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Trigonometry, 2ADV T3 2025 MET2 1 MC

A function that has a range of \([6,12]\) is

  1. \(f(x)=6+3 \cos (9 x)\)
  2. \(f(x)=6+6 \cos (3 x)\)
  3. \(f(x)=9-3 \cos (6 x)\)
  4. \(f(x)=9-6 \cos (3 x)\)
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\(C\)

Show Worked Solution

\(\text{By trial and error,}\)

\(\text{Consider option C:}\ \  f(x)=9-3 \cos (6 x)\)

\(\text{Since}\ \ -1 \leqslant \cos (6 x) \leqslant 1\)

\(\text{Range is} \ \ [-3+9,3+9]=[6,12]\)

\(\Rightarrow C\)

Trigonometry, 2ADV T3 2025 MET1 3

Let  \(f(x)=2 \cos (2 x)+1\)  over the domain \(x \in\left[0, 2 \pi \right]\).

  1. State the range of \(f(x)\).   (1 mark)

  2. Solve  \(f(x)=0\)  for \(x\).   (3 marks)

  3. Sketch the graph of  \(y=f(x)\)  for  \(x \in\left[\dfrac{\pi}{2}, \dfrac{3 \pi}{2}\right]\) on the axes below.
  4. Label the endpoints with their coordinates.   (2 marks)

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a.    \(\text{Range of } f(x):-1 \leqslant y \leqslant 3\)

b.    \(x=\dfrac{\pi}{3}, \dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{5 \pi}{3}\)

c.   

   

Show Worked Solution

a.    \(\text{Amplitude}=2 \ \ \text{about} \ \ y=1.\)

\(\text{Range of } f(x):\ -1 \leqslant y \leqslant 3\)
 

b.     \(2 \cos (2 x)+1\) \(=0\)
  \(\cos (2 x)\) \(=-\dfrac{1}{2}\)

 
\(\text{Base angle}=\dfrac{\pi}{3}\)

\(2x=\pi-\dfrac{\pi}{3}, \pi+\dfrac{\pi}{3}, \cdots\)

\(2x=\dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{8 \pi}{3}, \dfrac{10 \pi}{3}\)

  \(x=\dfrac{\pi}{3}, \dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{5 \pi}{3}\)
  

c.   

   

♦ Mean mark (c) 49%.

Trigonometry, 2ADV T3 EQ-Bank 8

A function is defined as  \(f(x)=-2\cos\Big( \dfrac{\pi x}{3} \Big). \)

  1. Determine the amplitude and period of the function.   (2 marks)
  2. Sketch the function over the domain  \( {0 \leq x \leq 2\pi}\). On your sketch, label any \(x\)-axis intersections.   (3 marks)   
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a.   \(\text{Amplitude}\ = 2\)

\(\text{Period}\ (T) = 6 \)

b.   
         

Show Worked Solution

a.   \(\text{Amplitude}\ = 2\)

\(\text{Period}\ (T) = \dfrac{2\pi}{n} = \dfrac{2\pi}{\frac{\pi}{3}} = 6 \)
 

b.   
             

Trigonometry, 2ADV T3 EQ-Bank 6

  1. Sketch the function \(y=\cos \left(\dfrac{x}{2}\right)\)  from  \(0 \leqslant x \leqslant 2 \pi\)   (1 mark)

     

  2. Find the value of \(x\) when  \(y=0.5\), for  \(0 \leqslant x \leqslant 2 \pi\)   (2 marks)

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a.    
           

b.   \(x=\dfrac{2\pi}{3}\)

Show Worked Solution

a.    
           

b. \(\quad y\) \(=\cos \Big(\dfrac{x}{2}\Big)\)
  \(0.5\) \(=\cos \Big(\dfrac{x}{2}\Big) \)
  \(\dfrac{x}{2}\) \(=\cos^{-1} (0.5) \)
  \(\dfrac{x}{2}\) \(=\dfrac{\pi}{3}, \ \dfrac{5\pi}{3}\)
  \(x\)  \(=\dfrac{2\pi}{3}\ \ (0 \leqslant x \leqslant 2 \pi) \)

Trigonometry, 2ADV T3 EQ-Bank 3

By drawing graphs on the number plane, show how many solutions exist for the equation  `cosx = |(x - pi)/4|`  in the domain  `(−∞, ∞)`  (3 marks)

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`text(4 solutions)`

Show Worked Solution

`text(Sketch:)`

`y` `= cos x`
`y` `= |(x – pi)/4|`

 
`text(Translate)\ pi\ text(units to the right: )`

`y=|x| \ => \ y=|x-pi|`

`text(Multiply by)\ 1/4 :`

`y=|x-pi| \ => \ y= 1/4 |x-pi| = |(x-pi)/4|`

`:.\ text(There are 4 solutions.)`

Trigonometry, 2ADV T3 SM-Bank 14

For the function  `f(x) = 5 cos (2 (x + pi/3)),\ \ \ -pi<=x<=pi`

  1. Write down the amplitude and period of the function  (2 marks)

  2. Sketch the graph of the function  `f(x)`  on the set of axes below. Label axes intercepts with their coordinates.

     

