The inverse sine function, denoted
${\text{sin}}^{\mathrm{-1}}$ or arcsin, and the inverse cosine function, denoted
${\text{cos}}^{\mathrm{-1}}$ or arccos, are defined on the domain
$D=\text{{}x|-1\le x\le 1\}$ as follows:
$\begin{array}{c}{\text{sin}}^{\mathrm{-1}}\left(x\right)=y\phantom{\rule{0.2em}{0ex}}\text{if and only if}\phantom{\rule{0.2em}{0ex}}\text{sin}\left(y\right)=x\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}-\frac{\pi}{2}\le y\le \frac{\pi}{2};\hfill \\ {\text{cos}}^{\mathrm{-1}}\left(x\right)=y\phantom{\rule{0.2em}{0ex}}\text{if and only if}\phantom{\rule{0.2em}{0ex}}\text{cos}\left(y\right)=x\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}0\le y\le \pi .\hfill \end{array}$
The inverse tangent function, denoted
${\text{tan}}^{\mathrm{-1}}$ or arctan, and inverse cotangent function, denoted
${\text{cot}}^{\mathrm{-1}}$ or arccot, are defined on the domain
$D=\left\{x\right|-\infty <x<\infty \}$ as follows:
$\begin{array}{c}{\text{tan}}^{\mathrm{-1}}\left(x\right)=y\phantom{\rule{0.2em}{0ex}}\text{if and only if}\phantom{\rule{0.2em}{0ex}}\text{tan}\left(y\right)=x\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}-\frac{\pi}{2}<y<\frac{\pi}{2};\hfill \\ {\text{cot}}^{\mathrm{-1}}\left(x\right)=y\phantom{\rule{0.2em}{0ex}}\text{if and only if}\phantom{\rule{0.2em}{0ex}}\text{cot}\left(y\right)=x\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}0<y<\pi .\hfill \end{array}$
The inverse cosecant function, denoted
${\text{csc}}^{\mathrm{-1}}$ or arccsc, and inverse secant function, denoted
${\text{sec}}^{\mathrm{-1}}$ or arcsec, are defined on the domain
$D=\left\{x\right|\left|x\right|\ge 1\}$ as follows:
$\begin{array}{c}{\text{csc}}^{\mathrm{-1}}\left(x\right)=y\phantom{\rule{0.2em}{0ex}}\text{if and only if}\phantom{\rule{0.2em}{0ex}}\text{csc}\left(y\right)=x\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}-\frac{\pi}{2}\le y\le \frac{\pi}{2},y\ne 0;\hfill \\ {\text{sec}}^{\mathrm{-1}}\left(x\right)=y\phantom{\rule{0.2em}{0ex}}\text{if and only if}\phantom{\rule{0.2em}{0ex}}\text{sec}\left(y\right)=x\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}0\le y\le \pi ,y\ne \pi \text{/}2.\hfill \end{array}$
To graph the inverse trigonometric functions, we use the graphs of the trigonometric functions restricted to the domains defined earlier and reflect the graphs about the line
$y=x$ (
[link] ).
Go to the
following site for more comparisons of functions and their inverses.
