Prove trigonometric relationships involving the — Paper 2 · IGCSE Mathematics - Additional 0606

10 (a) Show that + = 2 cosec x . [3] 1 + cos x sin x (b) Solve the following equations. (i) cot 2 y + cosec y - 5 = 0 for 0° G y G 360° [5] r 3(ii) cos 2z + =- for 0 G z G r radians [4] ` 4 j 2 Question 11 is printed on the next page.

12 marks

Mark scheme: 10(a) 2 2 B1 correct addition of fractions sin x + (1 + cos x ) LHS = sin x (1 + cos x ) 1 + 2cos x + 1 B1 expansion and use of identity = sin x (1 + cosx ) 2 (1 + cos x ) B1 factorisation and completion = = 2cosecx sinx (1 + cosx ) 10(b)(i) cosec 2 y −+1 cosecy − 5 = 0 M1 use of identity for cot2y to obtain quadratic in cosecy cosec2y + cosecy – 6 = 0 ( cosecy − 2 )( cosecy + 3 ) = 0 M1 solve 3 term quadratic for cosecy 1 1 M1 obtain values for siny sin y = , sin y = − 2 3 y = 30°, 150°, 199.5 °, 340.5 ° A2 A1 for 2 values 10(b)(ii) π 5π 7π M2 5π 2 z + = or (2.6…, 3.6…) M1 equate to 4 6 6 6 7π M1 equate to 6 7π 11π A2 A1 for 1 value z = or (0.916, 1.44) 24 24

This question in 0606/23 Oct/Nov 2017

7 (i) Show that - = 2 cosec x cot x . [4] 1 - cos x 1 + cos x 1 1 (ii) Hence solve the equation - = sec x for 0 G x G 2 r radians. [4] 1 - cos x 1 + cos x

8 marks

Mark scheme: 7(i) (1 + cos x ) − (1 − cos x ) M1 Taking common denominator (1 − cos x )(1 + cos x ) 2cosx A1 = 2 1 − cos x 2cosx M1 Using 1 − cos 2 x = sin 2 x = 2 sin x 2cosx 1 A1 Fully correct completion = × AG sinx sinx = 2cosecxcot x 7(ii) 2cosecxcotx = secx M1 2 1 A1 cot x = 2 0.955, 2.19, 4.10, 5.33 A2 A1 for 2 correct values A1 for further 2 correct values

This question in 0606/21 Oct/Nov 2018

8 (i) Show that - = 2 tan x sec x . [4] 1 - sin x 1 + sin x 1 1 (ii) Hence solve the equation - = cosec x for 0° G x G 360° . [4] 1 - sin x 1 + sin x

8 marks

Mark scheme: 8(i) (1 + sinx ) − (1 − sinx ) M1 (1 − sinx )(1 + sinx ) 2sin x A1 1 − sin 2 x 2sinx M1 cos 2 x 2sinx 1 A1 AG × = 2tan x secx cosx cosx 8(ii) M1 equate 2sec x tan x = cosecx 2 1 A1 tan x = 2 35.3°,144.7°, 215.3°, 324.7° 2 A1 two correct

This question in 0606/22 Oct/Nov 2018

11 (a) (i) Show that i i = . [4] sin i 1 + cos i cosec i - cot i 5 (ii) Hence solve = for 180° 1 i 1 360 ° . [2] sin i 2 1 r(b) Solve tan 3z- 4 =- for 0 G z G radians. [3] ` j 2 2

9 marks

Mark scheme: 11(a)(i) 1  1 cosθ M2 M1 for either −   cosecθ− cotθ 1  cosθ sinθ  sinθ sinθ  = cosecθ−   sinθ sinθ sinθ  cosecθ− cotθ 1  1  or =  − cotθ sinθ sinθ sinθ  1 − cosθ M1 1 − cos 2 θ 1 − cosθ 1 A1 = (1 − cosθ)(1 + cosθ) 1 + cosθ 11(a)(ii) awrt 233.1 B2 with no extras in range 3 B1 for cosθ=− soi 5 11(b) −1 1  M1 3φ− 4 = tan  −  soi  2  awrt 0.132, 1.18 A2 with no extras in range A1 for one correct

