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Trigonometry, 2ADV EQ-Bank 5

Find the exact value of

  1. \(\sin 240^{\circ}\)   (1 mark)

  2. \(\tan \left(-120^{\circ}\right)\)   (1 mark)

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a.    \(-\dfrac{\sqrt{3}}{2}\)

b.    \(\sqrt{3}\)

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a.    \(\sin 240^{\circ}=\sin (180+60)=-\sin 60^{\circ}=-\dfrac{\sqrt{3}}{2}\)

b.    \(\tan \left(-120^{\circ}\right)=\tan 240^{\circ}\)

\(\tan 240^{\circ}=\tan (180 + 60)=\tan 60^{\circ}=\sqrt{3}\)

Trigonometry, 2ADV EQ-Bank 4

Find the exact value of

  1. \(\tan 330^{\circ}\)   (1 mark)
  2. \(\cos 510^{\circ}\)   (1 mark)

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a.    \(-\dfrac{1}{\sqrt{3}}\)

b.    \(-\dfrac{\sqrt{3}}{2}\)

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a.    \(\tan 330^{\circ}=\tan (360-30)=-\tan 30^{\circ}=-\dfrac{1}{\sqrt{3}}\)

b.    \(\cos 510^{\circ}=\cos 150^{\circ}\)

\(\cos 150^{\circ}=\cos (180-30)=-\cos 30^{\circ}=-\dfrac{\sqrt{3}}{2}\)

Trigonometry, 2ADV EQ-Bank 3

Find the exact value of

  1. \(\cos \left(-150^{\circ}\right)\)   (1 mark)

  2. \(\sin \, 585^{\circ}\)   (1 mark)

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a.    \(-\dfrac{\sqrt{3}}{2}\)

b.    \(-\dfrac{1}{\sqrt{2}}\)

Show Worked Solution

a.    \(\cos \left(-150^{\circ}\right)=\cos 210^{\circ}\)

\(\cos 210^{\circ}=\cos (180+30)=-\cos 30^{\circ}=-\dfrac{\sqrt{3}}{2}\)
 

b.    \(\sin 585^{\circ}=\sin 225^{\circ}\)

\(\sin 225^{\circ}=\sin (180+45)=-\sin 45^{\circ}=-\dfrac{1}{\sqrt{2}}\)

Trigonometry, 2ADV EQ-Bank 1

Given  \(\cot \theta=-\dfrac{12}{5}\)  and  \(\dfrac{\pi}{2}<\theta<\pi\), find the exact values of

  1. \(\sec \theta\)   (2 marks)

  2. \(\sin \theta\)   (1 mark)

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a.    \(\sec \theta=-\dfrac{13}{12}\)

b.    \(\sin \theta=\dfrac{7}{13}\)

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a.  \(\cot \theta=-\dfrac{12}{7} \ \Rightarrow \ \tan \theta=-\dfrac{7}{12}\)

\(\text{Graphically, given} \ \ \dfrac{\pi}{2}<\theta<\pi:\)
 

\(x=\sqrt{7^2+12^2}=13\)

\(\sec \theta=-\dfrac{13}{12}\)
 

b.    \(\sin \theta=\dfrac{7}{13}\)

Trigonometry, 2ADV EQ-Bank 2

Given  \(\operatorname{cosec}\,\alpha=-\dfrac{17}{8}\)  and  \(\pi<\alpha<\dfrac{3 \pi}{2}\), find the exact values of

  1. \(\sec \alpha\)   (2 marks)

  2. \(\tan \alpha\)   (1 mark)

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a.    \(\sec \alpha=-\dfrac{17}{15}\)

b.    \(\tan \alpha=\dfrac{8}{15}\)

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a.    \(\operatorname{cosec} \alpha=-\dfrac{17}{8} \ \Rightarrow \ \sin \alpha=-\dfrac{8}{17}\)

\(\text{Graphically, given} \ \ \pi<\alpha<\dfrac{3 \pi}{2}:\)
 

\(x=\sqrt{17^2-8^2}=15\)

\(\sec \alpha=-\dfrac{17}{15}\)
 

b.    \(\tan \alpha=\dfrac{8}{15}\)

Trigonometry, 2ADV T2 2025 HSC 31

The equation  \(\cos \, p x=\dfrac{1}{2}\) has 2 solutions where  \(0 \leq x \leq 2 \pi\)  and  \(p>0\). 

Find all possible values of \(p\).   (3 marks)

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\(\dfrac{5}{6} \leqslant p<\dfrac{7}{6}\)

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\(\cos (p x)=\dfrac{1}{2}\ \ \Rightarrow\ \ px=\cos ^{-1}\left(\dfrac{1}{2}\right)=\dfrac{\pi}{3}, \dfrac{5 \pi}{3}, \dfrac{7 \pi}{3}\)

♦♦♦ Mean mark 27%.

