Activity 1

  1. ∆ MNP is a right-angled triangle. Write down the trig ratios for:
    1. sin α
    2. sin β (4)
    3. tan β
  2. cos α (3)
  3. If MP = 13 and NP = 5, calculate cos β.
    6
    [7]

Solutions

    1. sin α = MN(1)
                  MP
    2. sin β = NP(1)
                  MP 
    3. tan β = NP(1)
                  MN 
    4. cos α = NP(1) 
                   MP(4)
  1. MP = 13 and NP = 5, so we can find MP,
    MP2 = MN2 + NP2 ……….Pythagoras
    132 = MN2 + 52
    169 = MN2 + 25
    MN2 = 169 – 25
    MN2 = 144 
    ∴MN = 12
    cos β = MN = 12
                 MP    13(3)
    [7] 

Activity 2

  1. If sin θ is negative and cos θ is positive, then which statement is true?
    1. 0° < θ < 90°
    2. 90° < θ < 180°
    3. 180° < θ < 270°
    4. 270° < θ < 360° (1)
  2. If tan θ < 0 and cos θ < 0, then which statement is true?
    1. 0° < θ < 90°
    2. 90° < θ < 180°
    3. 180° < θ < 270°
    4. 270° < θ < 360° (1)
  3. Will the following trig ratios be positive or negative?
    1. sin 315°
    2. cos (–215°)
    3. tan 215°
    4. cos 390° (4)
      [6]

Solutions

  1. Sin θ is negative in 3rd and 4th quadrants; cos θ is positive in 1st and 4th quadrants.
    So θ is in the 4th quadrant. D. 270° < θ < 360° 3 (1)
  2. tan θ < 0 in 2nd and 4th quadrants; cos θ < 0 in 2nd and 3rd quadrants.
    So θ is in the 2nd quadrant. B. 90° < θ < 180° 3 (1)
    1. sin 315° is in 4th quadrant so it is negative. 3 (1)
    2. cos (–215°) is in 2nd quadrant so it is negative. 3 (1)
    3. tan 215° is in 3rd quadrant, so it is positive. 3 (1)
    4. cos 390° is the same as cos 30° in the 1st quadrant, so it is positive. 3 (1)
      [6] 

Activity 3
If cos β = p/√5 where p < 0 and β ∈ [180°; 360°], determine, using a diagram, an expression in terms of p for:

  1. tan β
  2. 2 cos2β – 1
    [6]

Solutions

  1. cos β =p/√5 x/r = ; so x = p and r = √5
    By Pythagoras, y2 = r2 – x2
    ∴ y2 = (√5)2 – p2
    = 5 – p2
    ∴ y = ± √5 – p2 
    ∴ y = – √5 – p2 since β is in quadrant 3, y is negative
    ∴ tan β = –√5 – p2 (4)
                         p
    11
  2. 2 cos2β – 1 = 2 ( p/√5)2 - 1
    2p 2 -1 (2) [6] 
     5

Activity 4
Without using a calculator, determine the value of:

  1. cos 150°
  2. sin (–45°)
  3. tan 480°
    [7]

Solutions

  1. cos 150° rewrite as (180 – ?)
    = cos(180° – 30°) quadrant II, cos θ negative
    = –cos 30° special ratios
    = –√3/2 (2)
  2. sin(–45°) sin(–θ) = –sin θ; quadrant IV, sin θ negative
    = –sin 45° special ratios
    = – 1/√2 (2)
  3. tan 480° write as an angle in the first rotation of 360°
    = tan (480° – 360°)
    = tan 120°  quadrant II, rewrite as (180 – ?)
    = tan (180° – 60°) tan θ negative
    = –tan 60° special ratios
    = –√3  (3)
    [7] 

Activity 5
Write the trig ratios as the trig ratios of their co-functions:

  1. sin 50°
  2. cos 70°
  3. sin 100°
  4. cos 140°
    [4]

Solutions

  1. sin 50° = sin(90° – 40°) = cos 40° 
  2. cos 70° = cos(90° – 20°) = sin 20° 
  3. sin 100° = sin(90° + 10) = cos 10° 
  4. cos 140° = cos(90° + 50°) = –sin 50° 
    [4] 

Summary
Any angle (obtuse or reflex) can be reduced to an acute angle by using:

  • Convert negative angles to positive angles
  • Reduce angles greater than 360°
  • Use reduction formulae
  • Use co-functions

Activity 6
Simplify without using a calculator:

  1. sin(180° + x). cos 330°.tan 150° (4)
                        sin x
  2.       cos 750°.tan 315°.cos(–θ)      (8)
    cos(360°- θ).sin300°.sin(180°- θ)
  3. tan 480°.sin 300°.cos 14°.sin(–135°) (9)
                sin104°.cos225°
  4.     cos 260°.cos 170°   (7)
    sin10°.sin190°.cos350° 
    [28]

20

Activity 7
Simplify the following expressions.

