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Graphs of the inverse trigonometric functions Watch

    • Thread Starter

    Okay, in the book it says that to sketch the graph of arcsinx, for example, you need to restrict the domain of sinx so that it is one-to-one, otherwise it would not be a function. However, I don't seem to understand this, as I've understood sinx and cosx to both be many-to-one functions, which are functions. Can someone explain?

    I would suggest looking more into bijections

    Also think about why, if you were making an inverse function why you would want to restrict the domain of the original function if it is a many-to-one function
    • PS Helper

    Any function has to give just one number out when you put one number in. As such, any vertical line must intersect the graph of a function at at most one point; otherwise it's not a function. With \sin x, if you plug one number in, you get one number out, and this is reflected in the fact that when you draw a vertical line anywhere, it only intersects the graph at one point.

    However, \sin x is periodic (and hence not injective), so two numbers going in might give the same number coming out (e.g. \sin 0 = \sin \pi = \sin 2\pi = \cdots = 0). This means that if you were to look at the set of numbers that give, say, 0, then you'd get more than one value out. As such, if we want to have a well-defined function, we need to restrict it so that only one number comes out. It's for this reason that we restrict it so that it's bijective.

    We do a similar thing with the square root function, which is the inverse of the function x^2. By convention, we restrict \sqrt{x} so that 0 \le \sqrt{x} < \infty. We could just as well take the negative root, or take the positive root of all integers and the negative root of all other numbers, or whatever you like... but by convention we take the positive root. It's the same type of convention that means we take \dfrac{-\pi}{2} < \sin^{-1} x \le \dfrac{\pi}{2}.
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