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Math Help - [SOLVED] Inverse Trig Functions

  1. #1
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    [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
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  2. #2
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    Quote Originally Posted by OVechkin8 View Post
    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
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  3. #3
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    Quote Originally Posted by OVechkin8 View Post
    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 \cos^{-1} is only the exact inverse of \cos x for x\in [0;\pi]. Since \frac{5\pi}{3}\notin [0;\pi] we have to do some juggling to get the correct answer:

    \cos^{-1}\left(\cos\frac{5\pi}{3}\right)=\cos^{-1}\left(\cos \left(-\frac{5\pi}{3}\right)\right)
    =\cos^{-1}\left(\cos\left(2\pi-\frac{5\pi}{3}\right)\right)=\cos^{-1}\left(\cos \frac{\pi}{3}\right)=\frac{\pi}{3}
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  4. #4
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    Quote Originally Posted by Debsta View Post
    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 \sin(x)=\sin(\pi/4) has infinitely many solutions, but ask any calculator out there what the value of \sin^{-1}\left(\sin(\pi/4)\right) is, and you will get the same, unique answer.
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  5. #5
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    Quote Originally Posted by Failure View Post
    I would agree that an equation like \sin(x)=\sin(\pi/4) has infinitely many solutions, but ask any calculator out there what the value of \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.
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    Quote Originally Posted by Debsta View Post
    Yes but what do calculators know!!! They don't tell you the whole story.
    Ok, ok, but I take it that \sin^{-1}(x) is just another way of writing \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 \sin^{-1}(c), which is either undefined or has a unique value; and asking for the solution(s) x of the equation \sin(x)=c, which may have none or infinitely many.
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  7. #7
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    Quote Originally Posted by Failure View Post
    Ok, ok, but I take it that \sin^{-1}(x) is just another way of writing \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 \sin^{-1}(c), which is either undefined or has a unique value; and asking for the solution(s) x of the equation \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.
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  8. #8
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    Quote Originally Posted by Debsta View Post
    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 \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 \ln(e^5), I am sure to answer that this equals 5 without adding to that basic answer infinitely many different integral multiples of 2\pi\cdot i.
    Not so, if someone asks me to determine the solution to an equation like e^z=e^5.

    In the case of \sin^{-1} this has the advantage of being a bit more specific as to what the solutions of \sin(x)=c might be. They are x=\sin^{-1}(c)+n\cdot 2\pi, and x=\pi-\sin^{-1}(c)+n\cdot 2\pi, n\in\mathbb{Z}.
    If, on the other hand, one takes \sin^{-1}(c) to be the entire set of solutions, then all one can write is 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).
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  9. #9
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    Quote Originally Posted by Failure View Post
    .. 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 \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 \ln(e^5), I am sure to answer that this equals 5 without adding to that basic answer infinitely many different integral multiples of 2\pi\cdot i.
    Not so, if someone asks me to determine the solution to an equation like e^z=e^5.

    In the case of \sin^{-1} this has the advantage of being a bit more specific as to what the solutions of \sin(x)=c might be. They are x=\sin^{-1}(c)+n\cdot 2\pi, and x=\pi-\sin^{-1}(c)+n\cdot 2\pi, n\in\mathbb{Z}.
    If, on the other hand, one takes \sin^{-1}(c) to be the entire set of solutions, then all one can write is 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.
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  10. #10
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    would the answer for c be pi/4 then?
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