User interface language: English | Español

Date November 2008 Marks available 2 Reference code 08N.1.sl.TZ0.7
Level SL only Paper 1 Time zone TZ0
Command term Show that Question number 7 Adapted from N/A

Question

Let  \(f(x) = {\sin ^3}x + {\cos ^3}x\tan x,\frac{\pi }{2} < x < \pi \) .

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

[2]
a.

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

[5]
b.

Markscheme

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]

a.

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]

b.

Examiners report

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.

a.

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.

b.

Syllabus sections

Topic 3 - Circular functions and trigonometry » 3.3 » The Pythagorean identity \({\cos ^2}\theta + {\sin ^2}\theta = 1\) .

View options