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.