A community for students.
Here's the question you clicked on:
 0 viewing
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

This Question is Closed

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

ZeHanz
 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
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.