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Date	May 2021	Marks available	2	Reference code	21M.1.SL.TZ1.6
Level	Standard Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 1
Command term	Show that	Question number	6	Adapted from	N/A

Question

Show that $sin 2 x + cos 2 x - 1 = 2 sin x (cos x - sin x)$ $\sin 2 x + \cos 2 x - 1 = 2 \sin x (\cos x - \sin x)$ .

[2]

Hence or otherwise, solve $sin 2 x + cos 2 x - 1 + cos x - sin x = 0$ $\sin 2 x + \cos 2 x - 1 + \cos x - \sin x = 0$ for $0 < x < 2 π$ $0 < x < 2 π$ .

[6]

Markscheme

Note: Do not award the final A1 for proofs which work from both sides to find a common expression other than $2 sin x cos x - 2 {sin}^{2} x$ $2 \sin x \cos x - 2 \sin^{2} x$ .

METHOD 1 (LHS to RHS)

attempt to use double angle formula for $sin 2 x$ $\sin 2 x$ or $cos 2 x$ $\cos 2 x$ M1

LHS $= 2 sin x cos x + cos 2 x - 1$ $= 2 \sin x \cos x + \cos 2 x - 1$ OR

$sin 2 x + 1 - 2 {sin}^{2} x - 1$ $\sin 2 x + 1 - 2 \sin^{2} x - 1$ OR

$2 sin x cos x + 1 - 2 {sin}^{2} x - 1$ $2 \sin x \cos x + 1 - 2 \sin^{2} x - 1$

$= 2 sin x cos x - 2 {sin}^{2} x$ $= 2 \sin x \cos x - 2 \sin^{2} x$ A1

$sin 2 x + cos 2 x - 1 = 2 sin x (cos x - sin x) =$ $\sin 2 x + \cos 2 x - 1 = 2 \sin x (\cos x - \sin x) =$ RHS AG

METHOD 2 (RHS to LHS)

RHS $= 2 sin x cos x - 2 {sin}^{2} x$ $= 2 \sin x \cos x - 2 \sin^{2} x$

attempt to use double angle formula for $sin 2 x$ $\sin 2 x$ or $cos 2 x$ $\cos 2 x$ M1

$= sin 2 x + 1 - 2 {sin}^{2} x - 1$ $= \sin 2 x + 1 - 2 \sin^{2} x - 1$ A1

$= sin 2 x + cos 2 x - 1 =$ $= \sin 2 x + \cos 2 x - 1 =$ LHS AG

[2 marks]

attempt to factorise M1

$(cos x - sin x) (2 sin x + 1) = 0$ $(\cos x - \sin x) (2 \sin x + 1) = 0$ A1

recognition of $cos x = sin x \Rightarrow \frac{sin x}{cos x} = tan x = 1$ $\cos x = \sin x \Rightarrow \frac{\sin x}{\cos x} = \tan x = 1$ OR $sin x = - \frac{1}{2}$ $\sin x = - \frac{1}{2}$ (M1)

one correct reference angle seen anywhere, accept degrees (A1)

$\frac{π}{4}$ $\frac{π}{4}$ OR $\frac{π}{6}$ $\frac{π}{6}$ (accept $- \frac{π}{6}, \frac{7 π}{6}$ $- \frac{π}{6}, \frac{7 π}{6}$ )

Note: This (M1)(A1) is independent of the previous M1A1.

$x = \frac{7 π}{6}, \frac{11 π}{6}, \frac{π}{4}, \frac{5 π}{4}$ $x = \frac{7 π}{6}, \frac{11 π}{6}, \frac{π}{4}, \frac{5 π}{4}$ A2

Note: Award A1 for any two correct (radian) answers.
Award A1A0 if additional values given with the four correct (radian) answers.
Award A1A0 for four correct answers given in degrees.

