MATHEMATICS
Grade 11
MORE EXERCISES
Graphs of the trigonometric functions, sin, cos, tan : answers.
  
  
Answers  1
    
$$ \hspace*{2 mm}\mathrm{1.1\kern3mmy = 2\ sin\ x\ − 1\kern2mm\ } $$
           Vertical translation : q = minus; 1 and
           The graph oscillates about y = minus; 1.
           amplitude = 2
           maximum = 2 − 1 = 1
           minimum = − 2 − 1 = − 3
           Period = 360°
           The maximum is reached
           at x = 90° and x = -270°
           The minimum is reached
           x = − 90° and x = 270°

                                                                   [ Q 1.1 ]
    
$$ \hspace*{2 mm}\mathrm{1.2\kern3mmy = 1 − 2 sin\ x\kern2mm\ } $$
           Vertical translation : q = 1 and
           the graph oscillates about y = 1.
           amplitude = 2
           maximum = 3
           minimum = − 1
           Period = 360°
           The maximum is reached
           at x = − 270° and x = 90°
           The minimum is reached
           at x = −270° and x = 90°

                                                                   [ Q 1.2 ]
    
$$ \hspace*{2 mm}\mathrm{1.3\kern3mmy = 2\ cos\ x\ − 1\kern2mm\ } $$
           Vertical translation : q = − 1 and
           the graph oscillates about y = − 1.
           amplitude = 2
           maximum = 1
           minimum = − 3
           Period = 360°
           The maximum is reached
           at x = − 360°; x = 0° and x = 360°
           The minimum is reached
           at x = −180° and x = 180°

                                                                   [ Q 1.3 ]
    
$$ \hspace*{2 mm}\mathrm{1.4\kern3mmy = 1 − 2\ cos\ x\kern2mm\ } $$
           Vertical translation : q = 1 and
           the graph oscillates about y = 1.
           amplitude = 2
           maximum = 3
           minimum = − 1
           Period = 360°
           The maximum is reached
           at x = − 180° and x = 180°
           The minimum is reached
           at x = −360°; x = 0° and x = 360°

                                                                   [ Q 1.4 ]
    
$$ \hspace*{2 mm}\mathrm{1.5\kern3mmy = sin\ (x − 30°)\kern2mm\ } $$
           Vertical translation : q = 0 and
           the graph oscillates about y = 0, the X-axis.
           amplitude = 1
           maximum = 1
           minimum = − 1
           Period = 360°
           The maximum is reached
           at x = − 240° and x = 120°
           The minimum is reached
           at x = −60°; x = 0° and x = 300°

                                                                   [ Q 1.5 ]
    
$$ \hspace*{2 mm}\mathrm{1.6\kern3mmy = sin\ (x + 60°)\kern2mm\ } $$
           Vertical translation : q = 0 and
           the graph oscillates about y = 0, the X-axis.
           amplitude = 1
           maximum = 1
           minimum = − 1
           Period = 360°
           The maximum is reached
           at x = − 330° and x = 300°
           The minimum is reached
           at x = −150°; x = 210°

                                                                   [ Q 1.6 ]
    
$$ \hspace*{2 mm}\mathrm{1.7\kern3mmy = cos\ (x − 60°)\kern2mm\ } $$
           Vertical translation : q = 0 and
           the graph oscillates about y = 0, the X-axis.
           amplitude = 1
           maximum = 1
           minimum = − 1
           Period = 360°
           The maximum is reached
           at x = − 300° and x = 60°
           The minimum is reached
           at x = −120°; x = 240°

                                                                   [ Q 1.7 ]
    
$$ \hspace*{2 mm}\mathrm{1.8\kern3mmy = cos\ (x + 30°)\kern2mm\ } $$
           Vertical translation : q = 0 and
           the graph oscillates about y = 0, the X-axis.
           amplitude = 1
           maximum = 1
           minimum = − 1
           Period = 360°
           The maximum is reached
           at x = − 30° and x = 330°
           The minimum is reached
           at x = −210°; x = 150°

