Example 1. Solution: Chapter 6: Trigonometry 3 Example 2. Solution: - PDF document

SET 3 Chapter 6 Trigonometry لاتاـثلـثم Chapter 6: Trigonometry 1

6.1 Angles, Rotations and Degrees Measures تاـجردلاب سايقلا و نارودلا و اـياوزلا Chapter 6 : Trigonometry 2

Example 1. Solution: Chapter 6: Trigonometry 3

Example 2. Solution: Chapter 6 : Trigonometry 4

6. 2 Radian Measures يرـطق فصنلا سايقلا Chapter 6: Trigonometry 5

Example 3 . Solution: Chapter 6 : Trigonometry 6

Example 4. Solution: Chapter 6: Trigonometry 7

Example 5. Solution: Example 6. Solution: 6.3 Arc Length and Central Angles ةـيزكرملا ةـيوازـلا و سوقلا لوط Chapter 6 : Trigonometry 8

Example 7. Solution : Example 8. Solution: Chapter 6: Trigonometry 9

6. 4 The Trigonometric Ratios ةـيـثلـثملا بـسنلا Chapter 6 : Trigonometry 10

Example 9. Solution: Evaluating Trigonometric Ratios of Any Angles Chapter 6: Trigonometry 11

Scientific calculators are usually used to evaluate trigonometric ratios of any angles. The following examples were solved using a scientific calculator.     Example 1 0 . Evaluate sin 37 1 5 4 4 rounded to 4 decimal places.    15 44          Solution: sin 37 1 5 4 4 sin 37   60 3600   sin 37 . 26 2  0 . 6055 Note that scientific calculators allow entering angles in the form of DMS, and therefore using this feature is easier than dividing the minutes part of the angle by 60 and the seconds part by 3600 and then adding the tow results to the degrees part. Example 11. Evaluate the following rounded to 3 decimal places:             (a) sin 124 3 9 5 0 (b) cos 87 2 5 4 3 (c) tan 61 3 2 5 4      Solution: (a) sin 124 3 9 5 0 0 . 823      (b) cos 87 2 5 4 3 0 . 045      (c) tan 61 3 2 5 4 1 . 845 Example 12. Evaluate the following rounded to 3 decimal places:             (a) sec 48 1 1 3 5 (b) csc 76 5 2 8 (c) cot 9 1 0 1 3 Solution: Recalling that: 1 1 1       , sec csc and cot    cos sin tan 1       Then, (a) sec 48 1 1 3 5 1 . 500     cos 48 1 1 3 5 1       (b) csc 76 5 2 8 1 . 027     sin 76 5 2 8 1       (c) cot 9 1 0 1 3 6 . 195     tan 9 1 0 1 3 Example 13. Evaluate the following correct to 4 decimal places: sin  (b) cos  (c) tan  3 9 5 (a) 7 16 11  3  Solution: (a) sin 0 . 9749 7  9   (b) cos 0 . 1951 16  5  (c) tan 6 . 9551 11 Chapter 6 : Trigonometry 12

Example 14. Evaluate the following correct to 3 decimal places: sec  (b) csc  (c)  4 5 12 (a) cot 9 12 21  4 1   Solution: (a) sec 5 . 759  4 9 cos 9  5 1   (b) csc 1 . 035  5 12 sin 12  12 1    (c) cot 0 . 228  12 21 tan 21 Example 15. Determine the following acute angles in degrees and radians:    sin 1 cos 1 tan 1 (a) 0 . 354 (b) 0 . 548 (c) 2 . 537  sin 1 = 20.732 0 or 20 0 43’56” Solution: (a) 0 . 354  sin 1 0 . 354 = 0.362 radians  cos 1 = 56.77 0 or 56 0 46’12” (b) 0 . 548  cos 1 0 . 548 = 0.991 radians  tan 1 = 68.487 0 or 68 0 29’14” (c) 2 . 537  tan 1 2 . 537 = 1.195 radians Example 16. Determine the following acute angles in degrees:    sec 1 csc 1 cot 1 11 . 238 3 . 284 0 . 029 (a) (b) (c)     1  1 1   = 84.895 0 or 84 0 53’41” Solution: (a) sec 11 . 238 cos   11 . 238     1 1  1   = 17.728 0 or 17 0 43’43” (b) csc 3 . 284 sin   3 . 284     1 1  1   = 88.339 0 or 88 0 20’20” (c) cot 0 . 029 tan   0 . 029 Example 17. Evaluate the following expression , correct to 4 significant figures:      4 sec 32 1 0 2 cot 15 1 9     3 csc 63 8 tan 14 5 7       4 ( 1 . 1813 ) 2 ( 3 . 6512 ) 4 sec 32 1 0 2 cot 15 1 9  Solution:     3 csc 63 8 tan 14 5 7 3 ( 1 . 1210 )( 0 . 2670 )   4 . 7252 7 . 3024 2 . 5772   0 . 8979 0 . 8979   2 . 870 Chapter 6: Trigonometry 13

