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}.$
Given
$\text{\hspace{0.17em}}\mathrm{cos}(0.5)\approx \mathrm{0.8776,}$ write a relation involving the inverse cosine.
$\mathrm{arccos}(0.8776)\approx 0.5$
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
$\text{\hspace{0.17em}}{\mathrm{tan}}^{-1}\left(1\right)=\frac{\pi}{4}.$
Questions & Answers
so some one know about replacing silicon atom with phosphorous in semiconductors device?
In this morden time nanotechnology used in many field .
1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc
2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc
3- Atomobile -MEMS, Coating on car etc.
and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change .
maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.