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Math Problem

04/28/2011 4:22 PM

I'm reviewing my math skills from 30 years ago, and I was wondering if anyone could show me the math calculations for the values of Point P. If the diagram doesn't upload, I can email it to you as an Attachment. The diagram above is encoded with Windows Photo Gallery.

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#1

Re: Math Problem

04/28/2011 4:23 PM

No diagram... convert your diagram to a JPEG and reattempt to upload it... your question doesn't mean beans without the visual aid.

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#2
In reply to #1

Re: Math Problem

04/28/2011 6:34 PM

OK. I've just uploaded the problem to the Comments section of my Website: http://www.oberuch.org/Comments_WorldNews.html

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#3
In reply to #2

Re: Math Problem

04/28/2011 8:22 PM

??? I've read that problem a dozen times and can't figure out what it wants. The wording seems very odd. The math should be simple trig, but the wording is very confusing, especially the statement:

"Please show the math that gives the variables (√8/3, -1/3) of the circular function of the point P (0,1) of the unit circle above (x2 + y2 = 1)."

...reads as though it was written in some foreign language and then poorly translated into English. I.e.,

(√8/3, -1/3) -- This is a ordered pair of constants, not a variable.

"the circular function of the point P (0,1)" -- How can a point have a circular function? (Unless the circle's radius is zero.)

"of the unit circle above (x2 + y2 = 1)" -- I guess they're trying to say 'of the unit circle shown above (and is defined as x2 + y2 = 1)'.

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#4

Re: Math Problem

04/28/2011 9:42 PM

Neither the pic nor the Website worked for me; nor could I decipher the verbal description. Can you get this into a filetype that works with the CR4 posting window "Insert Graphic" camera?

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#5
In reply to #4

Re: Math Problem

04/29/2011 2:25 PM

OK. I tried to upload the diagram, but I was unable to upload to CR4 from my Windows Vista computer's Documents because they want a URL. They are combining Vector Algebra, Plane Trigonometry, and an introduction to Calculus in order to describe circular functions useful in electronics. The Point P in plane trigonometry can be described easily in terms of Radians or Degrees. However, in order to solve for the values shown, it needs to be solved in terms of applying either the Law of Sines or the Law of Cosines and solved from the formula for the Unit Circle shown.

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#6
In reply to #5

Re: Math Problem

04/29/2011 2:56 PM

To Usbport: Correct. They seem like an ordered pair of constants to me, but how does one calculate for the values shown for them given the information in the problem. If the values shown are wrong, anyone is welcome to show the work for the correct values. Thank you for your time.

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#8
In reply to #5

Re: Math Problem

04/29/2011 5:14 PM

Here we go. This should help.

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#7

Re: Math Problem

04/29/2011 4:39 PM

???我給那個問題讀了許多時期和不可以想出它想要的。措??? I've read that problem a dozen times and can't figure out what it wants. The wording seems very odd. The math should be simple trig, but the wording is very confusing, especially the statement:

"Please show the math that gives the variables (√8/3, -1/3) of the circular function of the point P (0,1) of the unit circle above (x2 + y2 = 1)."

...reads as though it was written in some foreign language and then poorly translated into English. I.e.,

(√8/3, -1/3) -- This is a ordered pair of constants, not a variable.

"the circular function of the point P (0,1)" -- How can a point have a circular function? (Unless the circle's radius is zero.)

"of the unit circle above (x2 + y2 = 1)" -- I guess they're trying to say 'of the unit circle shown above (and is defined as x2 + y2 = 1)'.

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起來很奇怪的。數學應該簡單瀟灑,但是措詞很在混淆不清,尤其是文告:
「請顯示那給其變數的數學(v8/3, - 1/3)點的循環的功能中 P(0,1) 單位的圈中在上面(x2」+「y2」=「1)。」
... 閱讀好像它以編寫一些外語,然後糟糕地譯成英語。即,
(v8/3, - 1/3) -- 這是一個有序副的常數,不是一個變數。
「點的循環的功能 P(0,1)」 -- 點可以有如何一種循環的功能?(除非圈的輻射光線是零。)
「單位的圈中在上面(x2」+「y2」=「1)」 -- 我猜測他們在嘗試說 ' 在上面被顯示的單位的圈中(和被定義為 x2 + y2 = 1)'。
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#9
In reply to #7

Re: Math Problem

04/29/2011 6:07 PM

This problem was taken out of a very reputable publishing company's math book with an ISBN and Library of Congress Number. They are leading up to more complex problems. This problem is a building block leading up to spherical trigonometry and its applications.

The graph of the equation shown is a circle in the rectangular coordinate system. This graph is called the unit circle and has its center at the origin and has a radius of 1 unit. Trigonometric functions are defined so that their domains are sets of angles and their ranges are sets of real numbers. Circular functions are defined such that their domains are sets of numbers that correspond to the measures (in radian units) of the angles of analogous trigonometric functions. The ranges of these circular functions, like their analogous trigonometric functions, are sets of real numbers. These functions are called circular functions because radian measures of angles are determined by the lengths of arcs of circles. In particular, trigonometric functions defined using the unit circle lead directly to these circular functions.

Begin with the unit circle shown. All six trigonometric functions exist on any point of an arc on the unit circle. Sin q and Cos q exist for each real number q because (sin q, cos q) are the coordinates of point P located on the unit circle, that correspond to an arc length of the absolute value of q. Because this arc lengh can be positive (counterclockwise) or negative (clockwise), the domain of each of these circular functions is the set of real numbers. The range is more restricted on the coordinate plane. The cosine and sine are the abscissa and ordinate of a point that moves around the unit circle, and they vary between -1 and 1. Therefore, the range of each of these functions is a set of real numbers z {-1,...1}.

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#10

Re: Math Problem

04/30/2011 1:02 AM

I don't see the point (√8/3, -1/3) on the diagram, but it is on the unit circle. (At about -19.47° or -0.34 radians.)

[(√8/3)2 + (-1/3)2 = 8/9 + 1/9 = 1.] [If that is what they were looking for.]

That doesn't sound like the world's best textbook....

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#11
In reply to #10

Re: Math Problem

04/30/2011 3:32 PM

Thank you for the response. I believe you need to take the cosine in the positive direction, and sine in the negative direction to arrive at the values shown. My calculations are close, but the text seems to leave step out in the Algebra used.

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#12
In reply to #11

Re: Math Problem

07/05/2011 2:34 PM

I figured it out. ThankyouThankyouThankyou for your help.

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Doorman (1); John Oberuch Jr (7); RVZ717 (1); Tornado (2); Usbport (1)

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