# Thread: [SOLVED] Inverse Trig Functions

1. ## [SOLVED] Inverse Trig Functions

Can someone check these for me?

a. sin(sin^-1(1/2)) = 1/2
b. cos(cos^-1(-1)) = -1
c. sin^-1(sin(pi/4)) = 3pi/4
d. cos^-1(cos(5pi/3) = 5pi/3

thanks

2. Originally Posted by OVechkin8
Can someone check these for me?

a. sin(sin^-1(1/2)) = 1/2
b. cos(cos^-1(-1)) = -1
c. sin^-1(sin(pi/4)) = 3pi/4
d. cos^-1(cos(5pi/3) = 5pi/3

thanks
The first two are correct. For the last two, what does the question say - there are actually an infinite number of related answers for these.
eg in Q(c) pi/4, 3pi/4, 9pi/4, 11pi/4, etc are all possible answers

3. Originally Posted by OVechkin8
Can someone check these for me?

a. sin(sin^-1(1/2)) = 1/2
b. cos(cos^-1(-1)) = -1
c. sin^-1(sin(pi/4)) = 3pi/4
d. cos^-1(cos(5pi/3) = 5pi/3

thanks
Your solutions of a. and b. are correct. c. and d. however are wrong.

c. is probably obvious. As to d.: you must remember that $\displaystyle \cos^{-1}$ is only the exact inverse of $\displaystyle \cos x$ for $\displaystyle x\in [0;\pi]$. Since $\displaystyle \frac{5\pi}{3}\notin [0;\pi]$ we have to do some juggling to get the correct answer:

$\displaystyle \cos^{-1}\left(\cos\frac{5\pi}{3}\right)=\cos^{-1}\left(\cos \left(-\frac{5\pi}{3}\right)\right)$
$\displaystyle =\cos^{-1}\left(\cos\left(2\pi-\frac{5\pi}{3}\right)\right)=\cos^{-1}\left(\cos \frac{\pi}{3}\right)=\frac{\pi}{3}$

4. Originally Posted by Debsta
The first two are correct. For the last two, what does the question say - there are actually an infinite number of related answers for these.
eg in Q(c) pi/4, 3pi/4, 9pi/4, 11pi/4, etc are all possible answers
I would agree that an equation like $\displaystyle \sin(x)=\sin(\pi/4)$ has infinitely many solutions, but ask any calculator out there what the value of $\displaystyle \sin^{-1}\left(\sin(\pi/4)\right)$ is, and you will get the same, unique answer.

5. Originally Posted by Failure
I would agree that an equation like $\displaystyle \sin(x)=\sin(\pi/4)$ has infinitely many solutions, but ask any calculator out there what the value of $\displaystyle \sin^{-1}\left(\sin(\pi/4)\right)$ is, and you will get the same, unique answer.
Yes but what do calculators know!!! They don't tell you the whole story.

6. Originally Posted by Debsta
Yes but what do calculators know!!! They don't tell you the whole story.
Ok, ok, but I take it that $\displaystyle \sin^{-1}(x)$ is just another way of writing $\displaystyle \arcsin(x)$, and if so then Inverse trigonometric functions - Wikipedia, the free encyclopedia will give you the rest of the story. A story that fits nicely with what I argued. So in this particular case, your average pocket calculator is absolutely right.

Moral: We must clearly distinguish between asking for the value of the term $\displaystyle \sin^{-1}(c)$, which is either undefined or has a unique value; and asking for the solution(s) $\displaystyle x$ of the equation $\displaystyle \sin(x)=c$, which may have none or infinitely many.

7. Originally Posted by Failure
Ok, ok, but I take it that $\displaystyle \sin^{-1}(x)$ is just another way of writing $\displaystyle \arcsin(x)$, and if so then Inverse trigonometric functions - Wikipedia, the free encyclopedia will give you the rest of the story. A story that fits nicely with what I argued. So in this particular case, your average pocket calculator is absolutely right.

