Show that $f(x) = \sin x$ .

Let $\sin x = \frac{2}{3}$ . Show that $f(2x) = - \frac{{4\sqrt 5 }}{9}$ .

changing $\tan x$ into $\frac{{\sin x}}{{\cos x}}$     A1

e.g. ${\sin ^3}x + {\cos ^3}x\frac{{\sin x}}{{\cos x}}$

simplifying     A1

e.g $\sin x({\sin ^2}x + {\cos ^2}x)$ , ${\sin ^3}x + \sin x - {\sin ^3}x$

$f(x) = \sin x$    AG     N0

[2 marks]

recognizing $f(2x) = \sin 2x$ , seen anywhere     (A1)

evidence of using double angle identity $\sin (2x) = 2\sin x\cos x$ , seen anywhere     (M1)

evidence of using Pythagoras with $\sin x = \frac{2}{3}$     M1

e.g. sketch of right triangle, ${\sin ^2}x + {\cos ^2}x = 1$

$\cos x = - \frac{{\sqrt 5 }}{3}$ (accept $\frac{{\sqrt 5 }}{3}$ )     (A1)

$f(2x) = 2\left( {\frac{2}{3}} \right)\left( { - \frac{{\sqrt 5 }}{3}} \right)$     A1

$f(2x) = - \frac{{4\sqrt 5 }}{9}$     AG     N0

[5 marks]

Not surprisingly, this question provided the greatest challenge in section A. In part (a), candidates were able to use the identity $\tan x = \frac{{\sin x}}{{\cos x}}$ , but many could not proceed any further.

Part (b) was generally well done by those candidates who attempted it, the major error arising when the negative sign "magically" appeared in the answer. Many candidates could find the value of cosx but failed to observe that cosine is negative in the given domain.