# inverse trig values and finding inverse

• Apr 5th 2009, 09:57 PM
deltemis
inverse trig values and finding inverse
I have been having problems with these 3 problems and I'm looking for the answer and how it was obtained so I can make sense of what I need to do and what I've been doing wrong

finding the exact values of following expressions:

problem 1: inverse cos(cos(-4pi/7))

for problem 1 I've tried using the sum and difference formulas but I simply can't get to 4pi/7

problem 2: inverse sec(root 2)

for problem 2 I can't understand how to find out what the proper values are

problem 3 - find inverse of sin(x+2) -1

once again, boggled by how to do so

I appreciate any help, explanations and hints go a long way

• Apr 5th 2009, 10:45 PM
Reckoner
Quote:

Originally Posted by deltemis
problem 1: inverse cos(cos(-4pi/7))

for problem 1 I've tried using the sum and difference formulas but I simply can't get to 4pi/7

What is the composition of a function with its inverse? You should always have $\displaystyle f^{-1}\left(f(x)\right)=x.$ In the case of $\displaystyle \arccos,$ the function is only an inverse on the interval $\displaystyle [0,\pi],$ so you need to make sure that the value falls in this range (if it does not, find an angle in the range that has an equal cosine).

Quote:

problem 2: inverse sec(root 2)

for problem 2 I can't understand how to find out what the proper values are
The secant is just the reciprocal of the cosine. Look at your unit circle: what common value of $\displaystyle \theta$ gives $\displaystyle \frac1{\cos\theta}=\sqrt2?$

Maybe this will make it more obvious:

$\displaystyle \frac1{\cos\theta}=\sqrt2\Rightarrow\cos\theta=\fr ac1{\sqrt2}=\frac{\sqrt2}2.$

Keep in mind that the range of $\displaystyle \text{arcsec}$ is defined as $\displaystyle \left[0,\frac\pi2\right)\cup\left(\frac\pi2,\pi\right]$ (usually, but your textbook or instructor may have defined it another way).

Quote:

problem 3 - find inverse of sin(x+2) -1

once again, boggled by how to do so
Let $\displaystyle f(x)=\sin(x+2)-1.$ Then the inverse, if it exists, satisfies

$\displaystyle x=\sin\left[f^{-1}(x)+2\right]-1$

$\displaystyle \Leftrightarrow\sin\left[f^{-1}(x)+2\right]=x+1$

$\displaystyle \Leftrightarrow f^{-1}(x)=\arcsin(x+1)-2,\;-2\leq x\leq0$
• Apr 5th 2009, 11:09 PM
deltemis
thanks reckoner, I understand problem 1 and 2 now but 3 still gets me

when I run through the steps you've provided here, logically I'm not understanding how it works.

when I graph the original function, it shows the periodicity of that function like a sine wave should appear with the changes, however when I graph the inverse with domain restricted to -2 <= x <= 0, I don't get why it is what it is

domain wise with the laws of functions and their inverses, the range of the original function is typically the domain of the inverse however when I graph the inverse graph, there's no switch between the original function's domain to the inverse's range

could you give a little bit more explanation as to what you did and what comes from each step?
• Apr 5th 2009, 11:48 PM
Reckoner
Quote:

Originally Posted by deltemis
when I run through the steps you've provided here, logically I'm not understanding how it works.

Given $\displaystyle y$ as a function of $\displaystyle x,$ the inverse can be obtained by swapping $\displaystyle x$ and $\displaystyle y,$ and then solving for $\displaystyle y.$ For example, take $\displaystyle f(x)=4x-3.$ We find the inverse as follows:

$\displaystyle y=4x-3$

$\displaystyle x=4y-3$

$\displaystyle \Rightarrow 4y=x+3$

$\displaystyle \Rightarrow y=\frac{x+3}4$

and $\displaystyle f^{-1}(x)=\frac{x+3}4.$

In your problem, $\displaystyle x$ needs to be restricted to $\displaystyle [-2,0]$ because $\displaystyle \arcsin(x+1)$ is only defined for $\displaystyle -1\leq x+1\leq1\Rightarrow-2\leq x\leq0.$

Quote:

when I graph the original function, it shows the periodicity of that function like a sine wave should appear with the changes, however when I graph the inverse with domain restricted to -2 <= x <= 0, I don't get why it is what it is
Graphically, a function's inverse is its reflection about the line $\displaystyle y=x.$ Look at the attached graph. Don't these look like reflections (with proper restrictions on $\displaystyle x$ and $\displaystyle y,$ of course)?

Quote:

domain wise with the laws of functions and their inverses, the range of the original function is typically the domain of the inverse however when I graph the inverse graph, there's no switch between the original function's domain to the inverse's range
The domain of the inverse $\displaystyle f^{-1}$ is $\displaystyle [-2,0]$ and its range is $\displaystyle \left[-\frac\pi2-2,\frac\pi2-2\right].$

So, the domain of $\displaystyle f$ is $\displaystyle \left[-\frac\pi2-2,\frac\pi2-2\right].$ On this interval, the range of $\displaystyle f$ is indeed $\displaystyle [-2,0],$ as you may verify.
• Apr 6th 2009, 12:04 AM
deltemis
thank you very much, I wasn't thinking about restricting the domain for the original function as well when I was reading their graphs

much appreciated