On these restricted domains, we can define the
inverse trigonometric functions .
The
inverse sine function$\text{\hspace{0.17em}}y={\mathrm{sin}}^{-1}x\text{\hspace{0.17em}}$ means
$\text{\hspace{0.17em}}x=\mathrm{sin}\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$ The inverse sine function is sometimes called the
arcsine function, and notated
$\text{\hspace{0.17em}}\mathrm{arcsin}x.$
The
inverse cosine function$\text{\hspace{0.17em}}y={\mathrm{cos}}^{-1}x\text{\hspace{0.17em}}$ means
$\text{\hspace{0.17em}}x=\mathrm{cos}\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$ The inverse cosine function is sometimes called the
arccosine function, and notated
$\text{\hspace{0.17em}}\mathrm{arccos}\text{\hspace{0.17em}}x.$
The
inverse tangent function$\text{\hspace{0.17em}}y={\mathrm{tan}}^{-1}x\text{\hspace{0.17em}}$ means
$\text{\hspace{0.17em}}x=\mathrm{tan}\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$ The inverse tangent function is sometimes called the
arctangent function, and notated
$\text{\hspace{0.17em}}\mathrm{arctan}\text{\hspace{0.17em}}x.$
The graphs of the inverse functions are shown in
[link] ,
[link] , and
[link] . Notice that the output of each of these inverse functions is a
number, an angle in radian measure. We see that
$\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}x\text{\hspace{0.17em}}$ has domain
$\text{\hspace{0.17em}}\left[\mathrm{-1},1\right]\text{\hspace{0.17em}}$ and range
$\text{\hspace{0.17em}}\left[-\frac{\pi}{2},\frac{\pi}{2}\right],$${\mathrm{cos}}^{-1}x\text{\hspace{0.17em}}$ has domain
$\text{\hspace{0.17em}}\left[\mathrm{-1}\mathrm{,1}\right]\text{\hspace{0.17em}}$ and range
$\text{\hspace{0.17em}}[0,\pi ],\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}{\mathrm{tan}}^{-1}x\text{\hspace{0.17em}}$ has domain of all real numbers and range
$\text{\hspace{0.17em}}\left(-\frac{\pi}{2},\frac{\pi}{2}\right).\text{\hspace{0.17em}}$ To find the
domain and
range of inverse trigonometric functions, switch the domain and range of the original functions. Each graph of the inverse trigonometric function is a reflection of the graph of the original function about the line
$\text{\hspace{0.17em}}y=x.$
Relations for inverse sine, cosine, and tangent functions
For angles in the interval
$\text{\hspace{0.17em}}\left[-\frac{\pi}{2},\frac{\pi}{2}\right],\text{\hspace{0.17em}}$ if
$\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}y=x,\text{\hspace{0.17em}}$ then
$\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}x=y.$
For angles in the interval
$\text{\hspace{0.17em}}\left[0,\pi \right],\text{\hspace{0.17em}}$ if
$\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}y=x,\text{\hspace{0.17em}}$ then
$\text{\hspace{0.17em}}{\mathrm{cos}}^{-1}x=y.$
For angles in the interval
$\text{\hspace{0.17em}}\left(-\frac{\pi}{2},\frac{\pi}{2}\right),\text{\hspace{0.17em}}$ if
$\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}y=x,\text{\hspace{0.17em}}$ then
$\text{\hspace{0.17em}}{\mathrm{tan}}^{-1}x=y.$
Writing a relation for an inverse function
Given
$\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{5\pi}{12}\right)\approx 0.96593,\text{\hspace{0.17em}}$ write a relation involving the inverse sine.
Use the relation for the inverse sine. If
$\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}y=x,\text{\hspace{0.17em}}$ then
$\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}x=y$ .
In this problem,
$\text{\hspace{0.17em}}x=0.96593,\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}y=\frac{5\pi}{12}.$
Finding the exact value of expressions involving the inverse sine, cosine, and tangent functions
Now that we can identify inverse functions, we will learn to evaluate them. For most values in their domains, we must evaluate the inverse trigonometric functions by using a calculator, interpolating from a table, or using some other numerical technique. Just as we did with the original trigonometric functions, we can give exact values for the inverse functions when we are using the special angles, specifically
$\text{\hspace{0.17em}}\frac{\pi}{6}\text{\hspace{0.17em}}$ (30°),
$\text{\hspace{0.17em}}\frac{\pi}{4}\text{\hspace{0.17em}}$ (45°), and
$\text{\hspace{0.17em}}\frac{\pi}{3}\text{\hspace{0.17em}}$ (60°), and their reflections into other quadrants.
