Core 2 - Ch 6, 8, 10 - 1 - Trigonometry - Solutions

8/13/2019 Core 2 - Ch 6, 8, 10 - 1 - Trigonometry - Solutions

1/29

1. sin2 x = 4 cos x M1

1 cos2 x = 4 cos x M1

cos2 x + 4 cos x 1 = 0 A1

cos x = 24164 +

M1

= 2420

, second root has no real solution for x A1, B1 x = 76.3 or 283.7 A1 A1 ft 8

[8]

2. a! Arc AB = 2a " 3

usin#r ) M1

$eri%eter ofS = 3 " 32 a

= 2a A1 2

&! Area of sector ABC = 3!2

21 2 a

= 2a2 3

B1

Area of trian#le ABC = 3sin!2

21 2 a

= a23 M1 A1

Area of se#%ent = 2a2 3

a23 M1Area ofS = 3 Area of se#%ent ABC ! + Area of trian#le ABC ! M1

= 2 a 2 3a23 +a23

= 2a2 3! A1 6[8]

3. a! Arc BD = r = 0.4 " 6 = 2.4 B1 1

&! 'osine (ule) AB2 = 62 + 122 2 " 6 " 12 " cos 0.4c! = 47.36* M1 A1 AB = 6.88** A1 3

c! $eri%eter = 6 + 6.88 + 2.4 = 1 .3 c%! 3 si#. fi#s! B1 ft 1[5]

Holland Park School 1


2/29

4. a! x 20 = 11 . *-r 244.08...

Any solution (awrt 116 or 244 ) B1360 candidates irst answer !1" 20 at correct sta#e !1

x = 136, 264 A1 4

&! 3

cossin

= 2 cos

$se o tan %

cossin

!1

3 sin = 2 cos2 = 2 1 sin2 !$se o cos 2 % 1 sin 2 !1

2 sin2 + 3 sin 2 = 0

3 ter& 'uadratic in sin % 0 A1

2 sin 1! sin + 2! = 0 Atte& t to sol e !1

sin =2 o solution!

sin = 21

At least 21

A1

/o = 30, 1 0 Both A1 6[10]

5. a!

1

1

3 6 0

/cales 1, 1 and 360! B1/ha e, osition B1



3/29

&! 0, 0. ! 1 0, 0! 330, 0! B1 B1 B1c! x + 30=! 210 or 330 B1

*ne o t+ese

x = 180 , 300 !, Su-tract 30. A, Bot+ !1 A1

[8]

6. a! B! = 72 + 242! = 2 ! B1

&! tan = 247

or e ui . and B!C = 2 , or cosine rule M1 A1

B!C = 0. 68 radians ! A1

c! AB! ) 21

14 " 24! = 168 %%2! or other a ro riate! B1

/ector) 21

2 2 " 0. 68! M1 A1

5otal) 168 + 168 + 177. = 13 %%2 or 14, or 10! M1 A1

d! olu%e = 13 " 8 %%3 M re uires unit con ersion! M1

= 44 c%3 A1[11]

7. a! 2

21

r

=2

21

r 1. = 1 M1 A1

r 2= 20 = 4 " ! r = 2 ! A1 3

&! r + 2r = 3 + 4 = 7 c% or 1 .7, or a.9.r.t 1 .6 *.! M1 A1 2

c! *AB ) sin

21 2r

= 10 sin 1. = . 74 *! M1/e#%ent area = 1 *AB = .02 c%2 M1 A1 3

[8]



4/29

8. 2cos2 cos 1 = 1 cos2 M1

3cos2 cos 2 = 0 A1

3cos + 2! cos 1! = 0 cos = 32

or 1 M1 A1

= 0, = 131.8 B1 A1 = 360 131.8 ! = 228.2 M1 A1 ft

[8]

9. a!'osine:/ine sha e, eriod 2 B1All features correct. B1 2

/or irst B1. i ull do&ain is not s+own. at least 0 to &ust -e s+own. and in t+is case t+ere &ust -e indication o scale on t+e x axis /or second B1. i#nore anyt+in# outside t+e 0 to 2 do&ain

&!

0,

4+

,0,4

,2

1,0

or, e.#. x = 0, y = 2

1

. B1 B1 B1 3 Allow 4 and 22 . allow any exact or& enalties, 5on exact. lose 1 &ar (&axi&u&)

!issin# 7eros. or coordinates wron# way round. lose 1&ar (&axi&u&)



5/29

c!

