# 6.3 Inverse trigonometric functions  (Page 4/15)

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## Evaluating compositions of the form f ( f−1 ( y )) and f−1 ( f ( x ))

For any trigonometric function, $\text{\hspace{0.17em}}f\left({f}^{-1}\left(y\right)\right)=y\text{\hspace{0.17em}}$ for all $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ in the proper domain for the given function. This follows from the definition of the inverse and from the fact that the range of $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ was defined to be identical to the domain of $\text{\hspace{0.17em}}{f}^{-1}.\text{\hspace{0.17em}}$ However, we have to be a little more careful with expressions of the form $\text{\hspace{0.17em}}{f}^{-1}\left(f\left(x\right)\right).$

## Compositions of a trigonometric function and its inverse

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{sin}\left({\mathrm{sin}}^{-1}x\right)=x\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}-1\le x\le 1\hfill \\ \mathrm{cos}\left({\mathrm{cos}}^{-1}x\right)=x\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}-1\le x\le 1\hfill \\ \text{\hspace{0.17em}}\mathrm{tan}\left({\mathrm{tan}}^{-1}x\right)=x\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}-\infty

Is it correct that $\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}\left(\mathrm{sin}\text{\hspace{0.17em}}x\right)=x?$

No. This equation is correct if $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ belongs to the restricted domain $\text{\hspace{0.17em}}\left[-\frac{\pi }{2},\frac{\pi }{2}\right],\text{\hspace{0.17em}}$ but sine is defined for all real input values, and for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ outside the restricted interval, the equation is not correct because its inverse always returns a value in $\text{\hspace{0.17em}}\left[-\frac{\pi }{2},\frac{\pi }{2}\right].\text{\hspace{0.17em}}$ The situation is similar for cosine and tangent and their inverses. For example, $\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}\left(\mathrm{sin}\left(\frac{3\pi }{4}\right)\right)=\frac{\pi }{4}.$

Given an expression of the form f −1 (f(θ)) where evaluate.

1. If $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is in the restricted domain of
2. If not, then find an angle $\text{\hspace{0.17em}}\varphi \text{\hspace{0.17em}}$ within the restricted domain of $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}f\left(\varphi \right)=f\left(\theta \right).\text{\hspace{0.17em}}$ Then $\text{\hspace{0.17em}}{f}^{-1}\left(f\left(\theta \right)\right)=\varphi .$

## Using inverse trigonometric functions

Evaluate the following:

1. ${\mathrm{sin}}^{-1}\left(\mathrm{sin}\left(\frac{\pi }{3}\right)\right)$
2. ${\mathrm{sin}}^{-1}\left(\mathrm{sin}\left(\frac{2\pi }{3}\right)\right)$
3. ${\mathrm{cos}}^{-1}\left(\mathrm{cos}\left(\frac{2\pi }{3}\right)\right)$
4. ${\mathrm{cos}}^{-1}\left(\mathrm{cos}\left(-\frac{\pi }{3}\right)\right)$
1. so $\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}\left(\mathrm{sin}\left(\frac{\pi }{3}\right)\right)=\frac{\pi }{3}.$
2. but $\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{2\pi }{3}\right)=\mathrm{sin}\left(\frac{\pi }{3}\right),\text{\hspace{0.17em}}$ so $\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}\left(\mathrm{sin}\left(\frac{2\pi }{3}\right)\right)=\frac{\pi }{3}.$
3. so $\text{\hspace{0.17em}}{\mathrm{cos}}^{-1}\left(\mathrm{cos}\left(\frac{2\pi }{3}\right)\right)=\frac{2\pi }{3}.$
4. but $\text{\hspace{0.17em}}\mathrm{cos}\left(-\frac{\pi }{3}\right)=\mathrm{cos}\left(\frac{\pi }{3}\right)\text{\hspace{0.17em}}$ because cosine is an even function.
5. so $\text{\hspace{0.17em}}{\mathrm{cos}}^{-1}\left(\mathrm{cos}\left(-\frac{\pi }{3}\right)\right)=\frac{\pi }{3}.$

Evaluate $\text{\hspace{0.17em}}{\mathrm{tan}}^{-1}\left(\mathrm{tan}\left(\frac{\pi }{8}\right)\right)\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}{\mathrm{tan}}^{-1}\left(\mathrm{tan}\left(\frac{11\pi }{9}\right)\right).$

$\frac{\pi }{8};\frac{2\pi }{9}$

## Evaluating compositions of the form f−1 ( g ( x ))

Now that we can compose a trigonometric function with its inverse, we can explore how to evaluate a composition of a trigonometric function and the inverse of another trigonometric function. We will begin with compositions of the form $\text{\hspace{0.17em}}{f}^{-1}\left(g\left(x\right)\right).\text{\hspace{0.17em}}$ For special values of $\text{\hspace{0.17em}}x,$ we can exactly evaluate the inner function and then the outer, inverse function. However, we can find a more general approach by considering the relation between the two acute angles of a right triangle where one is $\text{\hspace{0.17em}}\theta ,\text{\hspace{0.17em}}$ making the other $\text{\hspace{0.17em}}\frac{\pi }{2}-\theta .$ Consider the sine and cosine of each angle of the right triangle in [link] .

