HI,

I have completed a practice assessment task and I was wondering if you could go through my answers and see if there are any errors please I also have had some trouble with question 15 if you can explain it, that would be amazing. thank you so much

De Moivre Theorem Throughout this SAC, you will be using the following equations: cis(0) = cos(0) + isin(0) ... ...... ...... (1) cis(0) x cis() = cis(0 + p) .......... (2) (cis(@))" = cis(ne ) . . ..... . ... ... .. (3) where i = V-1 and n is a positive integer and all angls are in radians. In addition, you will need to use the following binomial expansions: (A + B) 3 = A3 + 3A2B + 3AB2 + B3 (A + B) 4 = A4 + 4A3B +6A2B2 + 4AB3 + B4 Question 1 Use equations (1) and (2) to prove the following addition formulas: sin(0 + 4) = sin(0) cos(4) + cos(0)sin() cos(0 + 4) = cos(0) cos(4) - sin(0) sin() (5 marks) CIs ( 0 ) = cos ( 0) + isin (e) / [1] CIs ( 0 + 0 ) = cis ( 8 ) XCIs ( $ ) [2] From 1 cis ( 6 + (D ) = cos ( 6 + Q ) +isin (6+ $ ) # 1 From ( 2) cis (@) x cis (p ) = cos ( 6 + d ) + isin(6 + d ) From 1 = D ( cis ( 0) + ism ( 61) x (cos ($ ) + isin ($1 ) = cas (6+ $ )+ isin (@+$) From [1] ED Cos ( 8 ) cos ( d ) ticos () sin (d ) + isin (@ cos ( $] + is sin (6) cos (Q) = cas (8+$)+ isin (@+b) = D cos ( B)cas ($)+if cas (onsin (@) + sin (0) cos ( $)/ -sin (/$) cos (6) = cos(6+ $)+ isin (6+d) : [coscojisin ( )- sin ( $1 ccs (8) ] + c [ cos ( 8)isIn ($) + sin ( 6 \\ ccs (4)] = cos ( 6+ $ ) + ism(07 $) On comparing real and imaginary part cos ( 6 + $ 1 = cos ( B ) cos ( d ) - since ) sin(Q ) sin ( A + 8 ) = sin Colces ( @) + cos (G) sin () provedlon 2 Write the expressions for a) sin(0 - 4) Sir (B) cos(d) -cos(@ )sin (cd) (1 mark) b) cos(0 - 4) (1 mark) cos ( 0 ) cos ( $ ) +sin ( @) sin ( cp ) Question 3 Use the results of Question 2 with suitable values of 0 & 4 to find the exact values of the following. Show your working and give answers with rational denominators. a) sin (7 sin ()= sin ( 150 ) ( 12 ) = sin is0 sin ( 150 ) = sin ( 45 - 30 ) (3 mar . 1 since - D ) = sin ( 0 ) co's () -cosco ) sin ( 4 ) 0 = 450 , 9 = 300 sin ( 150) = sin ( 45 - 30 ) = sin 450 cos, 300 - cos 4so sin 30 1 3 _ 1 13 2 12 2 213 2N2 = = N3 - 1 sin ($2 ) 2 N6 - NZ 2 1 LI b) cos (7 4 COs ( 72 ) = COS 150 = Cos ( 480 -300 ) (3 m : cos ( 6 - ($ ) = cos( @ ) cos ( $ ) + sin ( 8) sin ($ ) Plence , 8=450 , D = 300 cos ( 48 0 - 30 0 ) = cos 4So cas 30 0 + sin 4 So sin 300 D cos (72 ) : NE +NE 4 N 3 + 1 c) tan (7 tan ( 12 ) = tan ( Isa ) = _ sin ( 1so ] N 3 - ) ceo ( 150 ) 13 + 1 = 2 - N3 ban ($ 2 ) = 2 - 13Question 4 Use the results of Question 1 with suitable values of 0 & 4 to find the exact values of the following Show your working and give answers with rational denominators. a) sin = sin (750) = >in (450+300 ) (3 marks) . 