    Label endpoints of the graph with their coordinates.  (3 marks)

VCAA 2006 meth 4b

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  1. `text(Amplitude) = 5;\ \ \ text(Period) = pi`
  2.  
Show Worked Solution

a.   `text(Amplitude) = 5`

`text(Period) = (2 pi)/2 = pi`

 

b.   `text(Shift)\ \ y = 5 cos (2x)\ \ text(left)\ \ pi/3\ \ text(units).`

`text(Period) = pi`

`text(Endpoints are)\ \ (-pi, -5/2) and (pi,-5/2)`

Trigonometry, 2ADV T3 SM-Bank 12

State the range and period of the function

`h(x) = 4 + 3 cos ((pi x)/2).`  (2 marks)

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`text(Range:)\ \ 1<=y<= 7`

`text(Period) = 4`

Show Worked Solution
  `-1` `<= cos ((pi x)/2)<=1`
  `-3` `<=3cos ((pi x)/2)<=3`
  `1` `<= 4+ 3cos ((pi x)/2)<=7`

 
`:.\ text(Range:)\ \ 1<=y<= 7`

 
`text(Period) = (2pi)/n = (2 pi)/(pi/2) = 4`

Trigonometry, 2ADV T3 SM-Bank 9

Let   `f(x) = 2cos(x) + 1`  for  `0<=x<=2pi`.

  1. Solve the equation  `2cos(x) + 1 = 0`  for  `0 <= x <= 2pi`.  (2 marks)

  2. Sketch the graph of the function  `f(x)`  on the axes below. Label the endpoints and local minimum point with their coordinates.  (3 marks)

 

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a.    `(2pi)/3, (4pi)/3`

b.
     

Show Worked Solution
a.    `2cos(x) + 1` `= 0`
  `cos(x)` `= −1/2`

`=> cos\ pi/3 = 1/2\ text(and cos is negative)`

`text(in 2nd/3rd quadrant)`

`:.x` `= pi – pi/3, pi + pi/3`
  `= (2pi)/3, (4pi)/3`

 

b.    

Trigonometry, 2ADV T3 SM-Bank 2 MC

Let   `f(x) = 1 - 2 cos ({pi x}/2).`

The period and range of this function are respectively

  1. `4 and [−2, 2]`
  2. `4 and [−1, 3]`
  3. `1 and [−1, 3]`
  4. `4 pi and [−2, 2]`
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`B`

Show Worked Solution
`text(Period)` `= (2 pi)/n = (2pi)/(pi/2)=4`
   

`text(Amplitude = 2)`

`text{Graph centre line (median):}\ \ y=1.`

`:.\ text(Range)` `= [1 – 2, quad 1 + 2]`
  `= [−1, 3]`

`=>   B`

Trigonometry, 2ADV T3 2006 HSC 7b

A function  `f(x)`  is defined by  `f(x) = 1 + 2 cos x`.

  1. Show that the graph of  `y = f(x)`  cuts the `x`-axis at  `x = (2 pi)/3`.   (1 mark)

  2. Sketch the graph of  `y = f(x)`  for  `-pi <= x <= pi`  showing where the graph cuts each of the axes.   (3 marks)

  3. Find the area under the curve  `y = f(x)` between  `x = -pi/2`  and  `x = (2 pi)/3`.   (3 marks)

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a.    `text(Proof)\ \ text{(See Worked Solutions)}`

b.    
    

c.    `((7 pi)/6 + sqrt 3 + 1)\ text(u²)`

Show Worked Solution

a.    `f(x) = 1 + 2 cos x`

`f(x)\ text(cuts the)\ x text(-axis when)\ f(x) = 0`

`1 + 2 cos x` `= 0`
`2 cos x` `=-1`
 `cos x` `= -1/2`

`:.  x = (2 pi)/3\ …\ text(as required)`

 

b.     2UA HSC 2006 7b

 

c.    `text(Area)` `= int_(-pi/2)^((2 pi)/3) 1 + 2 cos x\ \ dx`
  `= [x + 2 sin x]_(-pi/2)^((2 pi)/3)`
  `= [((2 pi)/3 + 2 sin­ (2 pi)/3) – ((-pi)/2 + 2 sin­ (-pi)/2)]`
  `= ((2 pi)/3 + 2 xx sqrt 3/2) – ((-pi)/2 +2(- 1))`
  `= (2 pi)/3 + sqrt(3) + pi/2 + 2`
  `= ((7 pi)/6 + sqrt(3) + 2)\ text(u²)`