Because $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta =\frac{b}{c}=\mathrm{sin}\left(\frac{\pi }{2}-\theta \right),\text{\hspace{0.17em}}$ we have $\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}\left(\mathrm{cos}\text{\hspace{0.17em}}\theta \right)=\frac{\pi }{2}-\theta \text{\hspace{0.17em}}$ if $\text{\hspace{0.17em}}0\le \theta \le \pi .\text{\hspace{0.17em}}$ If $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is not in this domain, then we need to find another angle that has the same cosine as $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ and does belong to the restricted domain; we then subtract this angle from $\text{\hspace{0.17em}}\frac{\pi }{2}.$ Similarly, $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta =\frac{a}{c}=\mathrm{cos}\left(\frac{\pi }{2}-\theta \right),\text{\hspace{0.17em}}$ so $\text{\hspace{0.17em}}{\mathrm{cos}}^{-1}\left(\mathrm{sin}\text{\hspace{0.17em}}\theta \right)=\frac{\pi }{2}-\theta \text{\hspace{0.17em}}$ if $\text{\hspace{0.17em}}-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}.\text{\hspace{0.17em}}$ These are just the function-cofunction relationships presented in another way.

Given functions of the form $\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}\left(\mathrm{cos}\text{\hspace{0.17em}}x\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{\mathrm{cos}}^{-1}\left(\mathrm{sin}\text{\hspace{0.17em}}x\right),\text{\hspace{0.17em}}$ evaluate them.

1. If then $\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}\left(\mathrm{cos}\text{\hspace{0.17em}}x\right)=\frac{\pi }{2}-x.$
2. If then find another angle such that $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}y=\mathrm{cos}\text{\hspace{0.17em}}x.$
${\mathrm{sin}}^{-1}\left(\mathrm{cos}\text{\hspace{0.17em}}x\right)=\frac{\pi }{2}-y$
3. If then $\text{\hspace{0.17em}}{\mathrm{cos}}^{-1}\left(\mathrm{sin}\text{\hspace{0.17em}}x\right)=\frac{\pi }{2}-x.$
4. If then find another angle such that $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}y=\mathrm{sin}\text{\hspace{0.17em}}x.$
${\mathrm{cos}}^{-1}\left(\mathrm{sin}\text{\hspace{0.17em}}x\right)=\frac{\pi }{2}-y$

what is functions?
A mathematical relation such that every input has only one out.
Spiro
yes..it is a relationo of orders pairs of sets one or more input that leads to a exactly one output.
Mubita
Is a rule that assigns to each element X in a set A exactly one element, called F(x), in a set B.
RichieRich
If the plane intersects the cone (either above or below) horizontally, what figure will be created?
can you not take the square root of a negative number
No because a negative times a negative is a positive. No matter what you do you can never multiply the same number by itself and end with a negative
lurverkitten
Actually you can. you get what's called an Imaginary number denoted by i which is represented on the complex plane. The reply above would be correct if we were still confined to the "real" number line.
Liam
Suppose P= {-3,1,3} Q={-3,-2-1} and R= {-2,2,3}.what is the intersection
can I get some pretty basic questions
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Ama
is precalculus needed to take caculus
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Spiro
the solution doesn't seem right for this problem
what is the domain of f(x)=x-4/x^2-2x-15 then
x is different from -5&3
Seid
All real x except 5 and - 3
Spiro
***youtu.be/ESxOXfh2Poc
Loree
how to prroved cos⁴x-sin⁴x= cos²x-sin²x are equal
Don't think that you can.
Elliott
By using some imaginary no.
Tanmay
how do you provided cos⁴x-sin⁴x = cos²x-sin²x are equal
What are the question marks for?
Elliott
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For the first question, I got (3y-2)/15 Second one, I got Root 2 Third one, I got 1/(y to the fourth power) I dont if it's right cause I can barely understand the question.
Is under distribute property, inverse function, algebra and addition and multiplication function; so is a combined question
Abena
find the equation of the line if m=3, and b=-2
graph the following linear equation using intercepts method. 2x+y=4
Ashley
how
Wargod
what?
John
ok, one moment
UriEl
how do I post your graph for you?
UriEl
it won't let me send an image?
UriEl
also for the first one... y=mx+b so.... y=3x-2
UriEl
y=mx+b you were already given the 'm' and 'b'. so.. y=3x-2
Tommy
Please were did you get y=mx+b from
Abena
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Tommy
thanks Tommy
Nimo
0=3x-2 2=3x x=3/2 then . y=3/2X-2 I think
Given
co ordinates for x x=0,(-2,0) x=1,(1,1) x=2,(2,4)
neil
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so good
abdikarin
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
strategies to form the general term
carlmark
consider r(a+b) = ra + rb. The a and b are the trig identity.
Mike