=+

4

x

3

or 60 or 1.0 2 d. . at least! B1

-ther alue 3+

32

=

M1

/u&tract4

M1(/or t+e ! &ar s. allow wor in# in de#rees or radians to 2 dat least. -ut not de#rees and radians &ixed unless correctedlater )

x = 12

, x = 1217

A1 4 Allow 0 08 and 1 42 (2 d or -etter) 9#nore extras outside 0 to 2 n - :+ere are ot+er correct a roac+es or t+e irst !1. usin#

t+e sy&&etry o t+e #ra +[9]

10. a! 21

;21 2 =r

" 6. 2 " 0.8 = 16. a.9.r.t. if chan#ed to de#rees! M1 A1 2

&! sin 0.4 = +.6 x

, x = 6. sin 0.4 , 9here x is half of AB! M1, A1(n - 0 8 rad % 4 8 )

AB = 2 x = .06 a.9.r.t.! ! A1 3

Alternati e) AB2 = 6. 2 + 6. 2 2 " 6. " 6. cos 0.8


6/29

&! 2 = 21.8* ! At least 1 d. .! 'ould &e i% lied &? a correct !. B12 = + 180 or 2 = + 360 or 2 = + 40 -ne %ore solution! M1 = 10. , 100. , 1 0. , 280. M1) di ide &? 2!M1 A1ft A1A1ft) 2 correct, ft their ! A1) all 4 correct cao, at least 1 d . .!

c! 2sin

cossin

= 3, 2sin2 = 3cos M1, A12 1 cos2 ! = 3cos M12cos2 + 3cos 2 = 0

2cos 1! cos + 2! = 0 cos = 21

M) sol e 3 ter% uadraticu to cos = * or x = *! M1 A1

= 60, = 300 A1 6[14]

12. a! tan x = 38

or e@act e ui alent, or 3 s.f. or &etter! B1 1

&! tan x = 38

x = 6 .4 !, x = 24 .4 180 + ! M1 A1, A1ft 3

c! 3 1 cos2 y! 8 cos y = 0 3 cos2 y + 8 cos y 3 = 0 M1 A1

3 cos y 1! cos y + 3! = 0, cos y % ..., 31

or 3! M1 A1 y = 70. !, x = 28 . 360 ! A1 A1ft 6

[10]

13. a! cos A =( )

2627226

222

+

M1 A1

cos A = 21

A = 3

radians ! A1 3



7/29

&! r = 32

= 2.0 ! @act or at least 3 s.f.! M1 A1 2

c! /ector ABD) 32

21

21 22 =r

= ...0 4.2

32

M1

5rian#le ACB) 3sin6221 = 3 3 .1 6*! M1

5rian#le /ector = 32

33

= 3.1017 *! M1 A1 4 Allow 3 1 or a wrt 3 10

[9]

14. a! i! >sin# tan =

cos

sin

and cot =

sin

cos

or cot = tan

1

M1or%in# a sin#le fraction

CD/ =

cossincossin 22 +

or CD/ =

tantan1 2+

M1

(eachin# the e@ ression cossin1

A1>sin# sin 2 = 2sin cos M1

CD/ = 2sin2

cossin22 =

= 2cosec2 (D/ ! cso A1

ii! cos = 1312

M1 >se of sin2 + cos2 = 1 or ri#ht M1 A1an#led trian#le &ut acce t stated

&! >se of cos2 = cos2 sin2 or 1 2sin2 or 2cos2 1! M1

cos 2 = 1611

A1 4 c! >se of cos x + ! = cos x cos sin x sin M1

/u&stitutin# for sin and cos M112 cos x sin x + sin x = 6 12 cos x = 6! A1

x = 3

a9rt 1.0 M1 A1[14]



8/29

15. >sin# sin2 + cos2 = 1 to #i e a uadratic in cos . M1Atte% t to sol e cos2 + cos = 0 M1

cos = 0! = 2

, 23

B1, B1cos = 1! = B1

'andidate 9ho 9rites do9n 3 ans9ers onl? 9ith no 9orEin# scoresa %a@i%u% of 3!

[5]

16. a! = r + 2r , A = 21

r 2 B1, B1 2&! /u&stitutin# alue for r and e uatin# to A.