This question in 0606/21 May/June 2019

2 (i) Show that - = cosecx . [3] 1 - cos x cosecx - cot x (ii) Hence solve = 2 for 0° 1 x 1 180° . [2] 1 - cos x

5 marks

Mark scheme: 2(i) sin 1 − cossin xx M1 express in terms of sinx and cosx 1 − cos x (1 − cos x ) A1 rewrite not as a fraction within a fraction sin (1 − cos x ) 1 A1 correct completion = cosec x answer given sin x 2(ii)  1  B1 sin x = x = 30°    2  x = 150º nfww B1 no extra answers

This question in 0606/23 Oct/Nov 2019

8 (a) Solve 3 cot 2 x - 14 cosec x - 2 = 0 for 0° 1 x 1 360°. [5] sin 4 y - cos 4 y (b) Show that = tan y - 2 cos y sin y. [4] cot y

9 marks

Mark scheme: 8(a) 3(cosec 2 x − 1) − 14cosec x − 2 [ = 0 ] M1 3cosec2 x − 14cosec x −=5 0 A1 (cosecx – 5)(3cosecx + 1) M1 1 A1 sinx = nfww 5 11.5 and 168.5 nfww A1 8(b) Correct use of sin 2 y + cos 2 y = 1 B1 Factorises using the difference of 2 B1 squares 1 cos y B1 Uses = tan y or cot y = cot y sin y correctly Full and correct completion to given B1 answer: tan y − 2cos y sin y

This question in 0606/22 May/June 2020

3 (a) Show that + = 2 cot x cosec x . [4] sec x - 1 sec x + 1 1 1 (b) Hence solve the equation + = 3 sec x for 0° 1 x 1 360° . [4] sec x - 1 sec x + 1

8 marks

Mark scheme: 3(a) cos x cos x s ecx + 1 + sec x − 1 M1 + or 1 − cos x 1 + cos x sec 2 x − 1 cosx + cos 2 x + cos x − cos 2 x 2sec x A1 or 1 − cos 2 x tan 2 x 2cos x 2cos 2 x A1 or oe sin 2 x cos x sin 2 x Fully correct justification of given answer: 2cot xcosec x A1 3(b) 3tan 2 x = 2 oe or better, soi B1 or 5cos 2 x = 3 oe or better, soi or 5sin 2 x = 2 oe or better, soi 2 M1 FT an equation of the form tanx = [ ± ] oe or [±] 0.816[4…] 2 a tan x = b , a > 0, b > 0 3 or p sin 2 x = q or p cos 2 x = q 3 or cosx = [ ± ] oe or [±] 0.774[5…] where p > 0, q > 0 and p > q 5 2 or sinx = [ ± ] oe or [±] 0.632[4…] 5 39.2° or 39.2315… rot to 2 or more dp A2 no extras in range 140.8° or 140.7684… rot to 2 or more dp 219.2° or 219.2315… rot to 2 or more dp 320.8° or 320.7684… rot to 2 or more dp A1 for any two correct answers

This question in 0606/22 Oct/Nov 2021

5 (a) Show that + = 2 tan x sec x . [4] cosec x - 1 cosec x + 1 1 1 (b) Hence solve the equation + = 5 cosec x for 0° 1 x 1 360° . [4] cosec x - 1 cosec x + 1

8 marks

Mark scheme: 5(a) sin x sin x cosec x + 1 + cosec x − 1 M1 + or oe 1 − sin x 1 + sin x cosec 2 x − 1 sin x + sin 2 x + sin x − sin 2 x 2cosec x A1 or oe 1 − sin 2 x cot 2 x 2sin x 2sin 2 x A1 or oe cos 2 x sin x cos 2 x Fully correct justification of given answer: A1 2sin x 1 × = 2tan x sec x cos x cos x 1 or 2tan x × = 2tan x sec x cos x 2sin x or × sec x = 2tan x sec x cos x or equivalent 5(b) 2tan 2 x = 5 or better, soi B1 or 7cos 2 x = 2 or better, soi or 7sin 2 x = 5 or better, soi 5 M1 FT an equation of the form tan x = [ ± ] oe or [±] 1.58[1…] 2 a tan x = b a > 0, b > 0 2 or p sin 2 x = q or p cos 2 x = q 2 or cos x = [ ± ] oe or [±] 0.534[5…] where p > 0, q > 0 and p > q 7 5 or sin x = [ ± ] oe or [±] 0.845[1…] 7 57.7 or 57.6884… rot to 2 or more dp A2 no extras in range 237.7 or 237.6884… rot to 2 or more dp A1 for any two correct answers 122.3 or 122.3115… rot to 2 or more dp 302.3 or 302.3115… rot to 2 or more dp

