos x)^n <= cos (nx) if 0 <= x <= pi/2 and 0 < n < 1

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\cos^{n} x \le \cos(nx)\]

IsTim
 3 years ago
Best ResponseYou've already chosen the best response.0Not sure if I got this right, but have you used trial numbers?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0to test the inequality ?

IsTim
 3 years ago
Best ResponseYou've already chosen the best response.0Grade 12 Advanced Functions?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0f'(x)  g'(x) > 0 implies f'(x) > g'(x) right? i think i found a method to figure it out

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0f(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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ok 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok .. this becomes \[n \left( \sin(nx) + \cos^{p  1}(x) * \sin x \right)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[n \left( \sin (nx) + \frac{ \sin x }{ \cos^{1  p} (x) } \right)\] reversing the order of the exponent's difference and putting cos in the denominator

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok 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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0youre right, the question uses p for the exponent but i'm used to n to denote an exponent . or k

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wow i should seriously review what i type before i post it. lets just say p = n
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