  1.      cos (180°–x) sin (x –90°) – 1    (8)
      tan2(540° + x) sin(90°+x)cos(–x) 
  2. [sin(–θ) + cos(360° + θ)][cos(θ – 90°) + cos(180°+θ)] (3)
  3. cos2θ (1 + tan2θ) (3)
  4. 1 – cos2θ
    1– sin2θ (3)
    [17]

Solutions

  1.     cos (180°–x) sin (x –90°) – 1    (8)
      tan2(540° + x) sin(90°+x)cos(–x)  – use reduction formulae and co-functions
    =         (–cos x)(–cos x) –1         
    tan2(540°- 360°+ x) cos x. cos x
    – multiply out numerator and denominator reduction of angle > 360°
    =        cos2x–1      
    tan2(180°+ x).cos2x     
    – use trig identity format for cos2x – 1 reduction formula
    =    –(1 – cos2x)    
          tanx . cos2x   
    – use trig identities for 1 – cos2x and for tan x
    =   –sin2x     
    sin2xcos2x
    cos2x     1
    = sin2x = -1
       sin2
    – simplify
    (8)
  2. [sin(–θ) + cos(360° + θ)][cos(θ – 90°) + cos(180°+θ)] – reduce to angle < 90°
    =[–sin θ + cos θ][cos (–(90° – θ))+ (–cos θ)] – simplify; use co-functions
    =(–sin θ + cos θ)(sin θ – cos θ) – multiply out using FOIL
    = –sin2 θ + sin θ cos θ + cos θ sin θ – cos2 θ 
    = –(sin2θ + cos2θ) + 2 sin θ cos θ – use trig identity
    = –1 + 2 sin θ cos θ – use double angle identity
    = –1 + sin2θ (3)
  3. cos2θ (1 + tan2θ) – multiply out the bracket
    =cos2θ + cos2θ.tan2θ – use trig identity for tan θ
    = cos2θ + cos2θ . sin2θ – simplify
                       1        cos2θ.
    = cos2θ + sin2θ 3 = 1 – use trig identity sin2θ + cos2θ = 1 (3)
  4. 1 - cos2θ. – use trig identity sin2θ + cos2θ = 1
    1 - sin2θ
    = sin2θ – use trig identity for tan θ
       cos2θ
    = tan²θ (3)
    [17] 

Activity 8
Prove the following identities:

sin x ∙tan x + cos x =   1   
                                 cos x (4)

(sin x + tan x) (   sin x   ) = sin x. tan x (7)
                       1 + cos x

    1     =   cos x   + tan x (6)
cos x    1 + sin x 

   1   + tan x = tanx  (5)
tanx                sin2x
[22]

22
23 


Hints for solving trig identities:

  • Choose either the lefthand side or the righthand side and simplify it to look like the other side.
  • If both sides look difficult, you can try to simplify on both sides until you reach a point where both sides are the same.
  • It is usually helpful to write tan θ as sinθ
                                                            cosθ .
  • Sometimes you need to simplify sinθ  to tan θ.
                                                       cos θ
  • If you have sin2x or cos2x with +1 or –1, use the squares identities (sin2θ + cos2θ = 1).
  • Find a common denominator when fractions are added or subtracted.
  • Factorise if necessary – specify with examples i.e. common factor, DOPS, Trinomial, sum/diff of two cubes

Activity 9

  1. If cos 20° = p, determine the following ratios in terms of p:
    1. cos 380°
    2. sin 110°
    3. sin 200° (6)
  2. Determine the general solution for x in the following equations:
    1. 5 sin x = cos 320° (correct to 2 decimal places)
    2. 3 tan x + √3 = 0 (without using a calculator)
    3. tan x–1 = –3 (correct to one decimal place) (10)
           2
  3. Determine x for x ∈[–180°; 180°] if 2 + cos (2x – 10°) = 2,537 (6)
    [22]

Solutions

  1. cos 20° = p/1so x = p and r = 1
    By Pythagoras, y2 = r2 – x2
    y2 = 12 – p2 = 1– p2
    y = √1 – p2 first quadrant, so y is positive
    1. cos 380° = cos (360° + 20°) = cos 20°  = p  (2)
    2. sin 110° reduction formula
      = sin (180° – 70°)
      = sin 70° 3 co-function
      = sin (90° – 20°)
      = cos 20° 3 = p 3 (3)
    3. sin 200° = sin (180° + 20°)
      = –sin20° 
      = – √1 – p2 = –√1 – p2   (1) (6)
    1. 5 sin x = cos 320° 
      5 sin x = 0,766044
      sin x = 0,15320... 
      Ref angle = 8,81°
      x = 8,81° + k360° OR x = 180° – 8,81° + k360° 
      x = 171,19° + k360°  k ∈ Z (4)
      Calculator keys:
      cos 320 =
      ÷ 5 =
      SHIFT sin ANS =
    2. 3 tan x + √3 = 0
      3 tan x = – √3
      tan x = – √3/3 [special angle: tan 30° tan 30° = – √3/3]
      Ref angle = 30°
      x = 180° – 30° + k180° 
      x = 150° + k180° 3 k ∈ Z (3)
    3. tan x–1 = –3 multiply both sides by 2
          2
      tan x – 1 = –6
      tan x = –5 reference angle is 78,69…°
      ∴x = 180° –78,69…° + k180° 
      x = 101,31° + k180°; k ∈ Z (3) (10)
  2. 2 + cos (2x – 10°) = 2,537
    cos (2x – 10°) = 0,537
    Ref angle = 57,52…..°
    2x – 10° = 57,52…..° + k360° or 2x – 10° = 360° – 57,52° + k360°
    [solve equations]
    2x = 67,52….° + k360° or 2x = 312,48…° + k360° 
    [divide all terms on both sides by 2]
    x = 33,76° + k180° or x = 156,24° + k180° 3 k ∈ Z
    x ∈ [–180°; 180°]
    So for k = –1: x = 33,76° –180° = –146,24° or x = 156,24° – 180° = –23,76° 
    For k = 0: x = 33,76° or x = 156,24° 
    (For k = 1, x will be > 180°, so it is too big)
    Solution: x ∈ {–146,24°; –28,76°; 33,76°; 156,24°}  (6)
    [22] 

