User interface language: English | Español

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.

[3]
a.

Hence find the value of \(\arctan \left( {\frac{1}{2}} \right) + \arctan \left( {\frac{1}{5}} \right) + \arctan \left( {\frac{1}{8}} \right)\).

[3]
b.

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]

a.

\(\tan \left( {\arctan \left( {\frac{1}{2}} \right) + \arctan \left( {\frac{1}{5}} \right) + \arctan \left( {\frac{1}{8}} \right)} \right) = \frac{{\frac{1}{2} + \frac{1}{3}}}{{1 - \frac{1}{2} \times \frac{1}{3}}} = 1\)     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]

b.

Examiners report

Those candidates who used the addition formula for the tangent were usually successful.

a.

Some candidates left their answer as the tangent of an angle, rather than the angle itself.

b.

Syllabus sections

Topic 3 - Core: Circular functions and trigonometry » 3.3 » Compound angle identities.

View options