                                                                   [ Q 1.8 ]
    
$$ \hspace*{2 mm}\mathrm{1.9\kern3mmy = 1 − sin x\kern2mm\ } $$
           Vertical translation : q = 1 and
           the graph oscillates about y = 1.
           amplitude = 1
           maximum = 2
           minimum = 0
           Period = 360°
           The maximum is reached at x = − 90°
           The minimum is reached at x = 90°
                                                                   [ Q 1.9 ]
    
$$ \hspace*{2 mm}\mathrm{1.10\kern3mmy = 3\ cos\ x\ − 2\kern2mm\ } $$
           Vertical translation : q = − 2 and
           the graph oscillates about y = −2.
           amplitude = 3
           maximum = 3 × 1 − 2 = 1
           minimum = 3 × −1 − 2 = −5
           Period = 360°
           The maximum is reached at x = 0°
           The minimum is reached
           at x = − 180° and x = 180°
                                                                   [ Q 1.1 ]
    
$$ \hspace*{2 mm}\mathrm{1.11\kern3mmy = sin\ (x − 30°)\kern2mm\ } $$
           Vertical translation : q = 0 and
           the graph oscillates about y = 0, the X-axis.
           amplitude = 1
           maximum = 1
           minimum = −1
           Period = 360°
           The maximum is reached at x = 60°
           The minimum is reached at x = − 120°
                                                                   [ Q 1.11 ]
nr11M54sctg.html#qtrg0110     
$$ \hspace*{2 mm}\mathrm{1.12\kern3mmy = cos\ (2x − 60°)\kern2mm\ } $$
$$ \hspace*{12 mm}\mathrm{y = cos\ 2(x − 30°)\kern2mm\ } $$
           Vertical translation : q = 0 and
           the graph oscillates about y = 0, the X-axis.
           amplitude = 1
           maximum = 1
           minimum = −1
$$ \hspace*{9 mm}\mathrm{Period\ = \Big(\frac{360}{2}\Big)^{°} = 180°\kern2mm\ } $$

           The maximum is reached
           at x = −150° and x = 30°
           The minimum is reached
           at x = − 60° and x = 120°
                                                                   [ Q 1.12 ]
    
$$ \hspace*{2 mm}\mathrm{1.13\kern3mmy = cos\ (x + 30°)\ − 1\kern2mm\ } $$
           Vertical translation : q = − 1 and
           the graph oscillates about y = −1.
           amplitude = 1
           maximum = 1 − 1 = 0
           minimum = −1 −1 = −2
           k = 1 and period = 360°
           Horizontal translation : p = 30° to the left
           The maximum is reached at x = −30°
           The minimum is reached at x = 150°
                                                                   [ Q 1.13 ]
    
$$ \hspace*{2 mm}\mathrm{1.14\kern3mmy = sin\ (x + 60°)\ + 1\kern2mm\ } $$
           Vertical translation : q = 1 and
           the graph oscillates about y = 1.
           amplitude = 1
           maximum = 1 + 1 = 2
           minimum = 1 − 1 = 0
           k = 1 and period = 360°
           Horizontal transtion : p = 60° to the left
           The maximum is reached at x = 30°
           The minimum is reached at x = −150°
                                                                   [ Q 1.14 ]
    
$$ \hspace*{2 mm}\mathrm{1.15\kern3mmy = tan\ x\kern2mm\ } $$
           Asimptotes at x = -90° and x = 90°
           Period = 180°
           tan x = 1 at x = -135°; x = 45°.
           tan x = − 1 at x = -45° and at x = 135°
           Graph is increasing.
                                                                   [ Q 1.15 ]
    
$$ \hspace*{2 mm}\mathrm{1.16\kern3mmy = tan\ x +\ 1\kern2mm\ } $$
           Asimptotes at x = -90° and x = 90°
           Period = 180°
           Vertical translation : q = 1
           tan x = 1 at x = 0°.
           Graph is increasing.
                                                                   [ Q 1.16 ]
    
$$ \hspace*{2 mm}\mathrm{1.17\kern3mmy = tan\ (x + 30°)\kern2mm\ } $$
           Asimptotes at x = -120° and x = 60°
           Period = 180°
           Vertical translation : q = 0
           Horizontal transtion : p = 30° to the left
           tan x = 1 at x = −165° ; x = 15°.
           Graph is increasing.
                                                                   [ Q 1.17 ]
    
$$ \hspace*{2 mm}\mathrm{1.18\kern3mmy = tan\ 2x\kern2mm\ } $$
           Asimptotes at x = -45° and x = 45°
           Period = 90°
           Vertical translation : q = 0
           Horizontal transtion : p = 0°
           tan x = 1 at x = &inus;1650° ; x = 150°.
           Graph is increasing.
                                                                   [ Q 1.18 ]
  
