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Can we prove that 
\[ \arctan \frac{1}{2} + \arctan \frac{1}{3} = \frac{\pi}{4} \]
One way is to take the tangent and use the formula 
\[ \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
but this ends up looking messy.

Given the geometrical problem that suggests the problem in the first place, we might read the diagram differently and read off the problem as whether
\[ \arcsin \frac{1}{\sqrt 5} 
+ \arcsin \frac{1}{\sqrt {10}} = \frac{\pi}{4}\]
It is easy enough to use a calculator to check this, but can we give
an analytic proof? Similar to the case of tangent, we can use the formula
\[\sin(A+B) = \sin A \cos B + \cos A \sin B\]
We get off to a good start simplifying $\sin(\arcsin \frac{1}{\sqrt{5}})$ to $1 \over \sqrt{5}$ thus
\[\sin( \arcsin \frac{1}{\sqrt 5} 
+ \arcsin \frac{1}{\sqrt {10}}) = 
\frac{1}{\sqrt{5}} \cos(\arcsin \frac{1}{\sqrt{10}})+\frac{1}{\sqrt{10}}\cos(\arcsin\frac{1}{\sqrt{5}}) \]
But what next?

Staring hard at the diagram, we notice that the angle that we read off as in inverse sine could just as well be read off as an inverse cosine
\[ \arcsin\frac{1}{\sqrt{5}} = \arccos\frac{2}{\sqrt{5}} \]
and similarly
\[ \arcsin\frac{1}{\sqrt{10}} = \arcsin\frac{3}{\sqrt{10}} \]
Replace the inverse sines by inverse cosines, then collapse cosine of inverse cosine to the identity function, getting
\[\frac{1}{\sqrt{5}}\frac{3}{\sqrt{10}} + \frac{1}{\sqrt{10}}\frac{2}{\sqrt{5}} = \frac{5}{\sqrt{50}} = \frac{1}{\sqrt{2}}\]
Since the sine came out to $\frac{1}{\sqrt{2}}$ we infer that the angle must have been $\frac{\pi}{4}$

Thu, 30 Nov 2023 15:31 GMT