\(\text{Since there are 2 solutions in range} \ \ 0 \leq x<2 \pi:\)

\(p x=\dfrac{5 \pi}{3} \ \Rightarrow \ x=\dfrac{5 \pi}{3 p}\)

\(\dfrac{5 \pi}{3 p}\) \(\leqslant 2 \pi\)  
\(3 p\) \(\geqslant \dfrac{5}{2}\)  
\(p\) \(\geqslant \dfrac{5}{6}\)  

 
\(\text{Since}\ \ p x=\dfrac{7 \pi}{3} \ \ \text{cannot be a solution in the range} \ \ 0 \leq x \leq 2 \pi:\) 

\(p x=\dfrac{7 \pi}{3} \ \Rightarrow \ x=\dfrac{7 \pi}{3 p}\)

\(\dfrac{7 \pi}{3 p}\) \(>2 \pi\)  
\(p\) \(<\dfrac{7}{6}\)  

 
\(\therefore \dfrac{5}{6} \leqslant p<\dfrac{7}{6}\)

Trigonometry, 2ADV T2 2023 HSC 20

Find all the values of `theta`, where  `0^@ <=theta <= 360^@`, such that

`sin(theta-60^@)=-sqrt3/2`  (3 marks)

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`theta=0^@, 300^@ and 360^@`

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`sin60^@=sqrt3/2\ \ =>\ \ text{Base angle}\ =60^@`

`=>\ text{sin is negative in 3rd and 4th quadrants}`

`sin(theta-60^@)` `=180+60, 360-60`  
  `=240^@, 300^@`  

 
`theta-60^@=240^@\ \ =>\ \ theta=300^@`

`theta-60^@=300^@\ \ =>\ \ theta=360^@`

 
`text{Consider}\ theta = 0^@`

`sin(0-60^@)=sin(-60^@)=-sqrt3/2`

`:.theta=0^@, 300^@ and 360^@`

Trigonometry, 2ADV T2 SM-Bank 35

Solve the equation

`sin (2x + pi/3) = 1/2\ \ text(for)\ \ 0<= x <=pi`  (2 marks)

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`x = pi/4, (11 pi)/12`

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`sin (2x + pi/3) = 1/2`

`=>\ text(Base angle is)\ \ pi/6`

`(2x + pi/3)` `= pi/6, (5pi)/6, (13pi)/6, (17pi)/6, …`
`2x` `= – pi/6, pi/2, (11pi)/6, (15pi)/6, …`
`x` `= – pi/12, pi/4, (11pi)/12, (15pi)/12, …`

 
`:. x = pi/4, (11 pi)/12\ \ (0<=x<=pi)`

Trigonometry, 2ADV T2 EQ-Bank 9

Given  `sectheta = −37/12`  for  `0 < theta < pi`,

find the exact value of  `text(cosec)\ theta`.   (2 marks)

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`37/35`

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`sec\ theta = -37/12 => cos\ theta = -12/37`

`text(Graphically:)`

`x=sqrt(37^2-12^2)=35`
 

`:.\ text(cosec)\ theta` `= 1/(sintheta)`
  `= 1/(35/37)`
  `= 37/35`

Trigonometry, 2ADV T2 EQ-Bank 6

Given  `cot theta = −24/7`  for  `−pi/2 < theta < pi/2`, find the exact value of

  1.  `sec theta`   (2 marks)

  2.  `sin theta`   (1 mark)

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a.    `25/24`

b.    `-7/25`

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a.   `cot theta = −24/7\ \ =>\ \ tan theta=– 7/24`

`text(Graphically, given)\ −pi/2 < theta < pi/2:`

`x= sqrt(24^2 + 7^2)=25`

`sectheta= 1/(costheta)= 1/(24/25)= 25/24`
  

b.   `sintheta = −7/25`

Trigonometry, 2ADV T2 2016 HSC 11g

Solve  `sin (x/2) = 1/2`  for  `0 <= x <= 2pi.`  (2 marks)

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`x = pi/3,\ \ (5 pi)/3`

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`sin\ x/2 = 1/2\ \ text(for)\ \ 0 <= x <= 2pi`

`=>\ text(Reference angle) = pi/6,`

`x/2` `= pi/6, pi-pi/6, 2pi + pi/6, …`
  `= pi/6, (5 pi)/6, (13 pi)/6, …`

 

`:. x = pi/3,\ \ (5 pi)/3\ \ text(for)\ \ 0 <= x <= 2 pi`

Trigonometry, 2ADV T2 2007 HSC 4a

Solve  `sqrt 2\ sin\ x = 1`  for  `0 <= x <= 2 pi`.  (2 marks)

