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Advanced Trigonometry, SMB-031

Find all the values of \(\theta\), where  \(0^{\circ} \leq \theta \leq 360^{\circ}\), such that

\(\sin\,\theta=\dfrac{\sqrt{3}}{2}\)   (2 marks)

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\(\theta=60^{\circ}, 120^{\circ}\)

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\(\text{Reference angle:}\ \ \sin\,\theta=\dfrac{\sqrt{3}}{2}\ \ \Rightarrow \ \ \theta = 60^{\circ}\)

\(\text{sin is positive in 1st/2nd quadrants.}\)

\(\theta=60^{\circ}, (180-60)^{\circ}=60^{\circ}, 120^{\circ}\)

Advanced Trigonometry, SMB-030

Solve for all \(\theta\) in the range  \(0^{\circ} \leq \theta \leq 360^{\circ}\), that make the following equation correct

\(\sin\,\theta(\sin\,\theta+1)=0\)   (3 marks)

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\(\theta=0^{\circ},180^{\circ},270^{\circ},360^{\circ}\)

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\(\text{If}\ \ \sin\,\theta=0\ \ \Rightarrow\ \ \theta=0^{\circ}, 180^{\circ}, 360^{\circ}\)

\(\text{If}\ \ \sin\,\theta+1=0\ \ \Rightarrow\ \ \sin\,\theta=-1\ \ \Rightarrow\ \ \theta=270^{\circ}\)

\(\theta=0^{\circ},180^{\circ},270^{\circ},360^{\circ}\)

Advanced Trigonometry, SMB-029

Find all the values of \(\theta\), where  \(0^{\circ} \leq \theta \leq 360^{\circ}\), such that

\(\sin\,\theta=-\dfrac{1}{\sqrt{2}}\)   (2 marks)

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\(\theta=225^{\circ}, 315^{\circ}\)

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\(\text{Reference angle:}\ \ \sin\,\theta=\dfrac{1}{\sqrt{2}}\ \ \Rightarrow \ \ \theta = 45^{\circ}\)

\(\text{sin is negative in 3rd/4th quadrants.}\)

\(\theta=(180+45)^{\circ}, (360-45)^{\circ}=225^{\circ}, 315^{\circ}\)

Advanced Trigonometry, SMB-028

Find all the values of \(\theta\), where  \(0^{\circ} \leq \theta \leq 360^{\circ}\), such that

\(\sin\,\theta=-\dfrac{\sqrt{3}}{2}\)   (2 marks)

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\(\theta=240^{\circ}, 300^{\circ}\)

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\(\text{Reference angle:}\ \ \sin\,\theta=\dfrac{\sqrt{3}}{2}\ \ \Rightarrow \ \ \theta = 60^{\circ}\)

\(\text{sin is negative in 3rd/4th quadrants.}\)

\(\theta=(180+60)^{\circ}, (360-60)^{\circ}=240^{\circ}, 300^{\circ}\)

Advanced Trigonometry, SMB-027

Find all the values of \(\theta\), where  \(0^{\circ} \leq \theta \leq 360^{\circ}\), such that

\(\sin\,\theta=\dfrac{1}{2}\)   (2 marks)

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\(\theta=30^{\circ}, 150^{\circ}\)

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\(\text{Reference angle:}\ \ \sin\,\theta=\dfrac{1}{2}\ \ \Rightarrow \ \ \theta = 30^{\circ}\)

\(\text{sin is positive in 1st/2nd quadrants.}\)

\(\theta=30^{\circ}, (180-30)^{\circ}=30^{\circ}, 150^{\circ}\)

Advanced Trigonometry, 2ADV T2 2008 HSC 6a

Solve  \(2\sin^{2}\left( \dfrac{\theta}{3}\right) = 1\)  for  \(-180^{\circ} \leq \theta \leq 180^{\circ}\).   (3 marks)

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\(\theta=-135^{\circ}, 135^{\circ}\)

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\(2\sin^{2}\left( \dfrac{\theta}{3}\right)\) \(= 1\)
\(\sin^{2}\left( \dfrac{\theta}{3}\right)\) \(= \dfrac{1}{2}\)
\(\sin\left( \dfrac{\theta}{3}\right)\) \(= \pm \dfrac{1}{\sqrt{2}}\)