Answers  2
2.1   The graph of y = a  sin k(x + p)
                                                                    [ A 2.1 ]
    
$$ \hspace*{2 mm}\mathrm{2.1\kern3mmy = a\ sin\ k(x + p)\ +\ q\kern2mm\ } $$
           The graph oscillates about y = 0, the
           X-axis and thus q = 0..
           Amplitude = 1 so that a = 1.
$$ \hspace*{9 mm}\mathrm{period\ = 180°\ and\ k = \Big(\frac{360°}{180°}\Big) = 2\kern2mm\ } $$

           Horizontal translation is 30° to the left; p = +30°
           Equation : y = 1.sin 2x + 30°
                                  = sin 2(x + 15)°                    [ Q 2.1 ]
    
$$ \hspace*{2 mm}\mathrm{2.2\kern3mmy = a\ cos\ k(x + p)\ +\ q\kern2mm\ } $$
           The graph oscillates about y = 0, the
           X-axis and thus q = 0..
           Amplitude = 1 so that a = 1.
$$ \hspace*{9 mm}\mathrm{period\ = 120°\ and\ k = \Big(\frac{360°}{120°}\Big) = 3\kern2mm\ } $$

           Horizontal translation is 30° to the right; p = − 30°
           Equation : y = 1.sin 2x + 30°
                                  = sin 2(x + 15)°                    [ Q 2.2 ]
    
$$ \hspace*{2 mm}\mathrm{2.3\kern3mmy = a\ sin\ k(x + p)\ +\ q\kern2mm\ } $$
           The graph oscillates about y = 0, the
           X-axis and thus q = 0.
           Amplitude = 2 so that a = 2.
$$ \hspace*{9 mm}\mathrm{period\ = 120°\ and\ k = \Big(\frac{360°}{120°}\Big) = 3\kern2mm\ } $$

           At A(20° ; 0) : 3(20° + p) = 0°
           60° + 3p = 0°
           p = − 20°
           Equation : y = 2.sin 3(x − 20°)
                                 = 2 sin 3(x − 20)°                   [ Q 2.3 ]
    
$$ \hspace*{2 mm}\mathrm{2.4\kern3mmy = a\ cos\ k(x + p)\ +\ q\kern2mm\ } $$
           The graph oscillates about y = 0, the
           X-axis and thus q = 0..
           Amplitude = 3 and the graph is a cosine graph
           so that a = −3.
$$ \hspace*{9 mm}\mathrm{period\ = 180°\ and\ k = \Big(\frac{360°}{180°}\Big) = 2\kern2mm\ } $$

           At A(15° ; 0) : 2(15° + p) = 90°
                                      30° + 2p = 90°
                                                  p = 30°
           Equation : y = − 3 cos 2(x + 30°)
                                 = − 3 cos 2(x + 30)°
                                                                                 [ Q 2.4 ]
    
$$ \hspace*{2 mm}\mathrm{2.5\kern3mmy = a\ sin\ k(x + p)\ +\ q\kern2mm\ } $$
           The graph oscillates about y = − 2, the
           and thus q = − 2.
           Amplitude = 3 so that a = 3.
           Period = 360° and k = 1
           There is no horizontal translation and p = 0
           Equation : y = 3 sin (x + 0°) − 2
                                 = − 3 sin x − 2                         [ Q 2.5 ]
    
$$ \hspace*{2 mm}\mathrm{2.6\kern3mmy = a cos k(x + p) + q\kern2mm\ } $$
           The graph oscillates about y = − 2, the
           and thus q = − 2.
           Amplitude = 3 so that a = 3.
$$ \hspace*{9 mm}\mathrm{period\ = 180°\ and\ k = \Big(\frac{360°}{180°}\Big) = 2\kern2mm\ } $$

           There is no horizontal translation and p = 0°
           Equation : y = 3 cos 2(x + 0°) − 2
                                 = 3 cos 2x − 2                         [ Q 2.6 ]
  
