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 March 15th, 2017, 12:36 PM #1 Newbie   Joined: Mar 2017 From: Scotland Posts: 8 Thanks: 0 Finding minimum value of wave Please forgive me if this is in the wrong section of the forum, this is my first time posting here. Anyway, I am quite stuck on this problem and require a bit of help in solving it. Sorry for the low quality image, had to rescale it.
 March 15th, 2017, 12:58 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 1,785 Thanks: 920 do you know what the minimum of $\cos(x)$ is for any $x$ ? if so then what is the minimum of of $\sqrt{2} \cos(x)$ ? and the minimum of $3 + \sqrt{2}\cos(x)$ Assuming you do know the minimum of $\cos(x)$ for what value of $x$ does it occur? Let's say this value is $\theta^\circ$ then we know that $x+20^\circ = \theta^\circ$ $x = \theta^\circ - 20^\circ$ Thanks from Beams
 March 16th, 2017, 12:20 AM #3 Newbie   Joined: Mar 2017 From: Scotland Posts: 8 Thanks: 0 I appreciate the help but I still don't full get what is going on here. I understand the last two steps but the rest just seems to confuse me, sorry.
 March 16th, 2017, 12:27 AM #4 Senior Member   Joined: Jun 2015 From: England Posts: 763 Thanks: 219 Taking romsek's first line, adding the discussion in this thread might help. Graph of sine function Thanks from Beams
 March 16th, 2017, 11:40 AM #5 Newbie   Joined: Mar 2017 From: Scotland Posts: 8 Thanks: 0 Thanks for that studiot, I understand how to graph it but I would like some further help on this specific problem. Unfortunately I'm not a quick learner so I would appreciate detailed steps as to how to go about getting the minimum value. A solution with working to this would be of great help to use as a guide for my other questions.
 March 16th, 2017, 12:29 PM #6 Math Team   Joined: Jul 2011 From: Texas Posts: 2,722 Thanks: 1376 $y = 3 + \sqrt{2} \cdot \cos(x+20)$ note that for all values of $\theta$, $-1 \le \cos{\theta} \le 1 \implies 3-\sqrt{2} \le 3 + \sqrt{2} \cdot \cos(x+20) \le 3+\sqrt{2}$ the minimum value of $y$ will occur when $\cos(x+20)^\circ = -1$ to solve for $x$, note the fact that $\cos(180^\circ) = -1 \implies (x+20)^\circ = 180^\circ \implies x = 160^\circ$ Thanks from Beams
March 16th, 2017, 12:33 PM   #7
Newbie

Joined: Mar 2017
From: Scotland

Posts: 8
Thanks: 0

Quote:
 Originally Posted by skeeter $y = 3 + \sqrt{2} \cdot \cos(x+20)$ note that for all values of $\theta$, $-1 \le \cos{\theta} \le 1 \implies 3-\sqrt{2} \le 3 + \sqrt{2} \cdot \cos(x+20) \le 3+\sqrt{2}$ the minimum value of $y$ will occur when $\cos(x+20)^\circ = -1$ to solve for $x$, note the fact that $\cos(180^\circ) = -1 \implies (x+20)^\circ = 180^\circ \implies x = 160^\circ$
Thanks this is exactly what I needed.
You are a life saver, this will really help me answer my other questions

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