Date | May 2013 | Marks available | 3 | Reference code | 13M.1.hl.TZ2.10 |
Level | HL only | Paper | 1 | Time zone | TZ2 |
Command term | Find and Hence | Question number | 10 | Adapted from | N/A |
Question
Given that \(\arctan \left( {\frac{1}{5}} \right) + \arctan \left( {\frac{1}{8}} \right) = \arctan \left( {\frac{1}{p}} \right)\), where \(p \in {\mathbb{Z}^ + }\), find p.
Hence find the value of \(\arctan \left( {\frac{1}{2}} \right) + \arctan \left( {\frac{1}{5}} \right) + \arctan \left( {\frac{1}{8}} \right)\).
Markscheme
attempt at use of \(\tan (A + B) = \frac{{\tan (A) + \tan (B)}}{{1 - \tan (A)\tan (B)}}\) M1
\(\frac{1}{p} = \frac{{\frac{1}{5} + \frac{1}{8}}}{{1 - \frac{1}{5} \times \frac{1}{8}}}{\text{ }}\left( { = \frac{1}{3}} \right)\) A1
\(p = 3\) A1
Note: the value of p needs to be stated for the final mark.
[3 marks]
\(\tan \left( {\arctan \left( {\frac{1}{2}} \right) + \arctan \left( {\frac{1}{5}} \right) + \arctan \left( {\frac{1}{8}} \right)} \right) = \frac{{\frac{1}{2} + \frac{1}{3}}}{{1 - \frac{1}{2} \times \frac{1}{3}}} = 1\) M1A1
\(\arctan \left( {\frac{1}{2}} \right) + \arctan \left( {\frac{1}{5}} \right) + \arctan \left( {\frac{1}{8}} \right) = \frac{\pi }{4}\) A1
[3 marks]
Examiners report
Those candidates who used the addition formula for the tangent were usually successful.
Some candidates left their answer as the tangent of an angle, rather than the angle itself.