 

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

\(\text{When}\ \ \sin\left( \dfrac{\theta}{3}\right)= \dfrac{1}{\sqrt{2}}:\)

\(\dfrac{\theta}{3}\) \(= 45, 180-45=45^{\circ}, 135^{\circ}\)
\(\theta\) \(= 135^{\circ}, 405^{\circ}\)

 

\(\text{When}\ \ \sin\left( \dfrac{\theta}{3}\right)= -\dfrac{1}{\sqrt{2}}:\)

\(\dfrac{\theta}{3}\) \(= -45^{\circ}, -135^{\circ}\)
\(\theta\) \(= -135^{\circ}, -405^{\circ}\)

 

\(\therefore \theta=-135^{\circ}, 135^{\circ}\ \ \text{for}\ \ -180^{\circ} \leq \theta \leq 180^{\circ}\)

Advanced Trigonometry, 2ADV T2 2023 HSC 20v2

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

`sin(theta-30^@)=-1/2`   (3 marks)

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`theta=0^@, 240^@ and 360^@`

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`sin30^@=1/2\ \ =>\ \ text{Reference angle}\ =30^@`

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

`sin(theta-30^@)` `=180+30, 360-30`  
  `=210^@, 330^@`  

 
`theta-30^@=210^@\ \ =>\ \ theta=240^@`

`theta-30^@=330^@\ \ =>\ \ theta=360^@`

 
`text{Consider}\ theta = 0^@`

`sin(0-30^@)=sin(-30^@)=-1/2`

`:.theta=0^@, 240^@ and 360^@`

Advanced 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{Reference 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^@`

Advanced Trigonometry, 2ADV T2 SM-Bank 36

Solve the equation  \(\sin\left(\dfrac{x}{2}\right) = -\dfrac{1}{2}\)  for  \(360^{\circ} \leq x \leq 720^{\circ}\)   (2 marks)

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\(x =420^{\circ}, 660^{\circ}\)

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\(\sin\left(\dfrac{x}{2}\right) = -\dfrac{1}{2}\)

\(\text{Reference angle:}\ \sin\,30^{\circ}=\dfrac{1}{2}\)

\(\text{Since sin is negative in 3rd/4th quarter:}\)

\(\dfrac{x}{2}\) \(=180+30, 360-30, 360+180+30,\ …\)
  \(=210^{\circ}, 330^{\circ}, 390^{\circ},\ …\)
\(x\) \(=420^{\circ}, 660^{\circ}, 780^{\circ},\ …\)

 
\(\text{Given}\ \ 360^{\circ} \leq x \leq 720^{\circ}:\)

\(\therefore x =420^{\circ}, 660^{\circ}\)

Advanced Trigonometry, 2ADV T2 2016 HSC 11g

Solve  \(\sin\left(\dfrac{x}{2}\right)= \dfrac{1}{2}\)  for  \(0^{\circ} \leq x \leq 360^{\circ}\).   (2 marks)

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\(x = 60^{\circ}, 300^{\circ}\)

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\(\sin\left(\dfrac{x}{2}\right)= \dfrac{1}{2}\ \ \text{for}\ \ 0^{\circ} \leq x \leq 360^{\circ}\)

\(\Rightarrow\ \text{Reference angle}\ = 30^{\circ}\)

\(\dfrac{x}{2}\) \(= 30^{\circ}, 180-30, 360+30, …\)
  \(=30^{\circ}, 150^{\circ}, 390^{\circ},\ …\)

 

\(\therefore x = 60^{\circ}, 300^{\circ}\ \ \text{for}\ \ 0^{\circ} \leq x \leq 360^{\circ}\)

Advanced Trigonometry, 2ADV T2 2011 HSC 2b

Find the exact values of \(x\) such that  \(2\sin\,x =-\sqrt{3}\), where  \(0^{\circ} \leq x \leq 360^{\circ}\).   (2 marks)

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\(240^{\circ}, 300^{\circ}\)

