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
${\text{sin}}^{\mathrm{-1}}\left(\text{cos}\left(2\pi \text{/}3\right)\right)={\text{sin}}^{\mathrm{-1}}\left(\mathrm{-1}\text{/}2\right)=\text{\u2212}\pi \text{/}6.$
If you're using calculus to find the range, you have to find the extrema through the first derivative test and then substitute the x-value for the extrema back into the original equation.
Sand falling freely from a lorry form a conical shape whose height is always equal to one-third the radius of the base.
a. How fast is the volume increasing when the radius of the base is (1m) and increasing at the rate of 1/4cm/sec
Pls help me solve
also known as the y value so I could say y=2x or f(x)= 2x same thing just using functional notation
your next question is what is dependent and independent variables. I am Dyslexic but know math and which is which confuses me. but one can vary the x value while y depends on which x you use. also
Marc
up domain and range
Marc
enjoy your work and good luck
Marc
I actually wanted to ask another questions on sets if u dont mind please?
Inembo
I have so many questions on set and I really love dis app I never believed u would reply
Inembo
Hmm go ahead and ask you got me curious too much conversation here
Adri
am sorry for disturbing I really want to know math that's why
*I want to know the meaning of those symbols in sets*
e.g n,U,A', etc pls I want to know it and how to solve its problems
Inembo
and how can i solve a question like dis
*in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
next questions
what do dy mean by (A' n B^c)^c'
Inembo
The sets help you to define the function. The function is like a magic box where you put inside stuff(numbers or sets) and you get out the stuff but in different shapes (forms).
Adri
I dont understand what you wanna say by (A' n B^c)^c'
Adri
(A' n B (rise to the power of c)) all rise to the power of c
Inembo
Aaaahh
Adri
Ok so the set is formed by vectors and not numbers
Adri
A vector of length n
Adri
But you can make a set out of matrixes as well
Adri
I I don't even understand sets I wat to know d meaning of all d symbolsnon sets
Inembo
Wait what's your math level?
Adri
High-school?
Adri
yes
Inembo
am having big problem understanding sets more than other math topics
Inembo
So f:R->R means that the function takes real numbers and provides real numer. For ex. If f(x) =2x this means if you give to your function a real number like 2,it gives you also a real number 2times2=4
Adri
pls answer this question
*in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
If you have f:R^n->R^n you give to your function a vector of length n like (a1,a2,...an) where all a1,.. an are reals and gives you also a vector of length n... I don't know if i answering your question. Otherwise on YouTube you havr many videos where they explain it in a simple way
Adri
I would say 24
Adri
Offer both
Adri
Sorry 20
Adri
Actually you have 40 - 4 =36 who offer maths or physics or both.
Adri
I know its 20 but how to prove it
Inembo
You have 32+24=56who offer courses
Adri
56-36=20 who give both courses... I would say that
Adri
solution: In a question involving sets and Venn diagram, the sum of the members of set A + set B - the joint members of both set A and B + the members that are not in sets A or B = the total members of the set. In symbolic form n(A U B) = n(A) + n (B) - n (A and B) + n (A U B)'.
Mckenzie
In the case of sets A and B use the letters m and p to represent the sets and we have:
n (M U P) = 40; n (M) = 24; n (P) = 32; n (M and P) = unknown; n (M U P)' = 4
Mckenzie
Now substitute the numerical values for the symbolic representation
40 = 24 + 32 - n(M and P) + 4
Now solve for the unknown using algebra:
40 = 24 + 32+ 4 - n(M and P)
40 = 60 - n(M and P)
Add n(M and P), as well, subtract 40 from both sides of the equation to find the answer.
Mckenzie
40 - 40 + n(M and P) = 60 - 40 - n(M and P) + n(M and P)
Solution:
n(M and P) = 20
Mckenzie
thanks
Inembo
Simpler form:
Add the sums of set M, set P and the complement of the union of sets M and P then subtract the number of students from the total.