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November 12th, 2015, 09:58 AM   #1
har
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Is the sine or cosine inverse of improper fractions imaginary?

Greetings. I am Hariharan, a class 9 student. Of late, I've been working with Rene Descartes' method of geometric square root extraction (See SquareSquareRootConstruction). I tried to deduce an algebraic equation to find the square root of any number from this.

Working further, I got some results. In the 2nd figure given in the above mentioned website, if we join the P and the midpoint of AB, say M, we get a right triangle MOP. Taking angle OMP into consideration, the angle could be given as arccos(x-1/x+1). Finding the sin of this, we get an equation.

The equation says: sqrt(x)= sin{arccos(x-1/x+1)} * (x+1)/2
Say, we substitute 9 in place of x. We get 3 (the square root of 9).
But, if we substitute (-4) in place of x, we get
sqrt(-4) = sin{arccos(-5/-3)}* (-1.5)
=> sin{arccos(5/3)} = 2i/-1.5 (where i stands for imaginary unit)
=> sin{arccos(5/3)} = (-4/3)i

From this, we reach a contradiction. 'cos' refers to the ratio obtained by dividing the length of the adjacent side by that of the hypotenuse, in a right-angled triangle. 5/3 means that the adjacent side is longer the hypotenuse, which is practically not possible in the real number system. Perhaps, this could be possible in some other number system, where the right triangles are constructed with 'special' lengths.

Since sqrt(-4) is an imaginary number, sin{arccos(5/3)} is an imaginary number. Hence, the 'special' lengths I spoke of is an imaginary number. We can also say that cos inverse of (5/3) is imaginary, since sin function does not bring such an impact to the result. Also, since sin{arccos(x)}=cos{arcsin(x)}, arcsin(5/3) is also imaginary.

We observe that 5/3 is an improper fraction. This mechanism of sin inverse and cos inverse is seen for all improper fractions.

So I concluded, that sin inverse and cos inverse of proper fractions are real numbers and for improper fractions, they are imaginary numbers.

I am not very sure whether this idea of mine is right. I am just in class 9 and haven't started learning trigonometry in school, as yet. But, I ardently want to explore it. I want to know whether my approach is correct, and that I can proceed with it. So, all I ask for is please tell me where I'm wrong anywhere and how to rectify them....
Thanks!!!
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November 12th, 2015, 11:16 AM   #2
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$\displaystyle e^{ix} = \cos x + i\sin x$ (Euler's formula), so $\displaystyle \cos x = \left(e^{ix} + e^{-ix} \right)\!\large{/}2$.
Hence for $x = i\ln 3$, $\cos x$ = (1/3 + 3)/2 = 5/3.
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