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I need help calculating the following:$$\arctan \frac{21 \pi}{\pi^2-54}+\arctan \frac{\pi}{18} + \arctan \frac{\pi}{3}$$ I don't know how to start, can anyone give me any information what should I do? Thank you

By the way, I have to calculate it without any use of a calculator, and without use of Maclaurin series.

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  • $\begingroup$ And what does calculator says? $\endgroup$ – Aqua Aug 20 at 11:40
  • $\begingroup$ Note that $\arctan x + \arctan y = \arctan(\frac{x+y}{1-xy})$. What happens if you do this with the latter two terms? $\endgroup$ – астон вілла олоф мэллбэрг Aug 20 at 11:42
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The arctangent addition formula gives $$\arctan \frac{\pi}{18} + \arctan \frac{\pi}{3}=\arctan\frac{7\pi/18}{1-\pi^2/54}=\arctan\frac{21\pi}{54-\pi^2}$$ Thus the expression is of the form $\arctan z+\arctan-z$, and since $\arctan$ is an odd function the expression equals $0$.

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Using $$\arctan x+\arctan y\stackrel{!}{=}\arctan\tan(\arctan x+\arctan y)\\=\arctan\frac{\tan\arctan x+\tan\arctan y}{1-\tan\arctan x\tan\arctan y}=\arctan\frac{x+y}{1-xy}$$(the conditions under which $\stackrel{!}{=}$ is valid are left as an exercise, but they apply to the examples we'll consider), we have $$\arctan\frac{\pi}{18}+\arctan\frac{\pi}{3}=\arctan\frac{21\pi}{54-\pi^2}.$$Since $\arctan x$ is odd, the final sum is $0$.

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