Anonymous

# how do i graph Sec^-1?

What is the domain, range and zeros of the graph y= Sec^-1

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• Anonymous

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

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 x-y 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 Sec-1(theta) has no zeros.

Hope this helps

• 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.

• 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 x-ccordinate 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 x-ccordinate 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

• Anonymous