Can someone explain this cosine concept? "Find local max/min of y = x + 2sin(x)"

[ricecrispie]

Can someone explain this cosine concept? "Find local max/min of y = x + 2sin(x)"

Hi all !

The question asks:

Find the local max/min val of the function:
y = x + 2sin(x)

I said:

y' = 1 + 2cos(x)
This means cos(x) = -1/2

So x = 2pi/3

However the solution set says it's = to 4pi/3 as well.

I don't understand this? I know the cos wave will be -1/2 again at some point but how do we get 4pi/3 exactly? Other than learning the unit circle off by heart

Thanks to anyone who takes the time to explain to me

Sent from my LG-H840 using Tapatalk

 

[mmm4444bot]

… This means cos(x) = -1/2

So x = 2pi/3

However the solution set says it's = to 4pi/3 as well.

I don't understand this? I know the cos wave will be -1/2 again at some point but how do we get 4pi/3 exactly? Other than learning the unit circle off by heart

Understanding the unit circle and how to use it (i.e., reference angles, symmetry) is what you need, when working with trigonometric functions.

Do you remember that points on the unit circle at θ radians have x-coordinate cos(θ) and y-coordinate sin(θ)?

To recall angles between 0 and 2Pi radians where cos(θ) is -1/2, I first visualize the two points on the unit circle whose x-coordinate is -1/2. That is, I see the point in Quadrant II and the point in Quadrant III (each lies on the vertical line x=-1/2). Also, I've memorized cosine for the special angles in Quadrant I, so I already know the reference angle Pi/3 has cosine 1/2 (at 60°). Reflecting across the y-axis, I know from symmetry the point in Quadrant II must be at 2 times Pi/3 (x=-1/2; at 120°). Finally, also due to symmetry, I understand that moving from the point in Quadrant II to the point in Quadrant III means adding the reference angle to half a revolution (180°+60°=240°).

3Pi/3 + Pi/3 = 4Pi/3

2Pi/3 and 4Pi/3 are the two angles between 0 and 2Pi where cos(θ) = -1/2

Other people may reason a bit differently, but these kinds of approaches allow you to recall basics without needing rote memorization of every special angle within one revolution. It all becomes second-hand with practice, and then it literally takes less than five seconds. :cool:

 

[ricecrispie]

Understanding the unit circle and how to use it (i.e., reference angles, symmetry) is what you need, when working with trigonometric functions.

Do you remember that points on the unit circle at θ radians have x-coordinate cos(θ) and y-coordinate sin(θ)?

To recall angles between 0 and 2Pi radians where cos(θ) is -1/2, I first visualize the two points on the unit circle whose x-coordinate is -1/2. That is, I see the point in Quadrant II and the point in Quadrant III (each lies on the vertical line x=-1/2). Also, I've memorized cosine for the special angles in Quadrant I, so I already know the reference angle Pi/3 has cosine 1/2 (that's 60°). Reflecting across the y-axis, I know from symmetry the point in Quadrant II must be at 2 times Pi/3 (x=-1/2; 120°). Finally, also due to symmetry, I understand that moving from the point in Quadrant II to the point in Quadrant III means adding the reference angle to half a revolution (180°+60°=240°).

3Pi/3 + Pi/3 = 4Pi/3

2Pi/3 and 4Pi/3 are the two angles between 0 and 2Pi where cos(θ) = -1/2

Other people may reason a bit differently, but these kinds of approaches allow you to recall basics without needing rote memorization of every special angle within one revolution. It all becomes second-hand with practice, and then it literally takes less than five seconds. :cool:

Thank you so much!! I really needed that explanation, I kept coming across getting 2 answers for cosx in my questions and I didn't know how to get the other solutions. Thanks again!

Sent from my LG-H840 using Tapatalk

 

[Steven G]

Hi all !

The question asks:

Find the local max/min val of the function:
y = x + 2sin(x)

I said:

y' = 1 + 2cos(x)
This means cos(x) = -1/2

So x = 2pi/3

However the solution set says it's = to 4pi/3 as well.

I don't understand this? I know the cos wave will be -1/2 again at some point but how do we get 4pi/3 exactly? Other than learning the unit circle off by heart

Thanks to anyone who takes the time to explain to me

Sent from my LG-H840 using Tapatalk

I would think as follows. Cos (2pi/3) = Cos (-2pi/3). So -2pi/3 is also a solution. I guess you only want solutions between 0 and 2pi. Note that Cos (-2pi/3) = Cos (-2pi/3 + 2pi) = Cos(-2pi/3 + 6pi/3) = Cos(4pi/3)

 

[mmm4444bot]

… I guess you only want solutions between 0 and 2pi …

I chose to skip commenting on the exercise statement, but, since you brought it up …

That guess matches mine, Jomo. The given function has infinite local extrema over its domain. The exercise statement ought to be more specific. The OP was given the two solutions posted, so we know in hindsight what the author intended, but you're correct in noting that answer as incomplete for the exercise as worded. :cool:

Find the local max/min val of the function:
y = x + 2sin(x)