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              10. (a) By sine rule, sin sin 16 2sin sin 30 20 5 23.6 or 156.4 . (rejected) ABC BAC AC BC ABC ABC           The second solution is rejected because sum of all angles in a triangle must be 180 . (b) By sine rule, sin sin sin 53.110 sin 30 16.0 cm. DF EF DEF EDF DF      (c) By cosine rule, 2 2 2 2( )( )cosHI GH GI GH GI HGI    2 2 212 10 2(12)(10)cos30 36.154 36.154... (lengths are positive) 6.01 cm. HI HI         (d) By cosine rule, 2 2 2 2( )( )cosJK JL KL JL KL JLK    2 2 2 2 2 2 cos 2( )( ) 9 8 12 1 2(9)(8) 144 89.6 . JL KL JKJLK JL KL JLK       

National Junior College Mathematics Department 2016 Revision: Trigonometry Page 3 of 5 Intermediate Level Questions 1. Since 22sin cos 1 for all , 2 2 34sin 1 cos cos 55 12 5cos 1 sin sin 13 13 A A A B B B           Using the identity cos( ) cos cos sin sin ,A B A B A B   cos A sin B  cos AB 4 5 5 13 33 65 4 5 5 13 63 65 4 5 5 13 33 65 4 5 5 13 63 65 Extension Question: Identify which quadrant does the angle A + B lies in for each of the possible values of  cos AB . 2. (a) sinLHS 1 cos 2 A A    2 2 sin 1 1 2sin sin RHS.2sin A A A A    (b) LHS 2 2cos x 2 2 2 2 2cos 1 2 4cos RHS.2 x x      (c) 44LHS cos sin        2222 2 2 2 2 cos sin cos sin cos sin (1)(cos 2 ) RHS.             2. (d) RHS tan 4x   4 4 sin cos sin cos sin cos cos cos sin cos tan tan 1 tan tan 1 1 LHS. x x x x xx x xx x x x           (e) sin 5 sinLHS cos5 cos     2sin 3 cos 2 2sin 3 sin 2 RHS.     (f) LHS sin 3 cos 344xx              1 sin 6 sin22 RHS. x      3. Method 1: Double angle Formula sin 2 sin , where 0 2x x x       2sin cos sin sin 2cos 1 0 1sin 0 or cos 2 50, , 2 , , or .33 x x x xx xx x          Method 2: Factor Formula sin 2 sin 0,where 0 2x x x     32cos sin 022 3cos 0 or sin 022 3 3 5 , , or 0,2 2 2 2 2 5, , , 0, or 2 .33 xx xx xx x                                

National Junior College Mathematics Department 2016 Revision: Trigonometry Page 4 of 5 4. This is a quadratic equation in terms of sin 2 .x 2sin 2 sin 2 2, where 0 2x x x        2sin 2 sin 2 2 0 sin 2 2 sin 2 1 0 sin 2 1 or sin 2 2 (rejected, 1 sin 2 1). Since basic angle for sin 2 1 is , 2 372 or (as 0 2 4 )22 37 or .44 xx xx x x x x xx x                         5. 180 (72 55 ) 53 .XYZ       By Sine rule, 13.2 sin 53 sin 72 sin 55 XY YZ   XY 15.7 cm (to 3 s.f.), and YZ 13.5 cm (to 3 s.f.).   6. By EITHER using cosine rule three times to find the smallest angle among all three angles, OR observing* that the smallest angle of the triangle must be opposite the side of the triangle with the smallest length, 2 2 26 8 11 2(8)(11)cos 32.2 .         * This observation is proved in Q8. 7. (a) LHS 2cos cos3 cos5              2 2 2cos cos3 cos5 2cos 2cos( ) cos 4 2cos 1 cos 4 2 cos 2sin 2 4cos 2sin cos RHS.                      Alternatively,   2 2 (cos cos3 ) (cos cos5 ) 2sin2 sin( ) 2sin3 sin( 2 ) 2(sin2 )(sin sin3 ) 2(sin2 ) 2sin2 cos( ) 4sin 2 cos 4(2sin cos ) cos RHS.                                (b) 2 1f( ) cos sin 2 12       1 cos 2 1 sin 2 122 11cos 2 sin 222 2 1 1 1(cos 2 ) (sin 2 )22 22 21cos 2 (R-formula).2 4 2                   (c) Since   ve valued 2 1 2 1 f( ) ,22       we have     22 2 2 2 1 2 10 f( ) max , 22 3 2 20 f( ) . 4                      Remark: In the above solution, we define max{a, b} to be the maximum of the two numbers a and b. Thus if a > b, then max{a, b} = a. If a < b, then max{a, b} = b. Illustration: max{–3, –4} = –3.

National Junior College Mathematics Department 2016 Revision: Trigonometry Page 5 of 5 8. (i) From sine rule constant,sin sin sin a b c A B C   sin sin sin .a b c A B C     (ii) Given that 0 , , 90A B C   , we use the fact that the graph y = sin x is increasing in this interval to conclude that A > B > C. (iii) Since 180 ( )A B C   and B + C is acute with B + C > B (> C respectively),  sin sin 180 ( ) sin( ) sin ( sin resp.). A B C BC BC      Observe that A > B + C > B. In addition from (ii), B > C. Therefore A > B > C. 9. 31 18 13 .ATB     Using sine rule on triangle ATB , 20 sin13 sin18 20sin18 .sin13 TB TB    Using sine ru