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March 27th, 2017, 10:59 AM   #1
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Calculable tan(ϕ)

Hi,

Who can help me to sum below?
I can not get them on.

Given is the function : f(x)= -2cos(x) with Df = [0, π ]


Calculate tan(ϕ) as (ϕ) the size of the angle which is where the graph of f (x) intersects the X-axis
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March 27th, 2017, 11:28 AM   #2
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Quote:
Originally Posted by westerwolde View Post
Hi,

Who can help me to sum below?
I can not get them on.

Given is the function : f(x)= -2cos(x) with Df = [0, π ]


Calculate tan(ϕ) as (ϕ) the size of the angle which is where the graph of f (x) intersects the X-axis
$f(x) = -2\cos{x} = 0$ at $x = \dfrac{\pi}{2}$ for $x \in [0,\pi]$

$\tan{\phi} = f'\left(\dfrac{\pi}{2}\right)$
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March 27th, 2017, 09:43 PM   #3
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Quote:
Originally Posted by skeeter View Post
$f(x) = -2\cos{x} = 0$ at $x = \dfrac{\pi}{2}$ for $x \in [0,\pi]$

$\tan{\phi} = f'\left(\dfrac{\pi}{2}\right)$


Thanks for your comment.

How do I find the angle now being asked?
According to the book should answer tan(ϕ)=2 out of here.
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March 28th, 2017, 05:24 AM   #4
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$\phi = \arctan(2)$
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