9/29

18. a! 1 sin2 x! = 3 1 + sin x! M1 sin 2 x = 3 + 3sin x0 = sin2 x + 3sin x 2 ! A1 cso 2

&! 0 = sin x 2! sin x + 1!

sin x =,

+

2

1 &oth! A1 cso

sin x = +2

x = 23.6 = 23.6 or 1 6.4! B1, 1 6.4 180 ! M1

sin x = 1 x = = 270 i#nore e@tra solutions outside the ran#e! B1[7]

19. a! r = 8 " 0.7, = .6c&! M1, A1 2

&! BC 2 = 82 + 112 2 " 8 " 11 cos0.7 M1 BC = 7.0 8 A1 $eri%eter = a! + 11 8! + BC , = 1 .7 c&! M1, A1cao 4

c! = 21

a- sinc = 21

" 11 " 8 " sin 0.7, = AF(5 28.3 M1, A1

/ector = 21

r 2 = 21

" 82 " 0.7 M1, A1Area of ; = 28.34 *.. 22.4 = . 4 = . c&2! A1

[11]

20. a! r< = 4 < = 63,< = 1.4 ! M1A1 2 !1 is or a lyin# correct or&ula or 'uotin# and atte& tin# to use correct or&ula



10/29

&! Area of sector*AB = 21

21 2 = r

4 2 " 1.4 = 1417. ! M1A1

Area of trian#le*CD = 21

302 " sin1.4 = 443.4 ! M1A1/haded area = 1417. 443.4 *= 74 %2 cao A1

/or eac+ area !1 is or atte& tin# to use correct or&ula or co& lete&et+od in case o = (>)

A1 is or a nu&erically correct state&ent (answer is notre'uired ?ust t+ere as c+ec )

/inal A1 is or @ 4 onlye # s littin# trian#le into two trian#les,

/or #uidance

0 . 7

0 . 7

* *

D D

C C

1 , . 3 3

1 , . 3 3

3 0 3 0

2 2 . , 4

+ . 1 0

2 4 . , 0

2 , . + 6

[7]



11/29

21. a! arctan 23

= 6.3 = ! seen an?9here! B1 20 , 20! G 3 M1M1 + 180 = 236.3 !, 180 = 123.7 ! -ne of these! M1

x = 47. , 12.1, 72.1 A1A1 6

/irst !1, Su-tractin# (allow addin#) 20 ro& Second !1, Di idin# t+at result -y 3 (order ital )

[So 12.1 gains B1M1M1] :+ird !1, i in# a t+ird 'uadrant result

/irst A1 is or 2 correct solutions.Second A1 or t+ird correct solution

B1, Allow 0 @83 (rads) or 62 6 (#rad). and ossi-le !s -ut A0A0

H5(A

>sin# e@ ansion of tan 3 x + 20 ! = 23

Iettin# as far as tan 3 x = nu%&er 0.7348..! irst M1tan 3 x = 36.3 , 216.3 , 143.7 36.3 B1

x = 12.1 , 72.1 , 47. 5hird uad result 5hird M1Ji ide &? 3 /econd M1

Ans9ers as sche%e A1A1

&! 2sin2 x + 1 sin2 x! =10

or 2 1 cos2 x! + cos2 x =10

M1

sin2 x =1

or cos2 x =8

or tan2 x = 81

or sec2 x = 8 or cos 2 x =7

A1 x = 1 . , 1 . A1A1ft 4

!1 or use o sin 2 x " cos 2 x % 1 or sin 2 x and cos 2 x inter&s o cos2x

5ote, !ax deduction o 1 or not correctin# to 1 dec lace ;ecord as 0 irst ti&e occurs -ut t+en treat as t Answers outside #i en inter al. i#nore Extra answers in ran#e. &ax deduction o 1 in eac+ art F/inal &ar(i e 4 or &ore answers wit+in inter al in (a). 1 ro& any

#ained A &ar sG3 or &ore answers wit+in inter al in (-). 1 ro& any

#ained A &ar s[10]



12/29

22. a! x + 10 =! 60 B1120 M1(!, 180 or )

x = 0 x = 110 or 0.0 and 110.0! M1 A1 4(!, su-tract 10)

/irst !, !ust -e su-tractin# ro& 180 -e ore su-tractin# 10

&! 2 x =! 1 4.2 B1

Allow a wrt 1 4 or a wrt 2 6@ (radians)

20 .8 M1(!, 360 or 2 )

x = 77.1 x = 102. M1 A1 4 !, Di ide -y 2 /irst !, !ust -e su-tractin# ro& 360 -e ore di idin# -y 2. or di idin# -y 2 t+en su-tractin# ro& 180

[8]

Kn each art)@tra solutions outside 0 to 180) K#nore.e@tra solutions &et9een 0 and 180 ) A0.

