00_(3)_2016_-_2017_H2_Maths_Trigonometry_Practice_Questions_Solutions_(student)

National Junior College Mathematics Department 2016 Revision: Trigonometry Page 1 of 5 National Junior College 2016 – 2017 H2 Mathematics Revision: Trigonometry Tutorial Solutions Basic Mastery Questions 1. Consider the following right -angled triangle Noting that ,2  we have (a) sin cos ,a c (b) cos sin ,b c (c)   11tan cot . tanb a a b     Remark: Even though we have only justif ied the above results for an acute angle  , these relationships in fact hold for all values of . 2. We use the ASTC diagram. (a) sin(120 ) is positive as it lies in the second quadrant. Therefore n = 1. (b) The smallest positive integer n is 2, since nd rd 1: tan(120 ) 0 (2 quad) 2 : tan(240 ) 0 (3 quad) n n       (c) The smallest positive integer n is 3. This is because the cosine function gives negative values in the 2nd and 3rd quadrants, and cos(360 ) 1. Remark: If the smallest positive integer is n = k, you must also explain why n cannot be 1, 2, 3, … , k – 1. 3. Since the principal value of 1tan 1 is ,4    1cos tan 1 cos 4    2 .2 4. (a) By considering the principal range for 1tan , x and the graph of the tangent (or inverse tangent) function, 0. 2  (b)(i) 11cot . tan x  (b)(ii) From the above right-angled triangle, 21sec 1. cos x    Note: Do not give your solution as   1sec tan . x Alternatively, use the identity 22tan 1 sec and the fact that sec x is positive when  is an acute angle. (b)(iii) By the same right-angled triangle, 211cosec . sin x x   Alternatively, use the identity coseccot . sec   (Analogy: sintan cos   ) b a c   x 1 2 1x  

National Junior College Mathematics Department 2016 Revision: Trigonometry Page 2 of 5 5. tan tantan( ) 1 tan tan ABAB AB     11 23 11 231 1.    2 2 tantan(2 ) 1 tan AA A    1 2 21 2 2 1 4 .3    6. R =   221 3 2. 1 3tan . 13     Therefore, sin 3 cos 2sin . 3x x x     Presentation: For this question, do not merely give the value of R and , because this is not precisely what the question is asking for. Your solution should be an expression in the form sin ( ).Rx  Remark: R-formulae are related to the addition formulae. Observe that             2222 22 31 1 3 1 3 sin 3 cos 1 3 sin cos sin cos cos sin sin ( ). xx xx R x x Rx           7. Basic angle = 1 1cos . 23    1cos 2x 5or .33x  8. Basic angle = 1 1tan . 63    1tan 3 x 5 11or .66x  9. Since 0 2 ,x  we need to consider the interval 2 4 .4 4 4 x       Therefore, sin 2 0. 4x   Basic angle = 0 2 , 2 , 3 or 44 3 7 11 15, , or .8 8 8 8 x x             