Example 18. Evaluate rounded to 4 decimal places:      (a) (b) csc( 95 4 7 ) sec( 115 ) Solution: Positive angles are considered to be counterclockwise and negative angles as clockwise. Hence  115 o is actually the same as 245 o (i.e. 360 o  115 o )     (a) sec( 115 ) sec 245 1    2 . 3662  cos 245    (b) csc( 95 4 7 ) 1    1 . 0051    sin( 95 4 7 ) 6.5 Trigonometric Identities ةـيـثلـثملا تاـقباـطتـملا Trigonometric Identities A trigonometric identity is a relationship that is true for all values of the unknown variable. The following identities are the fundamental trigonometric identities that are used to prove more complicated trigonometric identities:   sin cos     tan , cot ,   cos sin 1 1 1       , , , sec csc cot    tan cos sin 2   2    2   2   2   2  sin cos 1 , 1 cot csc , 1 tan sec      sin 2 Example 19. Prove the identity cot sec sin     sin 2 Solution: LHS cot sec  cos 1     sin 2   sin cos    sin RHS   1 cot   Example 20. Prove the identity cot   1 tan   1 cot  Solution: LHS   1 tan     cos sin cos  1   sin sin       sin cos sin  1 cos cos     (sin cos ) cos       sin (cos sin )  cos     cot RHS  sin Chapter 6 : Trigonometry 14

 1 sin x   Example 21. Show that: Example 22. Prove that: sec x tan x  1 sin x       2 2 2 cos sin 1 2 sin   ( 1 sin x )( 1 sin x )  Solution: LHS     2 2   Solution: We know that sin cos 1 ( 1 sin x )( 1 sin x ) from which we have:  2 ( 1 sin x )     2 2  cos 1 sin  2 ( 1 sin x )     Hence, 2 2 LHS cos sin      2 2 2  2  2   2 ( 1 sin ) sin Since sin x cos x 1 then cos x 1 sin x      2 2 1 sin sin   2 2 ( 1 sin x ) ( 1 sin x )   LHS     2 RHS 1 2 sin  2 2 ( 1 sin x ) cos x  1 sin x 1 sin x    cos x cos x cos x    sec x tan x RHS 6.5 Trigonometric Equations ةـيـثلـثملا تلبداـعملا Trigonometric Equations Equations which contain trigonometric ratios are called trigonometric equations. There are usually an infinite number of solutions to such equations; however, solutions are often restricted to those between 0° and 360°. Knowledge of angles of any magnitude is essential in the solution of trigonometric equations and calculators cannot be relied upon to give all the solutions. The figure to the right shows a summary for angles of any magnitude.    Example 23. Solve the trigonometric equation 5 sin 3 0 for values of θ from 0° to 360°.    Solution: 5 sin 3 0    5 sin 3 3      sin 0 . 6 5 Since sin θ is negative  θ is in the third and the fourth quadrants.   sin 1   (shown as the angle α in the right figure below) The acute angle ( 0 . 6 ) 36 . 87 Hence θ = 180 o + 36.87 o = 216.87 o or θ = 360 o − 36.87 o = 323.13 o Chapter 6: Trigonometry 15