Alternati e or (-), (Dou-le an#le or&ula)

1 2sin2 x = 0. 2sin

2 x = 1. B1

sin x = +.0 M1 x = 77.1

x = 180 77.1 = 102. M1 A1

23. a! 7+.0sin

8sin =

x

or +.0sin7

sin8 =

x , sin x = 7+.0sin8

M1 A1ftsin x = 0. 48 A1 3

!, Sine rule atte& t (sidesHan#les ossi-ly t+eIwron# way roundJ) A1 t, ollow t+rou#+ ro& sidesHan#les are t+eIwron# way roundJ



13/29

&! x = 0. 8 ! B1(:+is &ar &ay -e earned in (a))

= 2. 6 M1 A1ft 3[6]

5oo %an? d. . #i en)Ma@i%u% 1 %arE enalt? in the co% lete uestion.Jeduct on first occurrence!.

24. a! tan< = B1 1 !ust -e seen ex licitly. e # tan< % tan l % 8 or e'ui is B0. unless tan< % is also seen

&! tan< = E ( )6 1tan =

M1 = 78.7, 2 8.7 Acce t a9rt! A1, A1ft 3

:+e ! &ar &ay -e i& lied -y wor in# in (a)

A1 t or 180 "K (a ) Answers in radians would lose -ot+ t+e A &ar s Extra answers -etween 0 and 360, Deduct t+e inal &ar Alternati e,

$sin# cos 2 < % 1 sin 2 < (or e'ui ) and roceedin# to sin < % (or e'ui ), !1 t+en A &ar s as in &ain sc+e&e

[4]

25. a! r< % 2.12 " 0.6 1.38 %! M1 A1 2 !1, $se o r< wit+ r % 2 12 or 1 86. and < % 0 6 . or e'ui&et+od or t+e an#le c+an#ed to de#rees (allow awrt 3 )

&!6+.012.2

21

21 22 = r

1.46 %2! M1 A1 2

!1, $se o 22

1r

wit+ r % 2 12 or 1 86.and < % 0 6 . or e'ui&et+od or t+e an#le c+an#ed to de#rees (allow awrt 3 )



14/29

c!6+.0

2

0. 2 radians! ( ) M1 A1 2

!1, Su-tractin# 0 6 ro& 2

. or su-tractin# awrt 3 ro& @0(de#rees). ( er+a s i& lied -y awrt 3)

An#le c+an#ed to de#rees wron#ly and used t+rou#+out (a). (-)and (c),

enalise L&et+od only once. so could score !0A0. !1 A0. !1 A0

d! ACD,!86.1!12.2

21

sinK Fith the alue ofK fro% art c!! M1

Area = 1.46 + 1. 7 , 3.03 %2! M1 A1 3 /irst !1, *t+er area &et+ods &ust -e ully correct

Second !1, Addin# answer to (-) to t+eir ACD /ailure to round to 2 d , enalise only once. on t+e irst occurrence. t+en acce t awrt 9 0 6 is ta en as de#rees t+rou#+out,*nly award &ar s in art (d)

[9]

26. a! tan = 23

>se of tan =

cossin

M1

= 6.3 cao A1= 236.3 ft 180 + their rinci le alue A1ft 3Ma@i%u% of one %arE is lost if ans9ers not to 1 deci%al lace

&! 2 cos< = 2 1 cos2 < ! >se of sin2 < + cos2 < = 1 M1

2cos2 < cos< A1

Allo9 this A1 if &oth cos< = 0 and cos< = 2

1

are #i encos< = 0 < = 0 , 270 M1 one solution M1 A1

cos< % 21

< = 60 , 300 M1 one solution M1 A1 6[9]

