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March 16th, 2019, 12:47 AM   #1
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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.
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March 16th, 2019, 09:03 AM   #2
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$\displaystyle f’(x)=2p\cos(x) -2\sin(2x)=0$ .
Now solve the equation $\displaystyle p\cos(x)=\sin(2x)=2\sin(x)\cos(x)$.
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Last edited by skipjack; March 16th, 2019 at 04:07 PM.
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March 16th, 2019, 12:52 PM   #3
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Yes, but this is posted in the Trigonometry Forum, not Calculus.

-Dan
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March 16th, 2019, 03:48 PM   #4
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$\displaystyle y_{max} = 2p-1$
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March 16th, 2019, 03:51 PM   #5
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Quote:
Originally Posted by skeeter View Post
$\displaystyle y_{max} = 2p-1$
Did you find it with calculus or how ?
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March 16th, 2019, 05:30 PM   #6
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Quote:
Originally Posted by idontknow View Post
Did you find it with calculus or how ?
Of course I did. Call me lazy.
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March 17th, 2019, 05:30 AM   #7
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How can it be calculated without calculus?
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March 17th, 2019, 06:04 AM   #8
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$1-2\sin^2x+2p\sin x$ has vertex $\frac p2$, so the function is at a maximum where $\sin x=\frac p2$, so it's greatest value is where $\sin x=1\implies x=\frac{\pi}{2}$.
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March 17th, 2019, 06:17 AM   #9
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Quote:
Originally Posted by greg1313 View Post
$1-2\sin^2x+2p\sin x$ 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?
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March 17th, 2019, 06:39 PM   #10
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Yes. That's the biggest the sine can get, which is what is desired.
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