Activity 10
Do NOT use a calculator to answer this question. Show ALL calculations. Prove that:

  1. cos 75° =√2 ( √3 –1) (5)
                          4
  2. Prove that cos(90° – 2x).tan(180° + x) + sin2(360° – x) = 3sin2x (7)
  3. Prove that (tan x – 1)(sin 2x – 2cos2x) = 2(1 – 2sin x cos x) (7)
    [19]
  1. Solutions
    1. LHS = cos 75° = cos(45° + 30°) 
    = cos45°.cos30° – sin45°.sin30° 
    = √2/2 . √3/√2 √2/2 . ½  
    = √2 . √3√2/4
            4
    =√2 (√3 –1 = RHS (5)
            4
  2. LHS = cos(90° – 2x).tan(180° + x) + sin2(360° – x) co-functions and reductions
    = sin2 x. tan x + sin2 x                                               double angle for sin 2x
                                                                                      trig identity for tan x
    = 2sin x.cos x. sin x + sin2 x simplify
                           cos x
    = 2 sin2 x + sin2 x
    = 3 sin2x = RHS (7)
  3. There are several ways to prove this. Here is one solution.
    LHS = (tan x – 1)(sin 2x – 2cos2x)
    = (sin x – 1 ) (2sin x. cos x – 2cos2x) double angle identity for sin 2x
        cos x
    = 2sin2 x – 2sin x. cos x – 2sin x. cos x + 2cos2x multiply out
    = 2 sin2 x – 4 sin x cos x + 2 cos2 x
    = 2(sin2 x – 2sin x. cos x + cos2x)              trig identity sin2 x + cos2 x = 1
    = 2(1 – 2sin x. cos x) = RHS (7)
    [19] 

Activity 11
Determine the general solution for x in the following:

  1. sin 2x. cos 10° – cos 2x. sin 10° = cos 3x (8)
  2. cos2 x = 3 sin 2x (11)
  3. 2sinx = sin(x + 30°) (5)
    [24]

Solutions

  1. sin 2x. cos 10° – cos 2x. sin 10° = cos 3x use compound angle identity
    ∴ sin (2x – 10°) = cos 3x use co-functions
    ∴ sin (2x – 10°) = sin (90° – 3x)
    ∴ 2x – 10° = 90°–3x + k360°3or 2x–10° = 180°–(90°–3x) + k360°3 k ∈ Z
    ∴ 5x = 100° + k360° 2x – 10° = 90° + 3x + k360°
    ∴ x = 20° + k72° –x = 100 + k360°3
    x = –100 – k360 3 k ∈ Z (8)
  2. cos2 x = 3 sin 2x use double angles for sin 2x
    cos2 x = 3(2 sin x.cos x)3 make LHS = 0
    cos2 x – 3(2 sin x.cos x) = 0 multiply out
    cos2 x – 6 sin x.cos x = 0 common factor
    cos x (cos x – 6 sin x) = 0
    ∴ cos x = 03 or cos x – 6 sin x = 0
    cos x = 0 or cos x 6sin x
                       cos x      cos x
    cos x = 0 or 1 = 6 tan x
    cos x = 0 or tan x = 1/6
    Reference angle = 90° or reference angle = 9,46°
    ∴ x = 90° + k360° or x = 360°–90° + k360° or x = 9,46° + k180° 3k ∈ Z
    x = 270° + k360° or x = 180° + 9,46° + k360°  k ∈ Z
    = 189,46° + k360°  k ∈ Z (11)
  3. 2 sin x = sin ( x + 30° )
    2 sin x = sin x. cos30° + cosx.sin30°3
    2 sin x = sin x. √3/2  + cos x. ½ multiply by 2
    4 sin x = √3 sin x + cos x divide by cos x
    4 tan x = √3 tan x + 1
    4 tan x − √3 tan x = 1
    tan x =    1     
               4 − √3
    x = 23,79° + k180°; k ∈ Z 3 (5)
    [24] 

 

Last modified on Friday, 03 September 2021 07:54