27. a! Arc is 6 " 1.2 >se ofr M1$eri%eter is 6 " 1.2 + 6 + 6 + 6 = 1 .2 c%! A1 2



15/29

&! Area is21

" 62 " 1.2 = 21.6 c%2! >se of 21

r 2< M1 A1 2 c!

a

-

x y

6

1 . 2

B1 x = 6 cos 1.2 2.174 ...! M1

y = 6 x 3.82 ...!

a = 6 + y .83 c%! M1 A1 4Ma@i%u% of one %arE is lost if ans9ers not to 3 si#. fi#s

[8]

28. a! cos B* A L = ++26++ 222

+

or M1

sin< %3

9ith use of cos2< % 1 2 sin2 < atte% ted

= 2+7

A1cso 2

&! B* A L = 1.2870022... radians 1.287 or &etter B1 1

c! /ector % 21

M 2 " - !, = 16.087. AF(5!16.1 M1 A1 2

d! 5rian#le =21 " 2 " sin - ! or 21 " 6 " 22 3+ M1/e#%ent = their sector! their trian#le dM1= sector fro% c! 12 = AF(5!4.1 ft their art c!! A1ft 3

[8]

a! M1 for a full %ethod leadin# tocosA*B < .B. >se of calculator is M0

usual rules a&out uotin# for%ulae!

&! >se of &! in de#rees is M0



16/29

d! 1st M1 for full %ethod for the area of trian#le A*B2n dM1 for their sector their trian#le. Je endent on 1st M1 in art d!.A1ft for their sector fro% art c! 12


17/29

MISREADS

x2 %isread as x3

21

2+

21

3

2+2+ +=+ x x

x

x

M1 A0

f x! = 3 x +

+

21

2

27

+ 21

27

x x

+C ! M1 A1ft

6 = 3 + 710

+ 4 +C M1

C = 717

, f x! = 3 x + 717

47

10 21

27

+ x xA0, A1

30. 2 1 sin2 x! + 1 = sin x M12sin2 x + sin x 3 = 02sin x 1! sin x + 3! = 0 M1

sin x = 21

M1, A1

x = 6+

,6

M1, M1, A1cso 6

>se of cos2 x = 1 sin2 x.'ondone in isi&le &racEets in first line if 2 2 sin2 x is resentor i% lied! in a su&se uent line.Must &e usin# cos2 x = 1 sin2 x. >sin# cos2 x = 1 + sin2 x is M0. M1

Atte% t to sol e a 2 or 3 ter% uadratic in sin x u to sin x = *>sual rules for sol in# uadratics. Method %a? &e factorisin#, for%ulaor co% letin# the s uare M1

'orrect factorisin# for correct uadratic and sin x = 21

./o, e.#. sin x + 3! as a factors sin x = 3 can &e i#nored. A1

Method for findin# an? an#le in an? ran#e consistent 9ith either of!their tri#. e uation s! in de#rees or radians e en if x not e@act!.


18/29

Method for findin# second an#le consistent 9ith either of! their tri#. e uation s! in radians.Must &e in ran#e 0 x O 2 Must in ol e usin# e.#. O *, 2 *! &ut * can &e [email protected] &e usin# the sa%e e uation as the? used to atte% t the 3rd M %arE.>se of %ust &e consistent 9ith the tri#. e uation the? are usin#

e.#. if usin# cos 1

then %ust &e usin# 2 *!Kf findin# &oth an#les in de#rees) %ethod for findin# 2nd an#lee ui alent to %ethod a&o e in de#rees and an atte% t to chan#e &oth an#les to radians. M1

6+

,6

c.s.o. (ecurrin# deci%als are oEa? instead of61

and 6+

!.'orrect deci%al alues corrected or truncated! &efore the final

ans9er of 6+

,6

is acce ta&le. A1 csoK#nore e@tra solutions outside ran#eP deduct final A %arE for e@trasolutions in ran#e./ ecial case

Ans9er onl? 6+

,6

M0, M0, A0, M1, M1 A1

Ans9er onl? 6

M0, M0, A0, M1, M0 A0indin# ans9ers &? tr?in# different alues e.#. tr?in# %ulti les ofP ! in2cos2 x + 1 = sin x) as for ans9er onl?.

