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

Chapter 6: Trigonometry

1

SET 3

Chapter 6

Trigonometry

لاتاـثلـثم

2

Chapter 6: Trigonometry

6.1 Angles, Rotations and Degrees Measures

تاـجردلاب سايقلا و نارودلا و اـياوزلا

Chapter 6: Trigonometry

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Example 1. Solution:

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Chapter 6: Trigonometry Example 2. Solution:

Chapter 6: Trigonometry

5 6.2 Radian Measures

يرـطق فصنلا سايقلا

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Chapter 6: Trigonometry Example 3. Solution:

Chapter 6: Trigonometry

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Example 4. Solution:

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Chapter 6: Trigonometry Example 5. Solution: Solution: Example 6.

6.3 Arc Length and Central Angles ةـيزكرملا ةـيوازـلا و سوقلا لوط

Chapter 6: Trigonometry

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Solution : Solution: Example 7. Example 8.

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Chapter 6: Trigonometry

6.4 The Trigonometric Ratios ةـيـثلـثملا بـسنلا

Chapter 6: Trigonometry

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Evaluating Trigonometric Ratios of Any Angles

Solution: Example 9.

12

Chapter 6: Trigonometry

Scientific calculators are usually used to evaluate trigonometric ratios of any angles. The following examples were solved using a scientific calculator. Example 10. Evaluate 4 4 5 1 37 sin     rounded to 4 decimal places. Solution:

6055 . 2 26 . 37 sin   

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) 5 9 3 124 sin     (b) 3 4 5 2 87 cos     (c) 4 5 2 3 61 tan    

Solution: (a)

823 . 5 9 3 124 sin      (b) 045 . 3 4 5 2 87 cos      (c) 845 . 1 4 5 2 3 61 tan     

Example 12. Evaluate the following rounded to 3 decimal places:

(a) 5 3 1 1 48 sec     (b) 8 2 5 76 csc     (c) 3 1 1 9 cot    

Solution: Recalling that:

  cos 1 sec 

,

  sin 1 csc  and   tan 1 cot  Then, (a) 500 . 1 5 3 1 1 48 cos 1 5 3 1 1 48 sec           (b) 027 . 1 8 2 5 76 sin 1 8 2 5 76 csc           (c) 195 . 6 3 1 1 9 tan 1 3 1 1 9 cot          

Example 13. Evaluate the following correct to 4 decimal places:

(a) 7 3 sin  (b) 16 9 cos  (c) 11 5 tan 

Solution: (a)

9749 . 7 3 sin   (b) 1951 . 16 9 cos    (c) 9551 . 6 11 5 tan  

       

     

3600 44 60 15 37 sin 4 4 5 1 37 sin

Chapter 6: Trigonometry

13 Example 14. Evaluate the following correct to 3 decimal places:

(a) 9 4 sec  (b) 12 5 csc  (c) 21 12 cot 

Solution: (a)

759 . 5 9 4 cos 1 9 4 sec     (b) 035 . 1 12 5 sin 1 12 5 csc     (c) 228 . 21 12 tan 1 21 12 cot     

Example 15. Determine the following acute angles in degrees and radians:

(a) 354 . sin 1



(b) 548 . cos 1



(c) 537 . 2 tan 1



Solution: (a)

354 . sin 1



= 20.7320 or 20043’56” 354 . sin 1



= 0.362 radians (b) 548 . cos 1



= 56.770 or 56046’12” 548 . cos 1



= 0.991 radians (c) 537 . 2 tan 1



= 68.4870 or 68029’14” 537 . 2 tan 1



= 1.195 radians

Example 16. Determine the following acute angles in degrees:

(a) 238 . 11 sec 1



(b) 284 . 3 csc 1



(c) 029 . cot 1



Solution: (a)

      

 

238 . 11 1 cos 238 . 11 sec

1 1

= 84.8950 or 84053’41” (b)       

 

284 . 3 1 sin 284 . 3 csc

1 1

= 17.7280 or 17043’43” (c)       

 

029 . 1 tan 029 . cot

1 1

= 88.3390 or 88020’20”

Example 17. Evaluate the following expression , correct to 4 significant figures:

7 5 14 tan 8 63 csc 3 9 1 15 cot 2 1 32 sec 4         

Solution:

) 2670 . )( 1210 . 1 ( 3 ) 6512 . 3 ( 2 ) 1813 . 1 ( 4 7 5 14 tan 8 63 csc 3 9 1 15 cot 2 1 32 sec 4            870 . 2 8979 . 5772 . 2 8979 . 3024 . 7 7252 . 4      

14

Chapter 6: Trigonometry

Example 18. Evaluate rounded to 4 decimal places:

(a) ) 115 sec(   (b) ) 7 4 95 csc(   

Solution: Positive angles are considered to be counterclockwise and negative

angles as clockwise. Hence 115o is actually the same as 245o (i.e. 360o  115o) (a)     245 sec ) 115 sec( 3662 . 2 245 cos 1     (b) ) 7 4 95 csc(    0051 . 1 ) 7 4 95 sin( 1      

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:

   cos sin tan 

,

   sin cos cot 

,

  cos 1 sec 

,

  sin 1 csc 

,   tan 1 cot  ,

1 cos sin

2 2

   

,

 

2 2

csc cot 1  

,

 

2 2

sec tan 1  

Example 19. Prove the identity Solution:

    cos 1 sin cos sin2    RHS    sin

Example 20. Prove the identity Solution:

RHS        

    

        

       

cot cos sin cos sin cos sin cos sin 1 sin cos 1

sin cos ) sin (cos cos sin ) cos (sin    sec cot sin2  LHS    cot tan 1 cot 1        sin sec cot sin2    tan 1 cot 1    LHS 6.5 Trigonometric Identities ةـيـثلـثملا تاـقباـطتـملا

Chapter 6: Trigonometry

15 Example 21. Show that:   

2 2 2

sin 2 1 sin cos    Solution: We know that

1 cos sin

2 2

    from which we have:  

2 2

sin 1 cos  

Hence,

RHS              

2 2 2 2 2

sin 2 1 sin sin 1 sin ) sin 1 (

Example 22. Prove that: Solution:

Since 1 cos sin

2 2

  x x

then

x x

2 2

sin 1 cos   RHS LHS             x x x x x x x x x x x tan sec cos sin cos 1 cos sin 1 cos ) sin 1 ( ) sin 1 ( ) sin 1 (

2 2 2 2

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 for values of θ from 0° to 360°. Solution:

Since sin θ is negative  θ is in the third and the fourth quadrants. The acute angle   



87 . 36 ) 6 . ( sin 1 (shown as the angle α in the right figure below) Hence θ = 180o + 36.87o = 216.87o or θ = 360o − 36.87o = 323.13o  

2 2

sin cos   LHS ) sin 1 )( sin 1 ( ) sin 1 )( sin 1 ( x x x x      LHS

x x x x tan sec sin 1 sin 1    

) sin 1 ( ) sin 1 (

2 2

x x   

6 . 5 3 sin 3 sin 5

3 sin 5

     

 

 



3 sin 5    6.5 Trigonometric Equations ةـيـثلـثملا تلبداـعملا

16

Chapter 6: Trigonometry

Example 24. Solve: for 0° ≤ x ≤360° . Solution:

2 . 1 5 . 1 8 . 1 tan 8 . 1 tan 5 . 1    x x Tangent is positive in the first and third quadrants. The acute angle   



19 . 50 ) 2 . 1 ( tan 1 Hence, x = 50.19o or x = 180o + 50.19o =230.19o

Example 25. Solve: for values of A in the range 0° ≤ A ≤360° . Solution:

5 . 4 2 cos 2 cos 4

2 2

   A A Hence Cosine is positive in quadrants one and four and negative in quadrants two and three. Thus in this case there are four solutions, one in each quadrant. The acute angle



 



45 ) 7071 . ( cos 1 Hence ,

, ,

and

The Law of Sines and Cosines

The Law of Sines states that in any triangle, the ratio of a side length to the sine of its opposite angle is the same for all three

• sides. As a formula:

C c B b A a sin sin sin  

The Law of Cosines states that in any triangle, given two sides and the included angle, the third side is given by the formula:

) (

cos . 2

2 2 2

C

ab b a c   

8 . 1 tan 5 . 1   x cos 4 2

2

  A

8 . 1 tan 5 . 1   x

cos 4 2

2

  A

7071 . 5 . cos    A   45 A      135 45 180 A      225 45 180 A      315 45 360 A

6.5 The Law of Sines and Cosines ماـمت بـيجلا و بـيجلا نوناـق

Chapter 6: Trigonometry

17 Example 26. For the triangle shown below, find the length of sides b and c. Solution: Solving for side b using the Law of Sines:

   57 sin 42 sin 30 b    57 sin 30 42 sin b    42 sin 57 sin 30 b 6 . 37  b m Solving for side c: Third angle = 180o – (42o + 57o) = 81o    81 sin 42 sin 30 c    81 sin 30 42 sin c 42 sin 81 sin 30  c 3 . 44  c m

Example 27. From a certain point, the angle of elevation of the top of a building is 48o. From

a point 30 meters nearer to the building, the angle of elevation is 74o. Find the height of the building.

Solution: Solving for θ:

θ = 180o – 74o = 106o Solving for α: α = 180o – (48o + 106o) α = 26o Solving for side x:

  

48 26 30

sin sin x

   48 sin 30 26 sin x x = 50.9 m Find the height of the building using the sine ratio:

x h  

74

sin

9 . 50 74

sin h    

74 9 . 50 sin

h

h = 48.9 m

c

570

420

b 48o 74o

x h α θ

30 m

Building

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Chapter 6: Trigonometry

Example 88. Solve the triangle below: Solution: Using the law of cosines:

) cos( . 2

2 2 2

 ac c a b    ) 63 cos( ) 36 )( 25 ( 2 ) 36 ( ) 25 (

2 2 2

    b 18 . 817 1296 625

2

   b 82 . 1103  b b = 33.22 cm Solving for α:  sin 25 63 sin 22 . 33   22 . 33 63 sin 25

sin



 

     

 



22 . 33 63 sin 25 sin 1    42 

Solving for γ:

α + β + γ = 180°

42° + 63° + γ = 180° γ = 180° − 105°  = 75°

Example 29. A ship travels 90 kilometres due west, then adjust its direction 20° southward.

After traveling 125 kilometres in that direction, how far is the ship from the point of departure?

Solution: Find α:

α = 180° − 20° = 160°

Solving for x using the law of cosines: x2 = 902 + 1252 – 2(90)(125) cos160° x2 = 8100 + 15625 + 21143 x2 = 44868 x = 44868 x = 211.8 km α

β = 63°

γ

b α

x

20° 90 km