User Name Remember Me? Password

 Trigonometry Trigonometry Math Forum

 March 16th, 2019, 12:47 AM #1 Senior Member   Joined: Jan 2012 Posts: 140 Thanks: 2 Max value of a Trig.Function Hi, How can we solve the problem: If p>2, then evaluate the maximum value of the function: f(x)= cos(2x) + 2p sin(x) Thx. March 16th, 2019, 09:03 AM #2 Senior Member   Joined: Dec 2015 From: somewhere Posts: 605 Thanks: 88 $\displaystyle f’(x)=2p\cos(x) -2\sin(2x)=0$ . Now solve the equation $\displaystyle p\cos(x)=\sin(2x)=2\sin(x)\cos(x)$. Thanks from happy21 Last edited by skipjack; March 16th, 2019 at 04:07 PM. March 16th, 2019, 12:52 PM #3 Math Team   Joined: May 2013 From: The Astral plane Posts: 2,258 Thanks: 929 Math Focus: Wibbly wobbly timey-wimey stuff. Yes, but this is posted in the Trigonometry Forum, not Calculus. -Dan Thanks from idontknow March 16th, 2019, 03:48 PM #4 Math Team   Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1588 $\displaystyle y_{max} = 2p-1$ Thanks from idontknow March 16th, 2019, 03:51 PM   #5
Senior Member

Joined: Dec 2015
From: somewhere

Posts: 605
Thanks: 88

Quote:
 Originally Posted by skeeter $\displaystyle y_{max} = 2p-1$
Did you find it with calculus or how ? March 16th, 2019, 05:30 PM   #6
Math Team

Joined: Jul 2011
From: Texas

Posts: 3,002
Thanks: 1588

Quote:
 Originally Posted by idontknow Did you find it with calculus or how ?
Of course I did. Call me lazy. March 17th, 2019, 05:30 AM #7 Senior Member   Joined: Jan 2012 Posts: 140 Thanks: 2 How can it be calculated without calculus? March 17th, 2019, 06:04 AM #8 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond $1-2\sin^2x+2p\sin x$ has vertex $\frac p2$, so the function is at a maximum where $\sin x=\frac p2$, so it's greatest value is where $\sin x=1\implies x=\frac{\pi}{2}$. Thanks from happy21 and topsquark March 17th, 2019, 06:17 AM   #9
Senior Member

Joined: Jan 2012

Posts: 140
Thanks: 2

Quote:
 Originally Posted by greg1313 $1-2\sin^2x+2p\sin x$ has vertex $\frac p2$, so the function is at a maximum where $\sin x=\frac p2$, so it's greatest value is where $\sin x=1\implies x=\frac{\pi}{2}$.
Thx but only issue is with p>2, so for this sin (angle) is not obtained. Is that limiting case? March 17th, 2019, 06:39 PM #10 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Yes. That's the biggest the sine can get, which is what is desired. Thanks from happy21 Tags max, trigfunction Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post eagle2020 Trigonometry 2 December 6th, 2016 08:05 AM hatchelhoff Trigonometry 2 March 27th, 2014 01:08 AM WWRtelescoping Complex Analysis 4 February 26th, 2014 11:48 AM jimooboo Calculus 3 February 26th, 2012 12:07 PM anthonye Trigonometry 11 February 10th, 2012 05:55 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.       