[6]

31. a! cos Q; ==

+21

662!3666 222

M1, A1

Q; = 32

A1 3



19/29

.B. a2 = - 2 " c 2 2-c cos A is in the for%ulae &ooE.>se of cosine rule for cos Q; Allo9 A. or other s?%&ol for an#le. M1

i! 36 !2 = 62 + 62 2 6 6 cos Q; ) A l? usual rulesfor for%ulae) a! for%ula not stated, %ust &e correct,

&! correct for%ula stated, allo9 one si#n sli 9hen su&stitutin#.

or ii! cos Q; = 662!3666 222

Also allo9 in isi&le &racEets


20/29

c! Area of = 32

sin6621

%2 M1

= 3 %2 A1cso 2

>se of Qr 2sin 9ith r = 6 and their a!. = cos 1 their Q; ! in de#rees or radiansMethod can &e i% lied &? correct deci%al ro ided deci%alis correct corrected or truncated to at least 3 deci%al laces!.1 . 884 727 M1

3 c.s.o. Must &e e@act, &ut correct a ro@. follo9ed &?3 is oEa? e.#. * = 1 . 884 = 3 ! A1cso

Alternati e usin# Q-+)Atte% t to find+ usin# tri#. or $?tha#oras and use this+ in Q-+

for% to find the area of trian#le Q; M13 c.s.o. Must &e e@act, &ut correct a ro@. follo9ed &?3

is oEa? e.#. * = 1 . 884 = 3 ! A1cso

d! Area of se#%ent = 12 3 %2 M1= 22.1 %2 A1 2

>se of area of sector area of or use of Qr 2 sin ! M1An? alue to 1 deci%al lace or %ore 9hich rounds to 22.1 A1

e! $eri%eter = 6 + 6 +

32

6

% M1= 24.6 % A1ft 26 + 6 +


21/29

32. a! 42 = 2 + 62 2 " " 6cos ! M1

cos = 6+246+ 222

+

A1

43

604+ =

=

! A1cso 3

M) Ks also scored for 2 = 42 + 62 2 " 4 " 6cos !or 62 = 2 + 42 2 " " 4cos !

or cos = 642+64 222

+

or cos = 4+264+ 222

+

.

1st A) (earran#ed correctl? and nu%ericall? correct ossi&l?unsi% lified!, in the for% cos ... or 60cos = 4 or e ui .in the for% cos = ' !.

Alternati e erification!)

42 = 2 + 62

436+2


22/29

33. a!1

0 . +

0 . +

1

y

x

2 1 2 2

/ine 9a e an?9here 9ith at least 2 turnin# oints. M1/tartin# on ositi e ySa@is, #oin# u to a %a@., then %in. &elo9

xSa@is, no further turnin# oints in ran#e, finishin# a&o e xSa@is at x = 2 or 360 . 5here %ust &e so%e indication of scale on the [email protected].#. 1, 1 or 0. ! A1 2K#nore arts of the #ra h outside 0 to 2 .n.&. Ii e credit if necessar? for 9hat is seen on an initial sEetch

&efore an? transfor%ation has &een erfor%ed!

&!

0,

611

,0,6

+,

21

,0

K#nore an? e@tra solutions! ot 1 0 , 330 ! B1, B1, B1 3

62and

6

are sufficient, &ut if &oth are

seen allo9 B1B0.5he Teros are not re uired, i.e. allo9 0. , etc. and also

coordinates the 9ron# 9a? round!.5hese %arEs are also a9arded if the e@act interce t aluesare seen in art a!, &ut if alues in &! and a! arecontradictor?, &! taEes recedence.



23/29

c! a9rt 0.71 radians 0.707 8...!, or a9rt 40. 40. 416...! ! B1

! 2.43...! or 180 ! if is in de#rees .

6 -5

M1

/u&tract6

fro% or fro% !!... or su&tract 30 if is in de#rees M10.18 or 0.06!, 1. 1 or 0.61! Allo9 a9rt A1, A15he 1st A %arE is de endent on Uust the 2nd M %arE!B1) if the re uired alues of is not seen, this %arE can &e

#i en &? i% lication if a final ans9er roundin# to 0.18or 0.1 or a final ans9er roundin# to 1. 1 or 1. 0! isachie ed. Also see re%ature a ro@. note !.