[6 marks]

Examiners report

[N/A]

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.6—Pythagorean identity, double angles

Show 23 related questions

16N.1.AHL.TZ0.H_13c:
Use the principle of mathematical induction to prove that

$\sin x + \sin 3x + \ldots + \sin (2n - 1)x = \frac{{1 - \cos 2nx}}{{2\sin x}},{\text{ }}n \in {\mathbb{Z}^ + },{\text{ }}x \ne k\pi$ where $k \in \mathbb{Z}$ .
16N.1.SL.TZ0.S_2a:
Find $\cos \theta$ .
SPM.1.SL.TZ0.6b:
Hence or otherwise solve ${\text{lo}}{{\text{g}}_3}\left( {{\text{2}}\,{\text{sin}}\,x} \right) = {\text{lo}}{{\text{g}}_9}\left( {{\text{cos}}\,2x + 2} \right)$ for $0 < x < \frac{\pi }{2}$ .
16N.1.AHL.TZ0.H_13b:
Show that $\frac{{1 - \cos 2x}}{{2\sin x}} \equiv \sin x,{\text{ }}x \ne k\pi$ where $k \in \mathbb{Z}$ .
17M.1.SL.TZ1.S_10b:
Given that $\tan \theta > 0$ , find $\tan \theta$ .
SPM.1.SL.TZ0.1a:
Given that ${\text{cos}}\,\hat A = \frac{5}{6}$ find the value of ${\text{sin}}\,\hat A$ .
19M.1.SL.TZ2.S_9a:
Find the value of ${\text{tan}}\,\theta$ .
17M.1.SL.TZ1.S_10c:
Let $y = \frac{1}{{\cos x}}$ , for $0 < x < \frac{\pi }{2}$ . The graph of $y$ between $x = \theta$ and $x = \frac{\pi }{4}$ is rotated 360° about the $x$ -axis. Find the volume of the solid formed.
SPM.1.SL.TZ0.6a:
Show that ${\text{lo}}{{\text{g}}_9}\left( {{\text{cos}}\,2x + 2} \right) = {\text{lo}}{{\text{g}}_3}\sqrt {{\text{cos}}\,2x + 2}$ .
22M.1.SL.TZ1.4b:
Solve the equation $(f \circ g) (x) = 2 \cos 2 x$ where $0 \leq x \leq π$ .
22M.1.SL.TZ1.9a.ii:
By using a suitable substitution for $a$ , show that $1 - 3 \cos 2 x + 3 \cos^{2} 2 x - \cos^{3} 2 x = 8 \sin^{6} x$ .
20N.1.SL.TZ0.S_2a:
Given that $A \hat{B} C$ is acute, find $\sin θ$ .
20N.1.SL.TZ0.S_2b:
Find $\cos (2 \times C \hat{A} B)$ .
SPM.1.SL.TZ0.1b:
Find the area of triangle ABC.
21M.1.SL.TZ2.3b:
Hence, solve the equation $2 \cos^{2} x + 5 \sin x = 4, 0 \leq x \leq 2 π$ .
17M.1.SL.TZ2.S_7:
Solve ${\log _2}(2\sin x) + {\log _2}(\cos x) = - 1$ , for $2\pi < x < \frac{{5\pi }}{2}$ .
18N.1.SL.TZ0.S_7:
Given that ${\text{sin}}\,x = \frac{1}{3}$ , where $0 < x < \frac{\pi }{2}$ , find the value of ${\text{cos}}\,4x$ .
16N.1.AHL.TZ0.H_13d:
Hence or otherwise solve the equation $\sin x + \sin 3x = \cos x$ in the interval $0 < x < \pi$ .
21M.1.AHL.TZ1.12a:
Show that $f'' (x) = - \frac{1}{4 \sqrt{{(1 + x)}^{3}}}$ .
16N.1.SL.TZ0.S_2b:
Find $\cos 2\theta$ .
17M.1.SL.TZ1.S_10a:
Show that $\cos \theta = \frac{3}{4}$ .
21M.1.SL.TZ1.6b:
Hence or otherwise, solve $\sin 2 x + \cos 2 x - 1 + \cos x - \sin x = 0$ for $0 < x < 2 π$ .
21M.1.SL.TZ2.3a:
Show that the equation $2 \cos^{2} x + 5 \sin x = 4$ may be written in the form $2 \sin^{2} x - 5 \sin x + 2 = 0$ .

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