User interface language: English | Español

Date November 2011 Marks available 3 Reference code 11N.1.sl.TZ0.6
Level SL only Paper 1 Time zone TZ0
Command term Find Question number 6 Adapted from N/A

Question

Let \(\sin \theta  = \frac{2}{{\sqrt {13} }}\) , where \(\frac{\pi }{2} < \theta  < \pi \) .

Find \(\cos \theta \) .

[3]
a.

Find \(\tan 2\theta \) .

[5]
b.

Markscheme

METHOD 1

evidence of choosing \({\sin ^2}\theta  + {\cos ^2}\theta  = 1\)     (M1)

correct working     (A1)

e.g. \({\cos ^2}\theta  = \frac{9}{{13}}\) , \(\cos \theta  =  \pm \frac{3}{{\sqrt {13} }}\) , \(\cos \theta  = \sqrt {\frac{9}{{13}}} \)

\(\cos \theta  = - \frac{3}{{\sqrt {13} }}\)    A1     N2

Note: If no working shown, award N1 for \(\frac{3}{{\sqrt {13} }}\) .

METHOD 2

approach involving Pythagoras’ theorem     (M1)

e.g. \({2^2} + {x^2} = 13\) ,


finding third side equals 3     (A1)

\(\cos \theta  = - \frac{3}{{\sqrt {13} }}\)     A1     N2

Note: If no working shown, award N1 for \(\frac{3}{{\sqrt {13} }}\) .

[3 marks]

a.

correct substitution into \(\sin 2\theta \) (seen anywhere)     (A1)

e.g. \(2\left( {\frac{2}{{\sqrt {13} }}} \right)\left( { - \frac{3}{{\sqrt {13} }}} \right)\)

correct substitution into \(\cos 2\theta \) (seen anywhere)     (A1)

e.g. \({\left( { - \frac{3}{{\sqrt {13} }}} \right)^2} - {\left( {\frac{2}{{\sqrt {13} }}} \right)^2}\) , \(2{\left( { - \frac{3}{{\sqrt {13} }}} \right)^2} - 1\) , \(1 - 2{\left( {\frac{2}{{\sqrt {13} }}} \right)^2}\)

valid attempt to find \(\tan 2\theta \)     (M1)

e.g. \(\frac{{2\left( {\frac{2}{{\sqrt {13} }}} \right)\left( { - \frac{3}{{\sqrt {13} }}} \right)}}{{{{\left( { - \frac{3}{{\sqrt {13} }}} \right)}^2} - {{\left( {\frac{2}{{\sqrt {13} }}} \right)}^2}}}\) , \(\frac{{2\left( { - \frac{2}{3}} \right)}}{{1 - {{\left( { - \frac{2}{3}} \right)}^2}}}\)

correct working     A1

e.g. \(\frac{{\frac{{(2)(2)( - 3)}}{{13}}}}{{\frac{9}{{13}} - \frac{4}{{13}}}}\) , \(\frac{{ - \frac{{12}}{{{{\left( {\sqrt {13} } \right)}^2}}}}}{{\frac{{18}}{{13}} - 1}}\) , \(\frac{{ - \frac{{12}}{{13}}}}{{\frac{5}{{13}}}}\)

\(\tan 2\theta  = - \frac{{12}}{5}\)     A1     N4

Note: If students find answers for \(\cos \theta \) which are not in the range \([ - 1{\text{, }}1]\), award full FT in (b) for correct FT working shown.

[5 marks]

b.

Examiners report

While the majority of candidates knew to use the Pythagorean identity in part (a), very few remembered that the cosine of an angle in the second quadrant will have a negative value.

a.

In part (b), many candidates incorrectly tried to calculate \(\tan 2\theta \) as \(2 \times \tan \theta \) , rather than using the double-angle identities.

b.

Syllabus sections

Topic 3 - Circular functions and trigonometry » 3.3 » Double angle identities for sine and cosine.
Show 23 related questions

View options