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.
- $\arctan{\frac{1}{2}}+\arctan{\frac{1}{3}} = \frac{\pi}{4}$
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, $$ \arctan{\frac{1}{2}}+\arctan{\frac{1}{3}} = \frac{\pi}{4}.□ $$
- $\arcsin{\frac{16}{65}}+\arcsin{\frac{5}{13}} = \arcsin{\frac{3}{5}}$
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, $$ \arcsin{\frac{16}{65}}+\arcsin{\frac{5}{13}} = \arcsin{\frac{3}{5}}.□ $$
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