This question in 0606/23 Oct/Nov 2021

7 In this question, all angles are in radians. (a) Solve the equation sec 2i = tan i + 3 for - r 1 i 1 r . [5] r tan z (b) Show that, for 0 1 z 1 , = sec z . [3] 2 1 - cos 2z 17 3 r (c) Given that cosecx =- and that 1 x 1 2 r , find the exact value of cotx. [2] 8 2

10 marks

Mark scheme: 7(a) Uses a valid Pythagorean identity to write B1 in terms of a single trig ratio e.g. 1 + tan 2 θ = tanθ + 3 Rearranges and factorises/solves M1 e.g. tan 2 θ − tan θ − 2 = 0 (tanθ − 2)(tanθ + 1) = 0 tanθ = 2 tanθ = −1 soi A1 π 3π A2 A1 for any 3 correct, ignoring extras 1.11, −2.03, − , and no extras in 4 4 range cao 7(b) sin φ M1 Use of tan φ = in a correct cosφ expression or correct working Use of 1 − cos 2 φ = sin 2 φ in a correct M1 expression or correct working 1 A1 nfww Completion with = secφ cosφ 7(c) 2 M1 cot x = [ − ] cosec x − 1 soi 8 8 or sin x = − and tan x = − soi 17 15 8 15 or sin x = − and cos x = soi 17 17 1 or  −1  8   tan  sin  −     17   15 A1 − or −1.875 cao, isw 8

This question in 0606/22 Feb/March 2022

7 (a) Show that + = 2 cosec x . [4] 1 - cos x sin x sin x 1 - cos x (b) Hence solve the equation + = 3 sin x - 1 for 0° 1 x 1 360° . [4] 1 - cos x sin x

8 marks

Mark scheme: 7(a) sin 2 x + (1 − cos x ) 2 M1 (1 − cos x )sin x sin 2 x (1 − cos x) 2 or + (1 − cos x )sin x (1 − cos x)sin x sin 2 x + 1 − 2cos x + cos 2 x A1 1 − cos 2 x + (1 − cos x ) 2 OR (1 − cos x )sin x (1 − cos x )sin x 1 + 1 − 2cos x A1 (1 − cos x )(1 + cos x ) + (1 − cos x) 2 OR (1 − cos x )sin x (1 − cos x )sin x 1 − cos 2 x + 1 − 2cos x + cos 2 x or (1 − cos x )sin x Fully correct justification of given A1 All steps correct and final step justified answer: 2(1 − cos x ) 1 + cos x + 1 − cos x = 2cosec x OR = 2cosec x (1 − cos x )sin x sin x 2 − 2cos x 2 or = = 2cosec x (1 − cos x )sin x sin x or equivalent Alternative sin x (1 + cos x ) (1 − cos x )sin x (M1) + (1 − cos x )(1 + cos x ) sin x sin x sin x (1 + cos x ) (1 − cos x )sin x or + 1 − cos 2 x sin 2 x sin x + sin x cos x sin x − cos x sin x (A1) + sin 2 x sin 2 x 2sin x (A1) sin 2 x Fully correct justification of given (A1) All steps correct and final step justified 2 answer: = 2cosec x sin x 7(b) 3sin 2 x − sin x − 2  = 0 soi B1 (3sinx + 2)(sinx – 1) [= 0] oe M1 2 A1 sin x = − , sin x = 1 3 90 A1 and no extras in range 221.8 or 221.81[03…] rot to 2 or more If B1 M1 A0 A0 allow SC1 for dp 221.8 or 221.81[03…] rot to 2 or 318.2 or 318.18[96…] rot to 2 or more more dp dp and 318.2 or 318.18[96…] rot to 2 or more dp and no extras in range

