Yahoo Answers: Answers and Comments for How do i graph Sec^1? [Homework Help]
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From Anonymous
enUS
Sun, 28 Sep 2008 16:32:38 +0000
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Yahoo Answers: Answers and Comments for How do i graph Sec^1? [Homework Help]
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From Sandra: This is a separable DEQ [ sec^2(y) 1 ] dy = d...
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Thu, 07 Apr 2016 03:26:34 +0000
This is a separable DEQ [ sec^2(y) 1 ] dy = dx ===>intgerate each side tan y  y = x +C we need more info to find the xccordinate dy/dx is vertical when denom of dy/dx = 0 dy/dx =1/ (sec^2y 1) does not exist when sec^2(y) = 1 sec y = +/ 1 cos y = +/ 1 y = (n pi) ===> to get xccordinate x = tan y  y +C x = 0  n*pi + C x = (n * pi) + C (need more to get C) deriv of dy/dx is (1)(sec^2 y  1)^2 * (2secysecytany * dy/dx) =2sec^2(y)tan(y) / (sec^2 y  1)^3

From Are you mad? No? Press ALT+F4.:
I dont know
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https://answers.yahoo.com/question/index?qid=20080928163238AAxxbVC
Sun, 28 Sep 2008 16:35:02 +0000
I dont know

From Anonymous: sec^1(x)=an angle, for our case, theta,
Le...
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Sun, 28 Sep 2008 18:24:27 +0000
sec^1(x)=an angle, for our case, theta,
Lets use the unit circle and basic definitions to determine exactly what we are talking about
lets let y equal the placement of the point on the unit circle projected on the y axis and x equal the projection of the point on the unit circle on the x axis, thus we can talk about a point on the unit circle as (x,y)
so, we can determine a few of the basic trigonometric terms using this new definition.
we will also use SOHCAHTOA to help (Sin = Opposite over Hypotenuse, Cos = Adjacent over Hypotenuse, Tan= Opposite Over Adjacent)
we will also use the Pythagorean theorem so that if x is the adjacent side length, and y is the opposite side length, the square root (sqrt) of the sum of each side squared will give us the length of the hypotenuse:
let z equal the length of the hypotenuse,
z^2=x^2+y^2
z=sqrt(x^2+y^2)
Using the definitions, we get
Sin = y/z
Cos = x/z
Tan = y/x
and using the trigonometric identities
Csc = 1/sin
Sec = 1/cos
Cot = 1/tan
we get
Csc = z/y
Sec = z/x
Cot = x/y
So, to answer your question,
Sec^1 is the angle that produces Sec
lets let the angle be theta
so, it is safe to determine that
sec (theta) = sqrt(x^2+y^2)/x
so we must find the theta that allows for sqrt (x^2+y^2) = 0
This can only occur when x and y are both zero, or where one of the variables is 1 and the other is i (aqrt (1)), seeing as i cannot be graphed on the xy coordinate system, the only safe answer is where x and y are zero. Unfortunately this means that we are dividing by zero, so in fact, it is impossible to get a zero for the graph of sec (theta), and likewise, we cannot find a value for theta that produces a zero.
Therefore I believe that Sec1(theta) has no zeros.
Hope this helps

From hayharbr: y = Sec ^ 1 x means y is the angle whose seca...
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Sun, 28 Sep 2008 16:58:18 +0000
y = Sec ^ 1 x means y is the angle whose secant is x. If its secant is x, its cosine is 1/x. So to graph it, use y = (cos^1 (1/x)). Secants only go from 1 up or 1 down so the domain is x ≥ 1 or x ≤ 1
Since the S in Sec is capitalized it's just the part which is a function. Its range is then from 0 to pi, just like that of inverse cosine.

From Anonymous: wen typing it into the graphing calculator, u ...
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Sun, 28 Sep 2008 16:35:52 +0000
wen typing it into the graphing calculator, u would need to convert secant into sin, cos, tan version , it'll make it alot easier