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Prove: \arctan{\frac{1}{2}}+\arctan{\frac{1}{3}} = \frac{\pi}{4}, \arcsin{\frac{16}{65}}+\arcsin{\frac{5}{13}} = \arcsin{\frac{3}{5}}

We prove identities involving inverse trigonometric functions. Sum formulas from trigonometry are exploited.

  1. \arctan{\frac{1}{2}}+\arctan{\frac{1}{3}} = \frac{\pi}{4} 1

    Suppose that \arctan{\frac{1}{2}}+\arctan{\frac{1}{3}} \neq \frac{\pi}{4}. Then, \begin{align} \tan{\left(\arctan{\frac{1}{2}}+\arctan{\frac{1}{3}}\right)} &\neq \tan{\frac{\pi}{4}}\\ \frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\cdot\frac{1}{3}} &\neq 1 \\ \frac{5}{6-5} &\neq 1 \\ 1 &\neq 1, \end{align} which is a contradiction. Thus, \arctan{\frac{1}{2}}+\arctan{\frac{1}{3}} = \frac{\pi}{4}.□

  2. \arcsin{\frac{16}{65}}+\arcsin{\frac{5}{13}} = \arcsin{\frac{3}{5}} 1

    Suppose that \arcsin{\frac{16}{65}}+\arcsin{\frac{5}{13}} \neq \arcsin{\frac{3}{5}}. Then, \begin{align} \sin{\left(\arcsin{\frac{16}{65}}+\arcsin{\frac{5}{13}}\right)} &\neq \frac{3}{5}\\ \frac{16}{65}\cos{\left(\arcsin{\frac{5}{13}}\right)}+\cos{\left(\arcsin{\frac{16}{65}}\right)}\cdot\frac{5}{13} &\neq \frac{3}{5}\\ \frac{16}{65}\cos{\left(\arccos{\frac{\sqrt{13^2-5^2}}{13}}\right)}+\cos{\left(\arccos{\frac{\sqrt{65^2-16^2}}{65}}\right)}\cdot\frac{5}{13} &\neq \frac{3}{5}\\ \frac{16}{65}\cdot\frac{12}{13}+\frac{63}{65}\cdot\frac{5}{13} &\neq \frac{3}{5}\\ \frac{507}{65\cdot13} &\neq \frac{3}{5}\\ \frac{3}{5} &\neq \frac{3}{5}, \end{align} which is a contradiction. Thus, \arcsin{\frac{16}{65}}+\arcsin{\frac{5}{13}} = \arcsin{\frac{3}{5}}.□

References

1 University Tutorial Problems.

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