MSE2183 Assessment 4 2026-
DUE 26 AUGUST 2026
QUESTION 1: FORUM 5
!:] Fo,;,m 5 (Discussion• Lessons 7 and 8)
Forums
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Rr. Fanni 5: leuon 7 & I Dbcuuion Forwn
by,- Monday, 17 August 2026. 3:52 AM
I agree wlttl your pcMOI that inlentand1ng the concepts Is better IMO onty memoriseng fomua5. I af.o bked )'OW� of how tngonometry an be
used to md the hegh1 or a UN or butkting VOAJ/ ul)laNuon of c:arcle geometry was cleat. e:sc,ecialy the differtnt thootems about angles. chotdt and
t.inoema. I al:,o �ee lha1 these 1oi,.cs art useful., rHI life and help us� QOOd �no and k,glc.al thlnlung w,i..
QUESTION 2: LESSON 7 - TRIGONOMETRY
2.1 Self-Assessment 7.1
Question 2a: cos 150° (3 marks)
cos 150 ° = cos(180° - 30 °) = -cos 30 ° = -✓ 3/2
(UNISA, 2026, Lesson 7, p. 186) - shows reduction formula for cos(180° - 0)
(Serra, 2008, p. 185) - quadrant values and reduction
, Question 2b: sin(-45°)
sin(-45 °) = -sin 45 ° = -✓2/2
• (UNISA, 2026, Lesson 7, p. 187) - negative angle reduction formula
• (Serra, 2008, p. 184) - special angle table
Question 2c: tan 480°
tan 480 ° = tan(480 ° - 360 ° ) = tan 120 ° = tan(180 ° - 60 ° ) = -tan 60 ° = -✓3
• (UNISA, 2026, Lesson 7, p. 187) - quadrant Ill and IV reduction formulae
Question 2d: [sin(180 ° + x) x cos 330° x tan 150°] / sin x
sin(180 ° + x) = -sin x
cos 330 ° cos(360 ° - 30 ° ) cos 30 ° v3/2tan 150 °
tan(180 ° 30 ° ) = -tan 30 ° -1/✓3
[(-sin x) x (✓3/2) x (-1/✓3)]
/ sin x= [(-sin x) x (-1/2)]
I sin x
= (sin x/2) / sin x = 1/2
DUE 26 AUGUST 2026
QUESTION 1: FORUM 5
!:] Fo,;,m 5 (Discussion• Lessons 7 and 8)
Forums
"4of'CMY17�202t.l51AM
Aft.r...oirlQL-7.-.:ll,1 ....nedlhaltl�laaboul�.-.:1-'"of�W.-eo,oc,ecui,..,oct,•�--la,oenllOaolw
�foeo..,...,_CM_�liofrodlN'-Ol•ota11e,eWl1holll�lnQ'OWtrff6,'tCdy'
I..OINrnedflbo.Mac.i.OI0"'4'UY lll'ottpe111�1N�ol•cnie.llUChNNc«otN.1idiw.�chotdlt.cl� W.NO-carct.
""--10findmlN1ng.,,gi..-.:1 .......
Rr. Fanni 5: leuon 7 & I Dbcuuion Forwn
by,- Monday, 17 August 2026. 3:52 AM
I agree wlttl your pcMOI that inlentand1ng the concepts Is better IMO onty memoriseng fomua5. I af.o bked )'OW� of how tngonometry an be
used to md the hegh1 or a UN or butkting VOAJ/ ul)laNuon of c:arcle geometry was cleat. e:sc,ecialy the differtnt thootems about angles. chotdt and
t.inoema. I al:,o �ee lha1 these 1oi,.cs art useful., rHI life and help us� QOOd �no and k,glc.al thlnlung w,i..
QUESTION 2: LESSON 7 - TRIGONOMETRY
2.1 Self-Assessment 7.1
Question 2a: cos 150° (3 marks)
cos 150 ° = cos(180° - 30 °) = -cos 30 ° = -✓ 3/2
(UNISA, 2026, Lesson 7, p. 186) - shows reduction formula for cos(180° - 0)
(Serra, 2008, p. 185) - quadrant values and reduction
, Question 2b: sin(-45°)
sin(-45 °) = -sin 45 ° = -✓2/2
• (UNISA, 2026, Lesson 7, p. 187) - negative angle reduction formula
• (Serra, 2008, p. 184) - special angle table
Question 2c: tan 480°
tan 480 ° = tan(480 ° - 360 ° ) = tan 120 ° = tan(180 ° - 60 ° ) = -tan 60 ° = -✓3
• (UNISA, 2026, Lesson 7, p. 187) - quadrant Ill and IV reduction formulae
Question 2d: [sin(180 ° + x) x cos 330° x tan 150°] / sin x
sin(180 ° + x) = -sin x
cos 330 ° cos(360 ° - 30 ° ) cos 30 ° v3/2tan 150 °
tan(180 ° 30 ° ) = -tan 30 ° -1/✓3
[(-sin x) x (✓3/2) x (-1/✓3)]
/ sin x= [(-sin x) x (-1/2)]
I sin x
= (sin x/2) / sin x = 1/2
