1. Estimation, Approximation and Errors Percentages Polynomials and Formulas Identities and Factorization 52

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1 ontents ommonly Used Formulas. Estmaton, pproxmaton and Errors. Percentages. Polynomals and Formulas 8. Identtes and Factorzaton. Equatons and Inequaltes Rate and Rato 8 7. Laws of Integral Indces and Surds 9 8. asc Geometry Symmetry and Transformaton 0. Trgonometry 6. Mensuraton 60. oordnate Geometry 80. Probablty and Statstcal Dagrams 96. Measures of entral Tendency 0 Revson Test

2 0 Trgonometry Smart Revew. Trgonometrc Ratos. In the fgure, s a rght-angled trangle, where a s the opposte sde of, b s the adjacent sde of, c s the hypotenuse of. The trgonometrc ratos of sne, cosne and tangent are defned as follows. Opposte sde of (a) sn = Hypotenuse of = a c djacent sde of (b) cos = Hypotenuse of = b c Opposte sde of (c) tan = djacent sde of = a b For example: z () In the fgure, fnd the values of sn, cos and tan. sn =, cos =, tan =. () In the fgure, fnd the values of sn z, cos z and tan z. sn z =, cos z =, tan z =. Hypotenuse c djacent sde b Opposte sde a Let s Try. In the fgure, fnd, correct to decmal place. 7. In the fgure, fnd the value of cos. 8. In the fgure, fnd the length of, correct to sgnfcant fgures. cm 7. Trgonometrc ratos of specal angles 6 Referrng to the two rght-angled trangles on the rght. Trgonometrc rato sn cos tan 0 60 c m c m c m 0

3 0 Trgonometry + =? Worked Examples onventonal Questons Secton (). In the fgure, fnd the bearng of from. (Gve the answer correct to sgnfcant fgures f necessary.) ( marks) N 70 m 0 m E! Mnd the Trap Reference: HKEE 0 I Q anddates should let the requred angle be before dong the calculaton. Otherwse, mark wll be deducted for undefned symbol. Soluton Let =. 0 cos = 70. (cor. to sg. fg.) N 70 m 0 m E [M] []! Solvng Strategy anddates should only round off the numercal answer to the requred sgnfcant fgures n the fnal step. The queston requres to fnd the bearng of from, thus mark a cross at. Mnd the Trap a =.9 ` The bearng of from s N.9 E. [] Snce no specfc bearng s mentoned n the queston, canddates can gve 0.9 or N.9 E as the answer. Instant Drll In the fgure, fnd the bearng of from. (Gve the answer correct to sgnfcant fgures f necessary.) ( marks) 00 m N 60 m E 9

4 HKDSE Exam Seres Integrated Exam Revson Exercses for Mathematcs (Junor Secondary Topcs) (Upgraded Edton) Instant Drll In the fgure, Katy measured the angle of elevaton of the top of a buldng from the top of another buldng as. She also measured the angle of depresson of the bottom of the same buldng as 6. The heght of the measured buldng s 90 m. Fnd the dstance between the two buldngs. (Gve the answer correct to sgnfcant fgures.) ( marks) 6 90 m Multple-choce Questons Secton 6. tan^90c - h = cos. sn. cos. sn cos Soluton tan^90c - h cos = tancos cos = # sn cos = sn M Shortcut Reference: HKEE 08 II Q lternatve Method (for multple-choce questons only): Set any value for and put t nto each expresson and then compare the results. For example, let = 0, then the value of the gven expresson s.9. Opton : Opton : Opton :.9. Opton D: The answer s. Instant Drll sn^90c - h = tan^90c - h. sn. cos. sn cos

5 0 Trgonometry Mock Questons (In the followng questons, unless otherwse specfed, gve the answer correct to sgnfcant fgures f necessary.) onventonal Questons Secton () cos. Smplfy. ( marks) tan ^90c - h. Smplfy sn cos +. ( marks) cos sn. Smplfy sn^90c- h cos0c- cos. ( marks). Smplfy cos^90c- htan^90c- h. ( marks) cos. Smplfy cos^90c - h. ( marks) tan 6. Smplfy + sn -cos^90c - h. ( marks) 7. Smplfy - sn. ( marks) ^90c - h 8. Smplfy sn 0c sn 0c - - sn^90c- h + sn^90c- h. ( marks) tan^90c - h 9. Prove that /. ( marks) sn ^90c - h sn 0. Prove that sn + cos ^90c- htan ^90c - h /. ( marks) - sn. Prove that cos + sn / cos- sn. ( marks) cos c. Wthout usng a calculator, fnd the value of sn 0 tan. ( marks) c- 60c. Wthout usng a calculator, fnd the value of tanc- sn c. ( marks) tan 0c