Moral: We must clearly distinguish between asking for the value of the term $\displaystyle \sin^{-1}(c)$, which is either undefined or has a unique value; and asking for the solution(s) $\displaystyle x$ of the equation $\displaystyle \sin(x)=c$, which may have none or infinitely many.
Wikipedia???- bit of a dodgy source!
I've always understood that y = arcsin x or y = sin ^(-1) x is NOT a function. Think of the graph y=sin x reflected in y=x.
I have seen it written Sin(-1) x with an upper case S to indicate a function and therefore the principle argument.
All comes down to definition I suppose.

8. Originally Posted by Debsta
Wikipedia???- bit of a dodgy source!
I've always understood that y = arcsin x or y = sin ^(-1) x is NOT a function. Think of the graph y=sin x reflected in y=x.
I have seen it written Sin(-1) x with an upper case S to indicate a function and therefore the principle argument.
All comes down to definition I suppose.
.. and convention. If someone wants to go against the convention, he can do that - if he is prepared to bear the consequences: of being misunderstood all the time.

If a function, like $\displaystyle \arcsin$ can be said to have "multiple branches", by convention one sticks to the "principal branch".
Same game with the natural logarithm: There are infinitely many branches of the "inverse of the natural exponential function" as well, but if someone writes $\displaystyle \ln(e^5)$, I am sure to answer that this equals $\displaystyle 5$ without adding to that basic answer infinitely many different integral multiples of $\displaystyle 2\pi\cdot i$.
Not so, if someone asks me to determine the solution to an equation like $\displaystyle e^z=e^5$.

In the case of $\displaystyle \sin^{-1}$ this has the advantage of being a bit more specific as to what the solutions of $\displaystyle \sin(x)=c$ might be. They are $\displaystyle x=\sin^{-1}(c)+n\cdot 2\pi$, and $\displaystyle x=\pi-\sin^{-1}(c)+n\cdot 2\pi, n\in\mathbb{Z}$.
If, on the other hand, one takes $\displaystyle \sin^{-1}(c)$ to be the entire set of solutions, then all one can write is $\displaystyle x=\sin^{-1}(c)$. I really do consider this a disadvantage of that other convention (that I am not using, and that I hope few others are using).

9. Originally Posted by Failure
.. and convention. If someone wants to go against the convention, he can do that - if he is prepared to bear the consequences: of being misunderstood all the time.

If a function, like $\displaystyle \arcsin$ can be said to have "multiple branches", by convention one sticks to the "principal branch".
Same game with the natural logarithm: There are infinitely many branches of the "inverse of the natural exponential function" as well, but if someone writes $\displaystyle \ln(e^5)$, I am sure to answer that this equals $\displaystyle 5$ without adding to that basic answer infinitely many different integral multiples of $\displaystyle 2\pi\cdot i$.
Not so, if someone asks me to determine the solution to an equation like $\displaystyle e^z=e^5$.

In the case of $\displaystyle \sin^{-1}$ this has the advantage of being a bit more specific as to what the solutions of $\displaystyle \sin(x)=c$ might be. They are $\displaystyle x=\sin^{-1}(c)+n\cdot 2\pi$, and $\displaystyle x=\pi-\sin^{-1}(c)+n\cdot 2\pi, n\in\mathbb{Z}$.
If, on the other hand, one takes $\displaystyle \sin^{-1}(c)$ to be the entire set of solutions, then all one can write is $\displaystyle x=\sin^{-1}(c)$. I really do consider this a disadvantage of that other convention (that I am not using, and that I hope few others are using).
Good points. It is so impotant then that teachers/lecturers/textbook writers etc all use the correct conventions even at a low level of instruction. As a teacher of senior students, one of my main issues is with having students "unlearn" misconceptions which have resulted from trying to keep instructions at a low level - eg when solving equations - "swap sides and swap signs" causes all sorts of issues later.
That's why I asked the origonal poster what the actual question said - seen many textbooks get it wrong.

10. would the answer for c be pi/4 then?