Trigonometry 2 Questions and Answers

Trigonometry 2 Questions

1.  Solve the equation: (2 mks)

for

2.  (a) Complete the table below, leaving all your values correct to 2 d.p. for the functions y = cos x and y = 2cos (x + 30)⁰   (2 mks)

 (b) For the function y = 2cos(x+30)⁰

  State:

1. The period (1 mk)
2. Phase angle (1 mk)

(c) On the same axes draw the waves of the functions y = cos x and y = 2cos(x+30)⁰ for. Use the scale 1cm rep 30⁰ horizontally and 2 cm rep 1 unit vertically      (4 mks)

(d) Use your graph above to solve the inequality  (2 mks)

3.  Find the value of x in the equation.

Cos(3x – 180^o) = √3  in the range O^o < x < 180^o (3 marks)

2

4.  Given that and ө is an acute angle, find without using tables cos (90 – ө)  (2mks)

5.  Solve for ө if -¼ sin (2x + 30) = 0.1607, 0 ≤ ө ≥ 360⁰  (3mks)

6.  Given that Cos  = ⁵/₁₃ and that 270⁰


 360⁰ , work out the value of Tan  + Sin  without using a calculator or mathematical tables. (3 marks)

7.  Solve for x in the range 0⁰
 x  180⁰ (4mks)

-8 sin²x – 2 cos x = -5.

8. If tan x^o = ¹²/₅ and x is a reflex angle, find the value of 5sin x + cos x without using a

calculator or mathematical tables

9.   Find  given that 2 cos 3 -1 = 0 for 0^o 

 360^o

10.  Without a mathematical table or a calculator, simplify: Cos300^o x Sin120^o
giving your answer in

Cos330^o – Sin 405^o
rationalized surd form.

11.  Express in surds form and rationalize the denominator.

  1

Sin 60^o Sin 45^o – Sin 45^o

12.  Simplify the following without using tables;

Tan 45 + cos 45sin 60  

13.  Simplify the following surds in the form of a + b c where a, b, and c are constants

5
2  

2 2 – 5 2 2 – 5

14.  John cycles from shopping centre A on a bearing of 120^o for 5 km to shopping centre B. He then

cycles on a bearing of 200⁰ for 7 km to the shopping centre C. Calculate to 1 decimal place.

a) The direct distance from A to C.

b) The bearing of A from C.

c) Bearing of B from C.

Trigonometry 2 Answers

1.  5 sin x + cos x

= 5 12 – 5

13 13

= 60 – 5 = 55

  13 13 13

= 12

  13

2.    2 cos 3θ = 1

  2 2

 Cos 3 θ = 0.5

 3 θ = Cos^-10.5

  3 θ = 60^o, 300^o, 420^o, 66^o, 78^o, 102^o

3 3 3 3 3 3 3

 ∴
 = 20^o, 100^o, 140^o, 220^o, 260^o, 340^o

3..  ½ X ^3/₂

^3/₂ X ¹/_2

^3/₄ X ^3/₂ + ¹/_2

^3/₂ – ¹/_2
^3/₂ +
¹/_2

³/₈ + ^3/_42 = ³/₈ + ^3/_42

³/₄ – ¹/₂
¹/₄

= ³/₂ + ^3/_2

4.  a) b² = a² +
c² – 2ac cos B

b² = 7² + 5² – 2.5.. 7 cos 100

= 74 – 70(-0.173648)

= 74 + 12.15537

b² = 86.15537

b = 9.28199

 AC = 9.3 km

 b) 9.3 = 5

 sin 100 sin θ

 Sin  = 5 sin 100 = 0.529466

  9.3

  = 31.9694


≃ 32⁰

  32 – 20 = 12⁰

Bearing = 360⁰ – 12⁰   = 348⁰

c) 020⁰

5.  

Sin 60 = 
³/₂ 1

Sin 45 = ¹/₂
3
1 – 1

2 2 2

= 1

3 – 1

22 2

= 6 –  2

1. 2
   

= 6 – 22

4

6.  

1 + 1 x √3

√ 2 2

1 +√3 x 2√2

2√ 2 2√2

1 + 2√6

1 4

4 + 2 √6

4

7.  √5 (2√2 + √5 )+ √2(2√2 – √5)

(2√2)2 – (√5)²

  2√10 + 5 + 4 – √10

8 – 5

  9 +√10

  3

3 + ¹/₃ √10

8. a)   b² = a² +
c² – 2ac cos B

b² = 7² + 5² – 2.5.. 7 cos 100

= 74 – 70(-0.173648)

= 74 + 12.15537

b² = 86.15537

b = 9.28199

AC = 9.3 km

b) 9.3 = 5

sin 100 sin θ

Sin  = 5 sin 100 = 0.529466

9.3

 = 31.9694


≃ 32⁰

32 – 20 = 12⁰

Bearing = 360⁰ – 12⁰

= 348⁰

c) 020⁰