6 HKDSE Exam Seres Integrated Exam Revson Exercses for Mathematcs (Junor Secondary Topcs) (Upgraded Edton) Multple-choce Questons Secton 8. In the fgure, cos+ tan =. a a c + b.. a b + c a.. b b + a a. b a + c b. 9. In the fgure, tan = c b Reference: HKEE 08 II Q a. sn 0c cos 60c - cos^90c - h cos^90c h. cos. cos tan. tan cos costan Reference: HKDSE II Q9. If and are both acute angles and + = 90, then = cos sn sn.. cos. cos Reference: HKEE 09 II Q 0. In the fgure, sn x = x 7 D Reference: HKEE 06 II Q. tan^90c - h = cos. sn. cos. sn cos Reference: HKEE 08 II Q. In the fgure, sn x =. If 0 < x < 90, whch of the followng must be correct? x 6 Reference: HKEE 0 II Q I. sn x + sn (90 x) > 0 II. cos x cos (90 x) > III. tan x tan (90 x) =. I only. II only. I and III only II and III only Reference: HKDSE II Q 8

7 HKDSE Exam Seres Integrated Exam Revson Exercses for Mathematcs (Junor Secondary Topcs) (Upgraded Edton) Revson Test Marks: /00 Date: onventonal Questons - ^xy h. Smplfy - and express the answer wth postve ndces. ( marks) xy. Make c as the subject of the formula + d - c = d. ( marks). Factorze (a) x - xy + 9y, (b) x - xy + 9y - x + y. ( marks). The cost of a watch s $ 00. If the watch s sold at a dscount of 0% of ts marked prce, the proft percentage s 0%. Fnd the marked prce. ( marks). The rato of the costs of a bottle of orange juce to a bottle of mlk s :. If the total cost of bottles of orange juce and 6 bottles of mlk s $76, fnd the cost of a bottle of mlk. ( marks) 6. In a polar coordnate system, the polar coordnates of ponts, and are (8, ), (7, ) and (6, 0 ) respectvely. (a) Let O be the pole. re, O and collnear? Explan your answer. (b) Fnd the area of. ( marks) 7. (a) Solve x - x + 9. (b) Fnd the number of negatve ntegers that satsfy the nequalty n (a) and wrte down the smallest one. ( marks) 8. In the fgure, E s a pont on D such that E = E Gven that // D, +D = 8 and +E = 0. Fnd x, y and z. 8 ( marks) F z x 0 E y D

8 Identtes and Factorzaton Identtes and Factorzaton Let s Try (p.). L.H.S. = (x ) x = x 6 R.H.S. = (6x ) = x 0 a L.H.S. R.H.S. ` (x ) x = (6x ) s not an dentty.. L.H.S. = (x + )(x + ) = x + x + x + = x + x + = R.H.S. ` (x + )(x + ) = x + x + s an dentty.. (6 7y)(6 + 7y) = 6 (7y) = 6 9y. (m )(m + 0m + 6) = (m )[(m) + (m)() + ] = (m) = m 6 Let s Try (p.). ab b = b(ab ). 6x y z + 8xy z 9y z = y z(x y + 6xz y) Let s Try (p.). omparng the lke terms and the constant terms on the two sdes, we have a = and b = 7. Gudelnes The common factor of 6, 8 and 9 s, whle the common factor of x y z xy z and y z s y z.. omparng the lke terms and the constant terms on the two sdes, we have t = 9 and s + t = 6,.e., t = and s = 9. Let s Try (p.). p + q + mp + mq = (p + q) + m(p + q) = (p + q)( + m) Gudelnes s the calculaton nvolved s easer, frst fnd the value of t. Then use the result to fnd s.. h jk hj + hk = h + hk jk hj = h(h + k) j(k + h) = (h + k)(h j) Let s Try (p.). ( + h) = + ()h + h = h + 6h + 9 Gudelnes Solutons are usually expressed n descendng order of the varable. Let s Try (p.). a a + a +a = ( + )a = a ` a a = (a )(a + )

9 HKDSE Exam Seres Integrated Exam Revson Exercses for Mathematcs (Junor Secondary Topcs) (Upgraded Edton) Soluton Gude. x x + x +x = ( + )x = x ` 6x x = (x )(x + ) oncept ulder (p.). False If an equaton holds for any values of the unknowns, then the equaton s an dentty.. m +n m +n +6mn +0mn = (6 + 0)mn = 6mn ` m + 6mn + n = (m + n)(m + n) Let s Try (p.). a + 8a + 6 = a + ()a + = (a + ). u 0uv + v = u (v)u + (v) = (u v). 9m 9n = (m) (7n) = (m + 7n)(m 7n). x 8 = x = (x )[x + x() + ] = (x )(x + x + ) ommon Mstakes Some canddates may confuse the dentty of sum of two cubes wth that of dfference of two cubes. In each of these two denttes, there s only one negatve sgn. In the dentty of dfference of two cubes, the frst +/ sgn s negatve; whle n the dentty of sum of two cubes, the second +/ sgn s negatve.. False L.H.S. = 7(x ) + 8 = x + R.H.S. = 8x + 6(x ) + = x 7 a L.H.S. R.H.S. ` 7(x ) + 8 = 8x + 6(x ) + s not an dentty.. False In any dentty, besdes the constant terms, the lke terms on the two sdes are equal.. False (7x + 0) = (7x) + (7x)(0) + 0. True 6. True x = ( x) = (x ) = 9x + 0x False x x + x +x = ( + )x = 7x Ths method can be used to factorze x 7x 0 only.. 7k + = (k) + = (k + )[(k) (k)() + ] = (k + )(9k k + )