This question in 0606/22 Oct/Nov 2022

5 (a) Show that + = 2 sec x . [4] 1 - sin x cos x i i cos 1 - sin 2 2 2 i (b) Hence solve the equation + = 8 cos for - 360° 1 i 1 360° . [4] i i 2 1 - sin cos 2 2

8 marks

Mark scheme: 5(a) cos 2 x + (1 − sin x ) 2 M1 Correctly takes common denominator (1 − sin x ) cos x cos 2 x (1 − sin x ) 2 or + (1 − sin x ) cos x (1 − sin x ) cos x cos 2 x + 1 − 2sin x + sin 2 x A1 1 − sin 2 x + (1 − sin x ) 2 OR (1 − sin x ) cos x (1 − sin x ) cos x 1 + 1 − 2sin x A1 (1 − sin x )(1 + sin x ) + (1 − sin x ) 2 (1 − sin x ) cos x OR (1 − sin x ) cos x 1 − sin 2 x + 1 − 2sin x + sin 2 x or (1 − sin x ) cos x 2(1 − sin x ) A1 All steps correct and final step justified = 2sec x (1 − sin x ) cos x 2 − 2sin x 2 or = = 2sec x 1 + sin x + 1 − sin x (1 − sin x ) cos x cosx OR = 2sec x cos x or equivalent Alternative Must work with LHS only (cos x )(1 + sin x ) (1 − sin x )cos x (M1) Forms fractions with common + (1 − sin x )(1 + sin x ) ( cos x ) cos x denominator in different form (cos x )(1 + sin x ) (1 − sin x )cos x (A1) Uses difference of two squares and + 2 cos x cos 2 x sin 2 x + cos 2 x = 1 to write fractions with a common denominator in the same form 2cos x (A1) Combine as a single fraction and 2 collects terms cos x 2 (A1) All steps correct and final step justified = 2sec x cos x 5(b) 3 1 B1 cos = 2 4 1 M1  2  cos = 3 their soi dep on starting with 2sec = 8cos 2 4 2 2 101.9 awrt A2 and no extras in range A1 for either, ignoring extras in range If A0 then SC1 for 102 with no extras in range

This question in 0606/23 Oct/Nov 2022

10 (a) By writing cotx and tanx in terms of cosx and sinx, show that sin x cos x + = sin x + cos x. [5] 1 - cot x 1 - tan x (b) Solve the equation 9 cot x + 3 cosec x = tan x , for 0° 1 x 1 360° . [5]

10 marks

Mark scheme: 10(a) Writes cotx and tanx in terms of sinx and M1 OR cosx:  sin x   cos x  sin x  1 −  + cos x  1 −  sin x cos x  cos x   sin x  + cos x sin x  cos x  sin x  1 − 1 −  1 −  1 −  sin x cos x  sin x  cos x  Simplifies denominator: A1 OR sin x cos x  cos x − sin x   sin x − cos x  + sin x   + cos x   sin x − cos x cos x − sin x  cos x   sin x  sin x cos x  sin x − cos x  cos x − sin x      sin x  cos x  Writes as two simple algebraic fractions: A1 OR writes as a single simple algebraic sin 2 x cos 2 x fraction: + 2 2 sin x (cos x − sin x ) + cos x (sin x − cos x ) sin x − cos x cos x − sin x (sin x − cos x )(cos x − sin x ) Writes as a difference with a common A1 sin 2 x (cos x − sin x ) − cos 2 x (cos x − sin x ) OR denominator: (sin x − cos x )(cos x − sin x ) sin 2 x cos 2 x − sin x − cos x sin x − cos x Correct simplification to given answer, e.g., A1 All steps correct and final step fully justified (sin x − cos x )(sin x + cos x ) by factorising = sin x + cos x (sin x − cos x ) or (sin x − cos x ) (sin x + cos x )  = sin x + cos x  (sin x − cos x ) 10(b) 10cos 2 x + 3cos x − 1[ = 0] B2 9cos x 3 sin x B1 for + = or better or sec 2 x − 3sec x − 10[ = 0] sin x sin x cos x 3tan x 2 or 9 + = tan x or better sin x OR M1 for one sign error in 10cos 2 x + 3cos x − 1[ = 0] or sec 2 x − 3sec x − 10[ = 0] ( 5cos x − 1)( 2cos x + 1) = 0  M1 FT their 3-term quadratic in cosx or secx or ( sec x − 5 )( sec x + 2 ) = 0  1 1 A2 A1 for any two correct angles [cosx = and cosx = − 5 2 1 1 [found using cosx = and cosx = − 5 2 OR OR secx = 5 and secx = –2 leading to] secx = 5 and secx = –2]; 78.5 or 78.46[30…] rot to 2 or more dp ignore extras 281.5 or 281.53[69…] rot to 2 or more dp 120 240 and no extras in range 0  x  360