1 0 = 450, 0= 306 -7 ) = SIn ( 45 + 30 0 ) = sin LIS cos3C + cas us sin36 iNE 2NZ 4 sin ( is ) = NG ANE 4 b) cos (7) = COs 780 = (8 (45+ 30) (3 marks) ces (o ta) = cos @ cosas - sinG sin cos ( SP ) = COS ( US + 3 0 - ) = COS 48 CO= 30 - SIn US SIn 30 -LX N3 8 - 1 213 c) tan 12 , 2 sin (12 (3 marks cas (2 121/2 1 3 - 1 N 3 - 1 1 3 + 1 2 N 2 tan ( 72 ) 2 2 -13Question 5 Use equations (1) and (3) with n = 2 to prove the following double-angle formulas: (1] CIs ( 8 ) = cos ( 6 ) + isin(B ) sin(20) = 2sin(0) cos (0) cos(20) = cos2 (0) - sin2(0) [= ] ( CIs ( 8 ) ) ~ x cis ( me ) (4 marks) 2 D ( cos ( 6 ) + isin ( 6 1 ) " = ca( he ) +isin ( ne ) For n= 2 , ( cos ( 0 ) + isin ( 6 ) ) 2 = ccs ( 26 ) + i sin ( 26 ) COs?G + 2 isIn ( 6 ) coo CGI+ (isin ( 6) ) ? = cas ( 26 ) + isin ( 20 ) cos 20 + 2 isin ( 6 ) cas ( 6 1 - sing ( 6 ) = cc> ( 261+ Csin ( 20] ( cos Isc + + 12- N3 ( further simplify (3 marks) 17 cos (28 ) CO> D= AN 2. cos is0 2+ 1 + ccs ( 300 ) = + 1 + Z 2 + N 3 cas ( 32 ) = 1 2 +13 LI ( further simplify ) 2 ton liso = sin ( ISO ) c) tan ces ( LSC ) (3 marks 1 2 - NZ - IN 2 +N 3 tan (1 = N2 - V3 1 2 + 13 7Question 9 Compare the results of Questions 3 & 8. (2 mark a) Show that the two expressions for sin () are equal. . From 98, sir ( ; ) = 12- V3 2 From 43, sin (iz ) = sin ( 9 - = ) = N 2 - N3 1 1 - N 3 2 1 = sin ( 3 Ycas ( Fil ) - cas ( 8 ) om (? ) 2 2 21 3 x 12 - 4 x VZ NG _ N3 / 16 - 1 2 = 12 ( J ( 13 - 1 \\ x ) 2 4 41 as ( 1) = ( 2 ) (2 ) 16 - NZ Both are equal (2 marks) b) Show that the two expressions for cos ( ) are equal. From N6 + 1 3 cos ( 12 ) =1 From cos iz 2 4 N 2 + 1 3 2 X N ( 2 ) 3 + 2 ( NZ ) + ( 2 ) 2 2 = N 2 x N 2 = 1 2 x 2 2 1 ) = 1 2 x ( N 3 + 1 ) 2 (2 ) = - 12 (13 + 1 ) N6 +N2 (2 ) 4 as $ 2 ) are same Two expression are equal 8Question 10 Use equations (1) and (3) with n = 3 to prove the following formulas: cos(30) = cos3 (0) - 3sin2(0) cos(0) sin(30) = 3sin(0) cos2(0) - sin3 (0) (6 marks) cos( B ) = cos( @)tisince) (cisco)in = cis (no) D ( cosco ) + isin (871 = cos (30 Hisin ( 36) n = 3 Real part ; (CIS ( 6 ) ] = 418 ( 36 ) Cos ( 36 )= Real ( cos ( 6 ) +isincon?" CIS ( 30 ) = Cos ( 36 ) + isin ( 3@ ) = Real ( costs + issimo + 3coc20 ( isin 8) + 3 (i sin 20 \\ costs* 1 = - 1 so , ( CIS ( 6 1 ) 3 = ( Gos ( 0 ) + i since) ) . Real ( coso - (isin 3 6 + 3 () cos 2 6 cos 0 - 3 sin- @) cos 6 From Formula C $ ( Real Real ( cos ( e ) + is in (on)" = cos ( ng )+ isin (no ) Imaginary ( cos contisin ( 8) ) = = cos ( 30 ) + isin ( 36 ) . . cos ( 30 ) = cos * cost 3 sin (6 )ccg ( 0 ) )+ ( Formula) Formula 2 : [ . (a + 6 ) = = 0 3 + 63 + 3a - 6 + 3ab2] sin ( MB ) = Imag ( cosco ) + (since) ) ? sin ( 86 ) = 1mag ( coso + ising ) 3 Drainsat 3sino ( cos 2 6 ) = cosse + 1 3 sin 36 + 3 ( cos