When evaluating an inverse trigonometric function, the output is an angle. For example, to evaluate
${\text{cos}}^{\mathrm{-1}}\left(\frac{1}{2}\right),$ we need to find an angle
$\theta $ such that
$\text{cos}\phantom{\rule{0.1em}{0ex}}\theta =\frac{1}{2}.$ Clearly, many angles have this property. However, given the definition of
${\text{cos}}^{\mathrm{-1}},$ we need the angle
$\theta $ that not only solves this equation, but also lies in the interval
$\left[0,\pi \right].$ We conclude that
${\text{cos}}^{\mathrm{-1}}\left(\frac{1}{2}\right)=\frac{\pi}{3}.$
We now consider a composition of a trigonometric function and its inverse. For example, consider the two expressions
$\text{sin}\left({\text{sin}}^{\mathrm{-1}}\left(\frac{\sqrt{2}}{\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}2}\right)\right)$ and
${\text{sin}}^{\mathrm{-1}}(\text{sin}(\pi )).$ For the first one, we simplify as follows:
The inverse function is supposed to “undo” the original function, so why isn’t
${\text{sin}}^{\mathrm{-1}}\left(\text{sin}\left(\pi \right)\right)=\pi ?$ Recalling our definition of inverse functions, a function
$f$ and its inverse
${f}^{\mathrm{-1}}$ satisfy the conditions
$f\left({f}^{\mathrm{-1}}\left(y\right)\right)=y$ for all
$y$ in the domain of
${f}^{\mathrm{-1}}$ and
${f}^{\mathrm{-1}}\left(f\left(x\right)\right)=x$ for all
$x$ in the domain of
$f,$ so what happened here? The issue is that the inverse sine function,
${\text{sin}}^{\mathrm{-1}},$ is the inverse of the
restricted sine function defined on the domain
$\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\text{.}$ Therefore, for
$x$ in the interval
$\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\text{,}$ it is true that
${\text{sin}}^{\mathrm{-1}}\left(\text{sin}\phantom{\rule{0.1em}{0ex}}x\right)=x.$ However, for values of
$x$ outside this interval, the equation does not hold, even though
${\text{sin}}^{\mathrm{-1}}(\text{sin}\phantom{\rule{0.1em}{0ex}}x)$ is defined for all real numbers
$x.$
What about
$\text{sin}({\text{sin}}^{\mathrm{-1}}y)?$ Does that have a similar issue? The answer is
no . Since the domain of
${\text{sin}}^{\mathrm{-1}}$ is the interval
$\left[\mathrm{-1},1\right],$ we conclude that
$\text{sin}({\text{sin}}^{\mathrm{-1}}y)=y$ if
$\mathrm{-1}\le y\le 1$ and the expression is not defined for other values of
$y.$ To summarize,
Evaluating
${\text{sin}}^{\mathrm{-1}}\left(\text{\u2212}\sqrt{3}\text{/}2\right)$ is equivalent to finding the angle
$\theta $ such that
$\text{sin}\phantom{\rule{0.1em}{0ex}}\theta =\text{\u2212}\sqrt{3}\text{/}2$ and
$\text{\u2212}\pi \text{/}2\le \theta \le \pi \text{/}2.$ The angle
$\theta =\text{\u2212}\pi \text{/}3$ satisfies these two conditions. Therefore,
${\text{sin}}^{\mathrm{-1}}\left(\text{\u2212}\sqrt{3}\text{/}2\right)=\text{\u2212}\pi \text{/}3.$
First we use the fact that
${\text{tan}}^{\mathrm{-1}}\left(\mathrm{-1}\text{/}\sqrt{3}\right)=\text{\u2212}\pi \text{/}6.$ Then
$\text{tan}\left(\pi \text{/}6\right)=\mathrm{-1}\text{/}\sqrt{3}.$ Therefore,
$\text{tan}\left({\text{tan}}^{\mathrm{-1}}\left(\mathrm{-1}\text{/}\sqrt{3}\right)\right)=\mathrm{-1}\text{/}\sqrt{3}.$
To evaluate
${\text{cos}}^{\mathrm{-1}}\left(\text{cos}\left(5\pi \text{/}4\right)\right),$ first use the fact that
$\text{cos}\left(5\pi \text{/}4\right)=\text{\u2212}\sqrt{2}\text{/}2.$ Then we need to find the angle
$\theta $ such that
$\text{cos}\left(\theta \right)=\text{\u2212}\sqrt{2}\text{/}2$ and
$0\le \theta \le \pi .$ Since
$3\pi \text{/}4$ satisfies both these conditions, we have
$\text{cos}\left({\text{cos}}^{\mathrm{-1}}\left(5\pi \text{/}4\right)\right)=\text{cos}\left({\text{cos}}^{\mathrm{-1}}\left(\text{\u2212}\sqrt{2}\text{/}2\right)\right)=3\pi \text{/}4.$
Since
$\text{cos}\left(2\pi \text{/}3\right)=\mathrm{-1}\text{/}2,$ we need to evaluate
${\text{sin}}^{\mathrm{-1}}\left(\mathrm{-1}\text{/}2\right).$ That is, we need to find the angle
$\theta $ such that
$\text{sin}\left(\theta \right)=\mathrm{-1}\text{/}2$ and
$\text{\u2212}\pi \text{/}2\le \theta \le \pi \text{/}2.$ Since
$\text{\u2212}\pi \text{/}6$ satisfies both these conditions, we can conclude that
${\text{sin}}^{\mathrm{-1}}\left(\text{cos}\left(2\pi \text{/}3\right)\right)={\text{sin}}^{\mathrm{-1}}\left(\mathrm{-1}\text{/}2\right)=\text{\u2212}\pi \text{/}6.$
Luke 17:21 nor will they say, See here or See there For indeed, the kingdom of God is within you.