Answers  3
    
                                 A(− 150°;0);  B(30°;0);C(0;− 0,5)
                                                                   [ Q 3.1 ]
    
$$ \hspace*{2 mm}\mathrm{3.2\kern3mmperiod = 360°\kern2mm\ } $$
                                                                   [ Q 3.2 ]
    
$$ \hspace*{2 mm}\mathrm{3.3\kern3mmsin\ (x - 30°) = cos\ 2x\kern2mm\ } $$
$$ \hspace*{29 mm}\mathrm{= sin\ (90° − 2x)\kern2mm\ } $$
$$ \hspace*{17 mm}\mathrm{x − 30° = 2x + n.360°\kern2mm\ } $$
$$ \hspace*{26 mm}\mathrm{x = 40° + n.120°;\ \ n \isin Z\kern2mm\ } $$
$$ \hspace*{33 mm}\mathrm{\bold{OR}\kern2mm\ } $$
$$ \hspace*{17 mm}\mathrm{x − 30° = 180° − (90° − 2x) + n.360°\kern2mm\ } $$
$$ \hspace*{17 mm}\mathrm{x − 30° = 180° − 90° + 2x + n.360°\kern2mm\ } $$
$$ \hspace*{23 mm}\mathrm{− x = 120° + n.360°\kern2mm\ } $$
$$ \hspace*{24 mm}\mathrm{ x = −120° − n.360°\kern2mm\ } $$
           Solution : x = 40°−1.120° ;  40° + 0.120° ;
                             x = 40° + 1.120° and − 120° − 0.360°;
           Solution : x = − 80°(E); 40°(F); 160°(G) ; − 120°(D)
                                                                                [ Q 3.3 ]
    
$$ \hspace*{2 mm}\mathrm{3.4\kern3mmf(x) < − 0,5\ \ if\ − 120° < x < 0°\kern2mm\ } $$          [ Q 3.4 ]
    
$$ \hspace*{2 mm}\mathrm{3.5\kern3mmf(x) ≥ 0,5\ \ if\ 60° ≤ x ≤ 180°\kern2mm\ } $$               [ Q 3.5 ]
  
Answers  4
     4.1  f:  amplitude = 2 and thus a = 2
          The cosine graph reaches 0 at 60° instead
          of at 90° so that the graph is shifted 30° to
          the left and thus b = 30°
          The sine graph, g, oscillates about y = 1 so
          that c = 1
$$ \hspace*{8 mm}\mathrm{period = 180°\ and\ thus\ d = \Big(\frac{360°}{180°}\Big) = 2\kern2mm\ } $$
                                                                                [ Q 4.1 ]
    
$$ \hspace*{2 mm}\mathrm{4.2\kern3mmf(x) = 2\ cos\ (x − 30°)\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{f(0) = 2\ cos\ (0 + 30°)\kern2mm\ } $$
$$ \hspace*{16 mm}\mathrm{= 2\ cos\ 30°\kern2mm\ } $$
$$ \hspace*{16 mm}\mathrm{= 2\Big(\frac{\sqrt3}{2}\Big) = \sqrt3\kern2mm\ } $$                             [ Q 4.2 ]


     4.3.1  f:  x = 45° and x = −135°                         [ Q 4.3.1 ]

     4.3.2  f and g intersect at (−90° ; 1) and
               at (13,5° ; 1,45): hence the solution :
               −90° ≤ x ≤ 13,5°                                   [ Q 4.3.2 ]

     4.4  f(x) reaches its maximum at − 30°.
            The axis is shifted 30° to the left,
            i.e. the graph is shifted 30° to the right
            and p = 30°
$$ \hspace*{10 mm}\mathrm{f(x) = 2 cos (x − 30°)\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{f(x) = 2 cos (x + 30° − 30°)\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{f(x) = 2 cos x\kern2mm\ } $$                                          [ Q 4.4 ]
  
