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

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

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

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

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

binarymimic
 one year ago
Best ResponseYou've already chosen the best response.1f(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
 one year ago
Best ResponseYou've already chosen the best response.0showing (cos(nx))' .>= (cos^n(x))' is sufficient after showing they begin at some point.

binarymimic
 one year ago
Best ResponseYou've already chosen the best response.1Ok 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

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

binarymimic
 one year ago
Best 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

binarymimic
 one year ago
Best ResponseYou've already chosen the best response.1ok 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

ZeHanz
 one year 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...)!

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

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