This question in 0606/22 Oct/Nov 2023

7 DO NOT USE A CALCULATOR IN THIS QUESTION. A You may use the following 60° trigonometrical ratios. 3 2 sin 60 ° = , sin 45 ° = 2 2 2 1 2 cos60 ° = , cos45° = 2 2 75° tan60° = 3, tan45° = 1 45° B C 3 + 3 6 + 2 (a) Given that the area of triangle ABC is , show that sin75° = . [5] 4 4 (b) Hence find the exact length of AC. [2]

7 marks

Mark scheme: 7(a) BC 2 M1 = oe sin60 sin 45 BC = 3 oe A1 1 3 + 3 M1 FT their BC providing it has not been  2  3  sin75 = found using the given result for sin75 2 4 OR  3 + 3  2 height =    oe    4  3 2(3 + 3) A1 dep on all previous marks being sin75 = oe awarded 4 6 Isolates sin 75 correctly or deals with surds on LHS of correct OR equation  3 + 3  2  1 + 3  1  3 + 3      6   e.g. 6 sin75 =      4  3 2 1 + 3 2 4 sin 75 = oe  = =    2  2 2 2      6 + 2 A1 must be convincing with an correct completion to given answer intermediate step if needed 4 Alternative methods (finding AC first) 1 3 + 3 (M1)  2  AC  sin60 = oe 2 4 2 2 3 + 3 (A1) Isolates AC correctly AC =   oe 2 3 4 2 + 6 (A1) Must be convinced no calculator is AC = being used 2 sin75 sin45 (M1) FT their AC = oe AC 2 + 6 2 May simplify to sin75 = before 2 2 inserting their AC 6 + 2 (A1) dep on all previous marks being correct completion to given answer awarded 4 must be convincing with an intermediate step if needed 7(b) AC 2 AC 3 M1 = or = or better 6 + 2 2 6 + 2 3 4 2 4 2 6 + 2 A1 nfww 2

This question in 0606/23 Oct/Nov 2023

10 (a) Show that ( tan x + sec x) 2 can be written as . [4] 1- sinx (b) Hence solve the equation ( tan 3 i + sec 3i ) 2 = 6 for 0° G i G 180 ° . [4]

8 marks

Mark scheme: 10(a) tan 2 x  2tan x sec x  sec 2 x M1  sin x 1  2 or     cos x cos x  sin 2 x sin x 1 1 A1  2    cos 2 x cos x cos x cos 2 x 1 2 or factorises sin x  1 oe 2  cos x (1  sin x ) 2 A1 1  sin 2 x (1  sin x )(1  sin x ) 1  sin x A1 must be fully justified  (1  sin x )(1  sin x ) 1  sin x or (1  sin x ) 2 1  sin x  (1  sin x )(1  sin x ) 1  sin x 10(b) 7sin3 5 B1 One correct value for 3 soi e.g. M1 45.58… 134.4… 405.5… 494.4… 15.2 or 15.19 to 15.195 A2 with no extras in range 44.8 or 44.80 to 44.81 135.2 or 135.19 to 135.195 A1 for any 2 correct, ignoring extras 164.8 or 164.80 to 164.81

This question in 0606/21 May/June 2024