Answers  5
     5.1  f:  amplitude = 1 and thus a = 1
          The cosine graph reaches 1 at 30° in stead
          of at 0° so that the graph is shifted 30° to
          the right and thus b = −30°
          The sine graph, g, has a period = 360°
          so that c = 1
                                                                                [ Q 5.1 ]
    
$$ \hspace*{2 mm}\mathrm{5.2.1\kern3mmRange\ :\ −0,5 \le y \le 1\kern2mm\ } $$                    [ Q 5.2.1 ]
    
$$ \hspace*{2 mm}\mathrm{5.2.2\kern3mm30° < x \le; 90°\kern2mm\ } $$                                    [ Q 5.2.2 ]
  
$$ \hspace*{2 mm}\mathrm{5.2.3\kern3mmRange−90° ≤ x ≤ −60°\ \ and\ \ x = 0°\kern2mm\ } $$
                                                                                [ Q 5.2.3 ]
    
$$ \hspace*{2 mm}\mathrm{5.3\kern3mmcos\ (x − 30°)\ =\ sin\ x\kern2mm\ } $$
$$ \hspace*{31 mm}\mathrm{= cos\ (90° − x)\kern2mm\ } $$
$$ \hspace*{16 mm}\mathrm{\therefore x − 30° = 90° − x + n\bold{.}360°\ \kern3mm\ n ∈ Z\kern2mm\ } $$
$$ \hspace*{23 mm}\mathrm{\therefore 2x = 120° + n\bold{.}360°\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{\therefore x = 60° + n\bold{.}180°\kern2mm\ } $$
$$ \hspace*{25 mm}\mathrm{\therefore x = 60°\kern2mm\ } $$
          If g(x) > f(x) then 60° < x ≤ 90°                [ Q 5.3 ]


     5.4  The graph of g(x) is shifted 60° to the
            left and therefore x changes to x + 60°
$$ \hspace*{10 mm}\mathrm{g(x) = sin x\kern2mm\ } $$
$$ \hspace*{10 mm}\mathrm{h(x) = sin (x + 60°)\kern2mm\ } $$                               [ Q 5.4 ]


     5.5  f(x) must be shifted 60° to the right
            to become g(x).                                         [ Q 5.5 ]

  
Answers  6
     6.1  Amplitude = 2 and hence c = 2 and
            period = 360° and thus d = 1.                 [ Q 6.1 ]

     6.2  Amplitude = 1 and hence a = 1;
            f(60°) = 1 and thus the graph is shifted
            60° to the right so that b = − 60°              [ Q 6.2 ]

     6.3  Range : − 2 ≤ y ≤ 2                                   [ Q 6.3 ]

     6.4  −120° ≤ x ≤ − 30°                                       [ Q 6.4 ]

     6.5  The Y-axis is shifted 30° to the left so
            that the original sine graph was shifted
            30° to the right and thus x changed to x − 30°
            y = 2 sin x became y = 2 sin (x − 30°).
                                                                                [ Q 6.5 ]

  
Answers  7
     7.1  The graph of sin x = 0 at 0°
            g(x) = 0 at − 30°. g(x) was shifted 30° to
            the left and thus b = 30°.                         [ Q 7.1 ]

     7.2  Period = 360°                                             [ Q 7.2 ]

     7.3  f(x) − g(x) = 0 at x = − 150° and
            x = 30°:                                                        [ Q 7.3 ]

     7.4.1  sin (90° − x) = cos x, thus f(x) > g(x)
               x ∈ (− 150° ; 30°)                                    [ Q 7.4.1 ]

     7.4.2  For f(x).g(x) < 0, f(x) and g(x) must have
               opposite signs. f(x) < 0 if x < − 90° and
               g(x) < 0 if x < − 30°. Thus f(x) and g(x) have
               opposite signs if −90° < x < −30°.
               Thus f(x).g(x) < 0 if −90° < x < −30°.
                                                                                [ Q 7.4.2 ]

     7.4.3  f(x).g(x) ≥ 0 if −180° ≤ x ≤ −90° and
               −30° ≤ x ≤ 90°.                                     [ Q 7.4.3 ]

     7.5  f(x) has a minimum of − 1 and a maximum
            of 1. All y-values becomes 3 greater,
            therefore the range becomes
            -1+3 ≤ y ≤ 1+3,    i.e. 2 ≤ y ≤ 4.
                                                                                [ Q 7.5 ]

     7.6  f(x) is shifted 3 units downwards, so that
            y = 0 − 3 = − 3. p(x) = cos x − 3                 [ Q 7.6 ]

     7.7  f(x) is shifted 30° to the left, so that
            p = 30°. Thus q(x) = cos (x − 30°)             [ Q 7.7 ]