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`pi/4 ,  (3 pi)/4`

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`sqrt 2\ sin\ x` `= 1` `\ \ \ \ \ \ \ 0 <= x <= 2 pi`
`sin\ x` `= 1/sqrt 2`  

`=>sin\ pi/4 = 1/sqrt 2\ \ text(and sin is positive in)`

`text(1st/2nd quadrants,)`

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

Trigonometry, 2ADV T2 2015 HSC 12a

Find the solutions of  `2 sin theta = 1`  for  `0 <= theta <= 2 pi`.   (2 marks)

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`pi/6, (5 pi)/6`

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`2 sin theta` `= 1,\ \ \ \ 0 <= theta <= 2 pi`
`sin theta` `= 1/2`

`=> sin (pi/6) = 1/2 \ \ text(and sin is positive)`

`text(in the 1st/2nd quadrants)`

`:. theta` `= pi/6, pi-pi/6`
  `= pi/6, (5 pi)/6`

Trigonometry, 2ADV T2 2005 HSC 2a

Solve  `cos\ theta =1/sqrt2`  for  `0 ≤ theta ≤ 2pi`.   (2 marks)

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`pi/4, (7pi)/4`

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`cos\ theta = 1/sqrt2,\ \ \ 0 ≤ theta ≤ 2pi`

`text(S)text(ince)\ cos\ pi/4 = 1/sqrt2,\ \ text(and cos)`

`text(is positive in 1st/4th quadrants)`

`theta` `= pi/4, 2pi-pi/4`
  `= pi/4, (7pi)/4`

 

Trigonometry, 2ADV T2 2008 HSC 6a

Solve  `2 sin^2 (x/3) = 1`  for  `-pi <= x <= pi`.   (3 marks)

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`x = -(3pi)/4, (3pi)/4`

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♦♦ Although exact data not available, markers specifically mentioned this question was poorly answered.
`2 sin^2 (x/3)` `= 1\ \ text(for)\ \ -pi <= x <= pi`
`sin^2 (x/3)` `= 1/2`
`sin (x/3)` `= +- 1/sqrt2`

 

MARKER’S COMMENT: Many students had problems adjusting their answer to the given domain, especially when dealing with negative angles.

`text(When)\ sin (x/3) = 1/sqrt2`

`x/3` `= pi/4, (3pi)/4`
`x` `= (3pi)/4, (9pi)/4`

 

`text(When)\ sin (x/3) = – 1/sqrt2`

`x/3` `= – pi/4, -(3pi)/4`
`x` `= -(3pi)/4,  -(9pi)/4`

 

`:.\ x = -(3pi)/4\ \ text(or)\ \  (3pi)/4\ \ text(for)\ \ -pi <= x <= pi` 

Trigonometry, 2ADV T2 2011 HSC 2b

Find the exact values of  `x`  such that  `2sin x =-sqrt3`, where  `0 <= x <= 2pi`.   (2 marks)

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 `x = (4pi)/3,\ (5pi)/3`

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MARKER’S COMMENT: Better responses found the reference angle and then identified the correct quadrants, as shown clearly in the worked solution.
`2sinx` `=- sqrt3\ \ text(where)\ \ 0 <= x <= 2pi`
`sin x` `= -sqrt3/2`
`sin (pi/3)` `= sqrt3/2`

 

`text(S)text(ince)\ sin x\ text(is negative in)\ 3^text(rd) // 4^text(th)\ text(quadrants)`

`x` `= pi + pi/3,\ 2pi-pi/3`
  `= (4pi)/3,\ (5pi)/3\ \ text(radians)`

Trigonometry, 2ADV T2 2012 HSC 6 MC

What are the solutions of  `sqrt3 tanx = -1`  for  `0<=x<=2 pi`? 

  1. `(2 pi)/3\ text(and)\ (4 pi)/3` 
  2. `(2 pi)/3\ text(and)\ (5 pi)/3`
  3. `(5 pi)/6\ text(and)\ (7 pi)/6` 
  4. `(5 pi)/6\ text(and)\ (11 pi)/6` 
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`D`

Show Worked Solution
`sqrt3 tanx` `= -1`
`tanx` `= -1/sqrt3`

 `text(When)\ tanx = 1/sqrt3,\ \ x=pi/6`

`text(S)text(ince)\ tanx\ text(is negative in)\ 2^text(nd) // 4^text(th)\ text(quadrant)`

`:. x` ` = pi\-pi/6,\ 2 pi\-pi/6,\ …`
  `= (5 pi)/6,\ (11 pi)/6`

`=>  D`