You've never 'touched' anything. The e-energy field created by your body has pushed other electricfields. even our religions tell us we're the gods. We live in energies connecting us all. Doa/higgsfield
Scott
if you have any calculus questions many of us would be happy to help and you can always learn or even invent your own theories and proofs. math is the laws of logic and reality. its rules are permanent and absolute. you can absolutely learn calculus and through it better understand our existence.
Scott
ya doubtless
Bilal
help the integral of x^2/lnxdx
Levis
also find the value of "X" from the equation that follow
(x-1/x)^4 +4(x^2-1/x^2) -6=0
please guy help
Levis
Use integration by parts. Let u=lnx and dv=x2dx Then du=1xdx and v=13x3.
∫x2lnxdx=13x3lnx−∫(13x3⋅1x)dx
∫x2lnxdx=13x3lnx−∫13x2dx
∫x2lnxdx=13x3lnx−19x3+C
Bilal
itz 1/3 and 1/9
Bilal
now you can find the value of X from the above equation easily
Bilal
Pls i need more explanation on this calculus
usman
usman
from where do you need help?
Levis
thanks Bilal
Levis
integrate e^cosx
Uchenna
-sinx e^x
Leo
Do we ask only math question?
or ANY of the question?
bilal kumhar you are so biased
if you are an expert what are you doing here lol😎😎😂😂 we are here to learn
and beside there are many questions on this chat which you didn't attempt
we are helping each other stop being naive and arrogance
so give me the integral of x^x
Levis
Levis
I am sorry
Bilal
Bilal
it okay buddy
honestly i am pleasured to meet you
Levis
x^x ... no anti derivative for this function... but we can find definte integral numerically.
Bilal
thank you Bilal Kumhar
then how we may find definite integral let say x^x,3,5?
Levis
evaluate 5-×square divided by x+2
find x as limit approaches infinity
if your asking for derivative
dy/dz=x^2/2(lnx-1/2)
Levis
iam sorry
f(x)=x^x
it means the output(range ) depends to input(domain) value of x by the power of x
that is to say if x=2 then x^x would be 2^2=4
f(x) is the product of X to the power of X
its derivatives is found by using product rule
y=x^x
introduce ln each side we have
lny=lnx^x
=lny=xlnx
A maxima in a curve refers to the maximum point said curve. The maxima is a point where the gradient of the curve is equal to 0 (dy/dx = 0) and its second derivative value is a negative (d²y/dx² = -ve).
Leo we don't just do like that buddy!!!
use first principle
y+∆y=x+∆x
∆y=x+∆x-y
∆y=(x+∆x)^2+(x+∆x)-x^2+x
on solving it become
∆y=3∆x+∆x^2 as ∆x_>0 limit=3 if you do by calculator say plugging any value of x=0.000005 which approach 0 you get 3