Given a “special” input value, evaluate an inverse trigonometric function.
Find angle
$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ for which the original trigonometric function has an output equal to the given input for the inverse trigonometric function.
If
$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is not in the defined range of the inverse, find another angle
$\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ that is in the defined range and has the same sine, cosine, or tangent as
$\text{\hspace{0.17em}}x,$ depending on which corresponds to the given inverse function.
Evaluating inverse trigonometric functions for special input values
Evaluating
$\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)\text{\hspace{0.17em}}$ is the same as determining the angle that would have a sine value of
$\text{\hspace{0.17em}}\frac{1}{2}.\text{\hspace{0.17em}}$ In other words, what angle
$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ would satisfy
$\text{\hspace{0.17em}}\mathrm{sin}(x)=\frac{1}{2}?\text{\hspace{0.17em}}$ There are multiple values that would satisfy this relationship, such as
$\text{\hspace{0.17em}}\frac{\pi}{6}\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}\frac{5\pi}{6},\text{\hspace{0.17em}}$ but we know we need the angle in the interval
$\text{\hspace{0.17em}}\left[-\frac{\pi}{2},\frac{\pi}{2}\right],\text{\hspace{0.17em}}$ so the answer will be
$\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}.\text{\hspace{0.17em}}$ Remember that the inverse is a function, so for each input, we will get exactly one output.
To evaluate
$\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}\left(-\frac{\sqrt{2}}{2}\right),\text{\hspace{0.17em}}$ we know that
$\text{\hspace{0.17em}}\frac{5\pi}{4}\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}\frac{7\pi}{4}\text{\hspace{0.17em}}$ both have a sine value of
$\text{\hspace{0.17em}}-\frac{\sqrt{2}}{2},\text{\hspace{0.17em}}$ but neither is in the interval
$\text{\hspace{0.17em}}\left[-\frac{\pi}{2},\frac{\pi}{2}\right].\text{\hspace{0.17em}}$ For that, we need the negative angle coterminal with
$\text{\hspace{0.17em}}\frac{7\pi}{4}:$${\text{sin}}^{-1}(-\frac{\sqrt{2}}{2})=-\frac{\pi}{4}.\text{\hspace{0.17em}}$
To evaluate
$\text{\hspace{0.17em}}{\mathrm{cos}}^{-1}\left(-\frac{\sqrt{3}}{2}\right),\text{\hspace{0.17em}}$ we are looking for an angle in the interval
$\text{\hspace{0.17em}}\left[0,\pi \right]\text{\hspace{0.17em}}$ with a cosine value of
$\text{\hspace{0.17em}}-\frac{\sqrt{3}}{2}.\text{\hspace{0.17em}}$ The angle that satisfies this is
$\text{\hspace{0.17em}}{\mathrm{cos}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)=\frac{5\pi}{6}.$
Evaluating
$\text{\hspace{0.17em}}{\mathrm{tan}}^{-1}\left(1\right),\text{\hspace{0.17em}}$ we are looking for an angle in the interval
$\text{\hspace{0.17em}}\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\text{\hspace{0.17em}}$ with a tangent value of 1. The correct angle is
$\text{\hspace{0.17em}}{\mathrm{tan}}^{-1}\left(1\right)=\frac{\pi}{4}.$
The anwser is imaginary
number if you want to know The anwser of the expression
you must arrange The expression and use quadratic formula To find the
answer
master
The anwser is imaginary
number if you want to know The anwser of the expression
you must arrange The expression and use quadratic formula To find the
answer
master
Y
master
X2-2X+8-4X2+12X-20=0
(X2-4X2)+(-2X+12X)+(-20+8)= 0
-3X2+10X-12=0
3X2-10X+12=0
Use quadratic formula To find the answer
answer (5±Root11i)/3
master
Soo sorry (5±Root11* i)/3
master
x2-2x+8-4x2+12x-20
x2-4x2-2x+12x+8-20
-3x2+10x-12
now you can find the answer using quadratic
Mukhtar
explain and give four example of hyperbolic function
I think the formula for calculating algebraic is the statement of the equality of two expression stimulate by a set of addition, multiplication, soustraction, division, raising to a power and extraction of Root. U believe by having those in the equation you will be in measure to calculate it
the Cayley hamilton Theorem state if A is a square matrix and if f(x) is its characterics polynomial then f(x)=0 in another ways evey square matrix is a root of its chatacteristics polynomial.