Prove that $\tan^{-1}{\frac{1}{5}} \approx \frac{\pi}{16}$ using complex number method.
Hint: Take $z=5+i$
What is meant by complex number method in this problem? I couldn't think of do the above proof using complex numbers. Any suggestions ?
Prove that $\tan^{-1}{\frac{1}{5}} \approx \frac{\pi}{16}$ using complex number method
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trigonometry
complex-numbers
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1In fact we have $\pi/16 = \arctan(1/5) - {1 \over 4} \arctan({1 \over 239})$. – 2017-02-19
2 Answers
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$(5+i)^4 = 476 + 480i $ Since $476$ is approximately $480$, the angle of the line from $0$ to $476 + 480i$ made with the x-axis is about $45$ degrees or $\pi/4$ radians.
So $\arctan (1/5) $ is about $\pi/4 \over 4$ or $\pi/16$.
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We may notice that $$ (5+i)^8 = -3824 + 456960 i \tag{1} $$ is a complex number relatively close to the imaginary axis. By applying $\text{arg}$ to both sides of $(1)$, $$ 8\arctan\frac{1}{5}\approx \frac{\pi}{2} \tag{2} $$ follows.