/ ecial case)

1+.0sin6+.06

sinsin6+.06

sin ==+=

+ x x x

x = arcsin0.1 = 0.1 0 6... and x = 0.1 0 6 = 2. B0M1M0A0A0!5his s ecial case %arE is also a aila&le for de#rees... 180 8.62...!@tra solutions outside 0 to 2 ) K#nore.@tra solutions &et9een 0 and 2 ) Coses the final A %arE.$re%ature a ro@i%ation)e.#. = 41 , 180 41 =13 , 41 30 = 11 and 13 30 = 10'han#in# to radians) 0.1 and 1. 05his 9ould score B1 re uired alue of not seen, &ut there isa final ans9er 0.1 or 1. 0!!, M1M1A0A0.

[10]

34. a! 3 sin2 2 cos2 = 13 sin2 2 1 sin2 ! = 1 M1) >se of sin2 + cos2 = 1! M13 sin2 2 + 2 sin2 = 1 sin2 = 3 cso AG A1 2

.B) AG P need to see at least one line of 9orEin# after su&stitutin# cos2 .

&! sin2 = +3

, so sin = N!0.6 M1Atte% t to sol e &oth sin = +.. and sin = *

%a? &e i% lied &? later 9orE! M1 = 0.768 a9rt = 0.8 de endent on first M1 onl?! A1 = 180V 0.768c !P = 12 .23* a9rt 12 .2V M1P A1ft


24/29

irst M1) >sin# sin2 = 3 to find alue for sin or


25/29

35.

'

7 0 0 %

+ 0 0 %

1 + V

A

B

a! BC 2 = 7002 + 002 2 " 00 " 700 cos 1 M1A1= 638 1. 2*!

BC = 2 3 a9rt A1 3Kf use cos 1 V = *.., then A1 not scored until 9ritten asB' 2 = * correctl?S littin# into 2 trian#les BAR and CAR , w+ere M is oot o er

ro& B to AC indin# alue for BR andCR and usin# $?tha#oras M1

BC 2 = 00sin 1 !2 + 700 00 cos 1 !2 A1 BC = 2 3 a9rt A1

&! BC

B

scandidateX

1+sin

700

sin =M1sin B = sin 1 " 700 :2 3c = 0.716.. and #i in# ano#t)se B

134.2 ! de on 1st M M1 = 180V candidateWs an#le B Je . on first M onl?, B can &e acute! M1 = 180 134.2 = 0!4 .8 allo9 46 or a9rt 4 .7, 4 .8, 4 . ! A1 4


26/29

iii! >sin# sine rule or cos rule! to findC first)'orrect use of sine or cos rule for ' M1, indin# alue for ' M1ither B =180 1 V + candidateWsC ! or = 1 V + candidateWsC ! M1

i ! 700cos1 = 00 + BC cos M2 Yfirst t9o Ms earned in this caseZ

/ol in# for P = 4 .8 allo9 46 or .7, 4 .8, 4 . ! M1PA1 ote) /.'. Kn %ain sche%e, if used in lace of B,

third M #ained i%%ediatel?P-ther t9o %arEs liEel? to &e earned, too, for correctalue of stated.

[7]

36. a! x 6!2 + y 4!2 =P 32 B1P B1 2

Allo9 for 32.

&! 'o% lete %ethod for ! , =22 !46!612 + M1

= 40 or a9rt 6.32 A1,ese +i&st t%o !a& s can #e sco&ed i+ seen as a&t o+ so$)tion +o& c'

'o% lete %ethod for cos< , sin< or tan< M1

e.#. cos = 40scandidateX3

M$M5 =

= 0.4743! = 61.683 !


27/29

c! 'o% lete %ethod for area:! P e.#. = sin403

21

M1

=31

23

= 8.3 16..! allo9 a9rt 8.3 A1Area sector! !:Q = 0. " 32 " 1.0766 = 4.8446*! M1Area: Q = candidateWs 8.3 16.. 4.8446..! M1= 3. 07 a9rt A1< ote) 3. 1 is A0

irst M1) alternati e!403

21

/econd M1) allo9 e en if candidateWs alue of< used.Jes ite &ein# #i en\!

[11]

37. a! r< = 7 " 0.8 = .6 c%! M1A1 2M) >se ofr< 9ith < in radians!, or e ui alent could &e

9orEin# in de#rees 9ith a correct de#rees for%ula!.

