- #1

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First, multiple the whole fraction by cosx. It becomes (cosx - i sin x) / (cosx + i sin x). We can find the modulus and argument easily by using the fact that "if z1=r cis a, z2=r cis b , then z1 / z2 = r1/r2 [cos(a-b) + i sin (a-b)] ". They are 1 and -2x respectively.

However, there exists another way to handle it. By multipling the whole fraction by (cosx - i sin x) / (cosx - i sin x), we can obtain (cos

^{2}x-sin

^{2}x-2cosxsinx i). The modulus is, of course, found to be 1. However, problem arose when I wanted to find the argument. Letting B be the argument, I set up "tanB = (-2cosxsinx) / (cos

^{2}x-sin

^{2}x). The equation can be written as tanB=-tan2x. Finally, both -2x and 180

^{o}-2x are found as the solutions. However, there should be two arguments for the same "polar form", right? I wonder what is wrong with that.