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Date May 2008 Marks available 7 Reference code 08M.2.hl.TZ1.10
Level HL only Paper 2 Time zone TZ1
Command term Find Question number 10 Adapted from N/A

Question

Find, in its simplest form, the argument of \({\left( {\sin \theta  + {\text{i}}(1 - \cos \theta )} \right)^2}\) where \(\theta \) is an acute angle.

Markscheme

\({\left( {\sin \theta + {\text{i}}(1 - \cos \theta )} \right)^2} = {\sin ^2}\theta - {(1 - \cos \theta )^2} + {\text{i}}2\sin \theta (1 - \cos \theta )\)     M1A1

Let \(\alpha \) be the required argument.

\(\tan \alpha = \frac{{2\sin \theta (1 - \cos \theta )}}{{{{\sin }^2}\theta - {{(1 - \cos \theta )}^2}}}\)     M1

\( = \frac{{2\sin \theta (1 - \cos \theta )}}{{(1 - {{\cos }^2}\theta ) - (1 - 2\cos \theta + {{\cos }^2}\theta )}}\)     (M1)

\( = \frac{{2\sin \theta (1 - \cos \theta )}}{{2\cos \theta (1 - \cos \theta )}}\)     A1

\( = \tan \theta \)     A1

\(\alpha = \theta \)     A1

[7 marks]

Examiners report

Very few candidates scored more than the first two marks in this question. Some candidates had difficulty manipulating trigonometric identities. Most candidates did not get as far as defining the argument of the complex expression.

Syllabus sections

Topic 1 - Core: Algebra » 1.5 » Complex numbers: the number \({\text{i}} = \sqrt { - 1} \) ; the terms real part, imaginary part, conjugate, modulus and argument.
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