&! 21

r 2< = 21

" 72 " 0.8 = 1 .6 c%2! M1A1 2

M) >se of21

r 2 < 9ith < in radians!, or e ui alent could &e

9orEin# in de#rees 9ith a correct de#rees for%ula!.a! and &!) 'orrect ans9ers 9ithout 9orEin# score &oth %arEs.

c! BD2 = 72 + their AD!2 2" 7 " their AD! " cos 0.8! M1 BD2 = 72 + 3 2 2 "7 " 3. " cos 0.8! or a9rt 46 for the an#le! A1 BD = .21!$eri%eter = their DC ! + .6 + .21 = 14.3 c%! Acce t a9rt! M1A1 4

1st M) >se of the correct! cosine rule for%ula to find BD2 or BDAn? other %ethods need to &e co% lete %ethods to find

BD2 or BD

2nd M) Addin# their DC to their arc BC and their BDBe9are) Kf 0.8 is used, &ut calculator is in de#ree %ode, this can

still earn M1 A1 for the re uired e@ ression!, &ut this#i es BD = 3. 0*so the eri%eter %a? a ear as 3. + .6 + 3. earnin# M1 A0!.



28/29

d! ] ABD = 21

" 7 " their AD! " sin 0.8or a9rt 46 for the an#le! ft their AD!M1A1ft= 8.78*!

Kf the correct for%ula21

a- sinC is uoted the use of an?

t9o of the sides of ] ABD as a and- scores the M %arE!.Area = 1 .6 8.78* = 10.8 c%2! Acce t a9rt! M1 A1 4

1st M) >se of the correct! area for%ula to find ] ABDAn? other %ethods need to &e co% lete %ethods to find] ABD

2nd M) /u&tractin# their ] ABD fro% their sector ABC

>sin# se#%ent for%ula21

r 2 < sin< ! scores no %arEs in art d!.

>nits c% or c%2! are not re uired in an? of the ans9ers.[12]

38. a! 4 ! 5his %arE can &e i% lied &? an ans9er 6 ! B1180 , Add 20 for at least one an#le! M1, M16 1 A1 4@tra solution s! in ran#e) Coses the A %arE.@tra solutions outside ran#e) K#nore 9hether correct or not!.'o%%on solutions)6 onl? correct solution! 9ill score B1 M0 M1 A0 2 %arEs!6 and 11 9ill score B1 M0 M1 A0 2 %arEs!44. or si%ilar! fora is B0, and 64. , 1 .01 or si%ilar! is A0.

&! 120 or 240 !) 5his %arE can &e i% lied &? an ans9er 40 or 80! B1

'ould &e achie ed &? 9orEin# 9ith 60, usin# 180 60 and:or 180 + 60!360 , 360 + or 120 + an an#le that has &een di ided &? 3! M1, M1Ji idin# &? 3 for at least one an#le! M140 80 160 200 280 320 irst A1) at least 3 correct A1A1 6

@tra solution s! in ran#e) Coses the final A %arE.@tra solutions outside ran#e) K#nore 9hether correct or not!.'o%%on solutions)40 onl? correct solution! 9ill score B1 M0 M0 M1 A0 A0 2 %arEs!40 and 80 onl? correct solutions! B1 M1 M0 M1 A0 A0 3 %arEs!40 and 320 onl? correct solutions! B1 M0 M0 M1 A0 A0 2 %arEs!



29/29

Ans9ers 9ithout 9orEin#)ull %arEs can &e #i en in &oth arts!, B and M %arEs &? i% lication.Ans9ers #i en in radians)Jeduct a %a@i%u% of 2 %arEs %isread! fro% B and A %arEs.Jeduct these at first and second occurrence.!

Ans9ers that &e#in 9ith state%ents such as sin x 20! = sin x sin 20 or

cos x = 61

, then #o on to find a alue of ^ W or ^ W, ho9e er &adl?, cancontinue to earn the first M %arE in either art, &ut 9ill score no further %arEs.

$ossi&le %isread) cos3 x = 21

, #i in# 20, 100, 140, 220, 260, 340'ould score u to 4 %arEs B0 M1 M1 M1 A0 A1 for the a&o e ans9ers.

[10]

Date post:	04-Jun-2018
Category:	Documents
Upload:	ahadh12345
View:	222 times
Download:	0 times

Core 2 - Ch 6, 8, 10 - 1 - Trigonometry - Solutions

Documents