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MARKER’S COMMENT: Better responses found the reference angle and then identified the correct quadrants, as shown.
\(2\sin\,x\) \(=-\sqrt{3}\ \ \text{where}\ \ 0^{\circ} \leq x \leq 360^{\circ}\)
\(\sin\,x\) \(= -\dfrac{\sqrt{3}}{2}\)
\(\sin\,60^{\circ}\) \(= \dfrac{\sqrt{3}}{2}\)

 

\(\text{Since}\ \sin\,x\ \text{is negative in 3rd/4th quadrants:}\)

\(x\) \(= 180+60,\ 360-60\)
  \(= 240^{\circ}, 300^{\circ}\)

Advanced Trigonometry, 2ADV T2 SM-Bank 43v4

Find the exact value of

\(\sin(-300^{\circ})\).   (2 marks)

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\(\dfrac{\sqrt{3}}{2}\)

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\(\sin(-300^{\circ})= \sin\,60^{\circ}\)

\(\text{Reference angle:}\ 60^{\circ}\)

\(\text{Since sin is positive in 1st quadrant:}\)

\(\sin(-300^{\circ})= \sin\,60^{\circ}=\dfrac{\sqrt{3}}{2}\)

Advanced Trigonometry, 2ADV T2 SM-Bank 43v3

Find the exact value of

\(\sin(-210^{\circ})\).   (2 marks)

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\(\dfrac{1}{2}\)

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\(\sin(-210^{\circ})= \sin\,150^{\circ}\)

\(\text{Reference angle:}\ 180-150=30^{\circ}\)

\(\text{Since sin is positive in 2nd quadrant:}\)

\(\sin(-210^{\circ})= \sin\,30^{\circ}=\dfrac{1}{2}\)

Advanced Trigonometry, 2ADV T2 2015 HSC 12a

Find the solutions of  \(2\sin\,\theta = 1\)  for  \(0^{\circ} \leq \theta \leq 360^{\circ}\).   (2 marks)

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\(30^{\circ}, 150^{\circ}\)

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\(2\sin\,\theta=1\ \ \Rightarrow\ \ \sin\,\theta=\dfrac{1}{2}\)

\(\text{Reference angle:}\ \sin\,30^{\circ}=\dfrac{1}{2}\)

\(\text{Since sin is positive in the 1st/2nd quadrants:}\)

\(\therefore\ \theta\) \(= 30^{\circ}, 180-30^{\circ}\)
  \(= 30^{\circ}, 150^{\circ}\)

Advanced Trigonometry, 2ADV T2 2007 HSC 4a

Solve  \(\sqrt{2}\,\sin\,x = 1\)  for  \(0^{\circ} \leq x \leq 360^{\circ}.\)   (2 marks)

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\(45^{\circ}, 135^{\circ}\)

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\(\sqrt{2}\,\sin\,x = 1\ \ \Rightarrow \ \ \sin\,x= \dfrac{1}{\sqrt{2}} \)

\(\text{Reference angle:}\ \sin\,45^{\circ} = \dfrac{1}{\sqrt{2}}\)

\(\text{Since sin is positive in 1st/2nd quadrants:}\)

\(\therefore x\) \(= 45^{\circ}, (180-45)^{\circ}\)
  \(= 45^{\circ}, 135^{\circ}\)

Advanced Trigonometry, SMB-008

  1. Determine the reference angle for \(315^{\circ}\).   (1 mark)

  2. Using part (a) or otherwise, give the exact value for  \(\sin 315^{\circ}\).   (2 marks)

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a.   \(45^{\circ}\)

b.   \(-\dfrac{1}{\sqrt{2}}\)

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a.   \(315^{\circ}\ \text{is in the 4th quadrant.}\)

\(\text{Reference angle:}\ \ 360-315=45^{\circ}\)
  

b.   \(\sin 45^{\circ} = \dfrac{1}{\sqrt{2}}\)

\(\Rightarrow \ \sin \theta<0\ \ \text{in 4th quadrant}\)

\(\therefore \sin 315^{\circ} = -\dfrac{1}{\sqrt{2}}\)