Write down an expression for

(i)     \(\sin 140^\circ \) ;

(ii)    \(\cos 70^\circ \) .

Find an expression for \(\cos 140^\circ \) .

Find an expression for \(\tan 140^\circ \) .

(i) \(\sin 140^\circ = p\)     A1     N1

(ii) \(\cos 70^\circ = - q\)     A1     N1

[2 marks]

METHOD 1

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

e.g. diagram, \(\sqrt {1 - {p^2}} \) (seen anywhere)

\(\cos 140^\circ = \pm \sqrt {1 - {p^2}} \)     (A1)

\(\cos 140^\circ = - \sqrt {1 - {p^2}} \)     A1     N2

METHOD 2

evidence of using \(\cos 2\theta = 2{\cos ^2}\theta - 1\)     (M1)

\(\cos 140^\circ = 2{\cos ^2}70 - 1\)     (A1)

\(\cos 140^\circ = 2{( - q)^2} - 1\) \(( = 2{q^2} - 1)\)     A1     N2

[3 marks]

METHOD 1

\(\tan 140^\circ = \frac{{\sin 140^\circ }}{{\cos 140^\circ }} = - \frac{p}{{\sqrt {1 - {p^2}} }}\)     A1     N1

METHOD 2

\(\tan 140^\circ = \frac{p}{{2{q^2} - 1}}\)     A1     N1

[1 mark]

This was one of the most difficult problems for the candidates. Even the strongest candidates had a hard time with this one and only a few received any marks at all.

Many did not appear to know the relationships between trigonometric functions of supplementary angles and that the use of \({\sin ^2}x + {\cos ^2}x = 1\) results in a \( \pm \) value. The application of a double angle formula also seemed weak.

This was one of the most difficult problems for the candidates. Even the strongest candidates had a hard time with this one and only a few received any marks at all. Many did not appear to know the relationships between trigonometric functions of supplementary angles and that the use of \({\sin ^2}x + {\cos ^2}x = 1\) results in a \( \pm \) value. The application of a double angle formula also seemed weak.