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# Sine Function

Hey guys, does anybody know if the sine function can give imaginary numbers as a result? If so, can anybody explain why? :)

Note by Happy Hunter
1 year, 1 month ago

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$$\text {Do you remember the famous Formula (Equation 1): } e^{i\theta}=\cos{\theta}+i\sin{\theta} \\ \text {Now let's Do a little manipulation, we have } \\ e^{i\theta}=\cos{\theta}+i\sin{\theta} \\ e^{-i\theta}=\cos{(-\theta)}+i\sin{(-\theta)} \\ e^{-i\theta}=\cos{\theta}-i\sin{\theta} \\ \text{Now add this equation and Equation 1 } \\ e^{i\theta}+e^{-i\theta}=2\cos{\theta} \\ cos{\theta}=\dfrac12(e^{i\theta}+e^{-i\theta}) \implies cos{\theta}=\dfrac{e^{2i\theta}+1}{2e^{i\theta}} \\ sin{\theta}=\sqrt{1-cos^2{\theta}} \\ sin{\theta}=\sqrt{1-\dfrac{e^{4i\theta}+1+2e^{2i\theta}}{4e^{2i\theta}}} \implies sin{\theta}=\sqrt{\dfrac{4e^{2i\theta}-e^{4i\theta}-1-2e^{2i\theta}}{4e^{2i\theta}}} \\ sin{\theta}=\sqrt{\dfrac{-(e^{2i\theta}-1)^2}{4e^{2i\theta}}}\implies sin{\theta}=\dfrac{i(e^{2i\theta}-1)}{2e^{i\theta}} \\$$
From this Result we can conclude that For only imaginary values of $$\theta$$ we get imaginary values of $$\sin{\theta}$$ · 1 year, 1 month ago

@Sambhrant Sachan

The statement $$\sin \theta = \sqrt{1-\cos^2 \theta}$$ is incorrect.

Plus, equation 1 is valid only for real numbers. For it to be valid for complex numbers, you first need to define $$\sin$$ and $$\cos$$ for complex values of $$\theta$$. · 1 year, 1 month ago

Thank you very much. :) And are there cases where theta has an imaginary value? · 1 year, 1 month ago

@Happy Hunter

Actually, we define $$\sin \theta$$ by the following equation for complex values of $$\theta$$:

$\sin \theta =\frac{e^{i\theta}-e^{-i\theta}}{2i}$

Similary, we define $$\cos \theta$$ as:

$\cos \theta =\frac{e^{i\theta}+e^{-i\theta}}{2}$

Note:

We define $$e^z$$ for all complex $$z$$ by the following equation:

$e^z =\sum_{r=0}^{\infty} \frac{z^r}{r!}$

Note that $$e^{z+w}=e^ze^w$$ for all complex $$z$$ and $$w$$. · 1 year, 1 month ago