Waste less time on Facebook — follow Brilliant.

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
4 months, 3 weeks ago

No vote yet
1 vote


Sort by:

Top Newest

\(\text {Do you remember the famous Formula (Equation 1): } e^{i\theta}=\cos{\theta}+i\sin{\theta} \\ \text {Now let's Do a little manipulation, we have } \\ e^{i\theta}=\cos{\theta}+i\sin{\theta} \\ e^{-i\theta}=\cos{(-\theta)}+i\sin{(-\theta)} \\ e^{-i\theta}=\cos{\theta}-i\sin{\theta} \\ \text{Now add this equation and Equation 1 } \\ e^{i\theta}+e^{-i\theta}=2\cos{\theta} \\ cos{\theta}=\dfrac12(e^{i\theta}+e^{-i\theta}) \implies cos{\theta}=\dfrac{e^{2i\theta}+1}{2e^{i\theta}} \\ sin{\theta}=\sqrt{1-cos^2{\theta}} \\ sin{\theta}=\sqrt{1-\dfrac{e^{4i\theta}+1+2e^{2i\theta}}{4e^{2i\theta}}} \implies sin{\theta}=\sqrt{\dfrac{4e^{2i\theta}-e^{4i\theta}-1-2e^{2i\theta}}{4e^{2i\theta}}} \\ sin{\theta}=\sqrt{\dfrac{-(e^{2i\theta}-1)^2}{4e^{2i\theta}}}\implies sin{\theta}=\dfrac{i(e^{2i\theta}-1)}{2e^{i\theta}} \\ \)
From this Result we can conclude that For only imaginary values of \(\theta\) we get imaginary values of \(\sin{\theta}\) Sambhrant Sachan · 4 months, 3 weeks ago

Log in to reply

@Sambhrant Sachan @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 \(\theta\). Deeparaj Bhat · 4 months, 3 weeks ago

Log in to reply

@Sambhrant Sachan Thank you very much. :) And are there cases where theta has an imaginary value? Happy Hunter · 4 months, 3 weeks ago

Log in to reply

@Happy Hunter @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:

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


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 \(w\). Deeparaj Bhat · 4 months, 3 weeks ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...