Sine Function

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

#Geometry

Note by Happy Hunter
5 years, 1 month ago

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Comments

Do you remember the famous Formula (Equation 1): eiθ=cos⁡θ+isin⁡θNow let’s Do a little manipulation, we have eiθ=cos⁡θ+isin⁡θe−iθ=cos⁡(−θ)+isin⁡(−θ)e−iθ=cos⁡θ−isin⁡θNow add this equation and Equation 1 eiθ+e−iθ=2cos⁡θcosθ=12(eiθ+e−iθ)  ⟹  cosθ=e2iθ+12eiθsinθ=1−cos2θsinθ=1−e4iθ+1+2e2iθ4e2iθ  ⟹  sinθ=4e2iθ−e4iθ−1−2e2iθ4e2iθsinθ=−(e2iθ−1)24e2iθ  ⟹  sinθ=i(e2iθ−1)2eiθ\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 θ\theta we get imaginary values of sin⁡θ\sin{\theta}

Sabhrant Sachan - 5 years, 1 month ago

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@Sambhrant Sachan

The statement sin⁡θ=1−cos⁡2θ\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⁡\sin and cos⁡\cos for complex values of θ\theta.

A Former Brilliant Member - 5 years, 1 month ago

Thank you very much. :) And are there cases where theta has an imaginary value?

Happy Hunter - 5 years, 1 month ago

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@Happy Hunter

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

sin⁡θ=eiθ−e−iθ2i\sin \theta =\frac{e^{i\theta}-e^{-i\theta}}{2i}

Similary, we define cos⁡θ\cos \theta as:

cos⁡θ=eiθ+e−iθ2\cos \theta =\frac{e^{i\theta}+e^{-i\theta}}{2}

Note:

We define eze^z for all complex zz by the following equation:

ez=∑r=0∞zrr!e^z =\sum_{r=0}^{\infty} \frac{z^r}{r!}

Note that ez+w=ezewe^{z+w}=e^ze^w for all complex zz and ww.

A Former Brilliant Member - 5 years, 1 month ago
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