March 16th, 2019, 12:47 AM  #1 
Senior Member Joined: Jan 2012 Posts: 133 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: 511 Thanks: 79 
$\displaystyle f’(x)=2p\cos(x) 2\sin(2x)=0$ . Now solve the equation $\displaystyle p\cos(x)=\sin(2x)=2\sin(x)\cos(x)$. 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: 2,156 Thanks: 878 Math Focus: Wibbly wobbly timeywimey stuff. 
Yes, but this is posted in the Trigonometry Forum, not Calculus. Dan 
March 16th, 2019, 03:48 PM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 2,924 Thanks: 1521 
$\displaystyle y_{max} = 2p1$

March 16th, 2019, 03:51 PM  #5 
Senior Member Joined: Dec 2015 From: somewhere Posts: 511 Thanks: 79  
March 16th, 2019, 05:30 PM  #6 
Math Team Joined: Jul 2011 From: Texas Posts: 2,924 Thanks: 1521  
March 17th, 2019, 05:30 AM  #7 
Senior Member Joined: Jan 2012 Posts: 133 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: 7,935 Thanks: 1129 Math Focus: Elementary mathematics and beyond 
$12\sin^2x+2p\sin x$ has vertex $\frac p2$, so the function is at a maximum where $\sin x=\frac p2$, so it's greatest value is where $\sin x=1\implies x=\frac{\pi}{2}$.

March 17th, 2019, 06:17 AM  #9 
Senior Member Joined: Jan 2012 Posts: 133 Thanks: 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: 7,935 Thanks: 1129 Math Focus: Elementary mathematics and beyond 
Yes. That's the biggest the sine can get, which is what is desired.


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