Olympiad Maths Prep

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Problem 1096

National olympiad, first round
Algebra Difficulty 6.1 Prove it

Example 10 Consider the following expression: 1k⋅1−(1k+1)2−1−(1k)2⋅1k+1=(k+1)2−1k(k+1)−\frac{1}{k} \cdot \sqrt{1-\left(\frac{1}{k+1}\right)^{2}}-\sqrt{1-\left(\frac{1}{k}\right)^{2}} \cdot \frac{1}{k+1}=\frac{\sqrt{(k+1)^{2}-1}}{k(k+1)}-
k2−1k(k+1)=(k+1)2−1−k2−1k(k+1). \frac{\sqrt{k^{2}-1}}{k(k+1)}=\frac{\sqrt{(k+1)^{2}-1}-\sqrt{k^{2}-1}}{k(k+1)} .

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

It can be written in the form of an inverse sine function as follows:
arcsin⁡(k+1)2−1−k2−1k(k+1)=arcsin⁡1k−arcsin⁡1k+1,  \arcsin \frac{\sqrt{(k+1)^{2}-1}-\sqrt{k^{2}-1}}{k(k+1)}=\arcsin \frac{1}{k}-\arcsin \frac{1}{k+1} \text {, }

We can obtain the following identity:
arcsin⁡32+arcsin⁡8−36+arcsin⁡5−812+⋯+arcsin⁡(n+1)2−1−n2−1n(n+1)=arccos⁡1n+1. (Hint: The above expression =arcsin⁡1−arcsin⁡1n+1=π2−arcsin⁡1n+1=arccos⁡1n+1 )  \begin{array}{l} \arcsin \frac{\sqrt{3}}{2}+\arcsin \frac{\sqrt{8}-\sqrt{3}}{6}+\arcsin \frac{\sqrt{5}-\sqrt{8}}{12}+\cdots+\arcsin \frac{\sqrt{(n+1)^{2}-1}-\sqrt{n^{2}-1}}{n(n+1)} \\ =\arccos \frac{1}{n+1} . \\ \text { (Hint: The above expression }=\arcsin 1-\arcsin \frac{1}{n+1}=\frac{\pi}{2}-\arcsin \frac{1}{n+1}=\arccos \frac{1}{n+1} \text { ) } \end{array}

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.