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Advanced Trigonometry, SMB-034

Find every angle, \(\theta\), between  \(-180^{\circ} \leq \theta \leq 180^{\circ}\), for which

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

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

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\(\text{Reference angle:}\ \ \cos\,\theta=\dfrac{\sqrt{3}}{2}\ \ \Rightarrow \ \ \theta = 30^{\circ}\)

\(\text{cos is positive in 1st/4th quadrants.}\)

\(\theta=30^{\circ}, (360-30)^{\circ}=30^{\circ}, 330^{\circ}\)

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

Advanced Trigonometry, SMB-036

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

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

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\(\theta=30^{\circ}, 330^{\circ}\)

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\(\text{Reference angle:}\ \ \cos\,\theta=\dfrac{\sqrt{3}}{2}\ \ \Rightarrow \ \ \theta = 30^{\circ}\)

\(\text{cos is positive in 1st/4th quadrants.}\)

\(\theta=30^{\circ}, (360-30)^{\circ}=30^{\circ}, 330^{\circ}\)

Advanced Trigonometry, SMB-035

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

\(\cos^{2}\theta-\cos\,\theta=0\)   (3 marks)

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\(\theta=0^{\circ},90^{\circ},270^{\circ},360^{\circ}\)

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\(\cos^{2}\theta-\cos\,\theta\) \(=0\)  
\(\cos\,\theta(\cos\,\theta-1)\) \(=0\)  

 
\(\text{If}\ \ \cos\,\theta=0\ \ \Rightarrow\ \ \theta=90^{\circ}, 270^{\circ}\)

\(\text{If}\ \ \cos\,\theta-1=0\ \ \Rightarrow\ \ \cos\,\theta=1\ \ \Rightarrow\ \ \theta=0^{\circ}, 360^{\circ}\)

\(\theta=0^{\circ},90^{\circ},270^{\circ},360^{\circ}\)

Advanced Trigonometry, SMB-033

Find every angle, \(\theta\), between  \(0^{\circ} \leq \theta \leq 360^{\circ}\), for which

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

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\(\theta=135^{\circ}, 225^{\circ}\)

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

\(\text{cos is negative in 2nd/3rd quadrants.}\)

\(\theta=(180-45)^{\circ}, (180+45)^{\circ}=135^{\circ}, 225^{\circ}\)

Advanced Trigonometry, SMB-032

Find every angle, \(\theta\), between  \(0^{\circ} \leq \theta \leq 360^{\circ}\), for which

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

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

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\(\text{Reference angle:}\ \ \cos\,\theta=\dfrac{1}{2}\ \ \Rightarrow \ \ \theta = 60^{\circ}\)

\(\text{cos is positive in 1st/4th quadrants.}\)

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

Advanced Trigonometry, 2ADV T2 SM-Bank 38

Solve  \(2\cos(2\theta) = -\sqrt{3}\)  for  \(\theta\), where  \(0^{\circ} \leq \theta \leq 180^{\circ}\).   (2 marks) 

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\(\theta=75^{\circ}, 105^{\circ}\)

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\(2\cos(2\theta)\) \(= -\sqrt{3}\)  
\(\cos(2\theta)\) \(=-\dfrac{\sqrt{3}}{2}\)  

 
\(\text{Reference angle:}\ \cos\,30^{\circ}=\dfrac{\sqrt{3}}{2}\)

\(\text{Since cos is negative in 2nd/3rd quadrants:}\)

\(2\theta\) \(= 180-30, 180+30, 360+180-30, 360+180+30,\ …\)
  \(=150^{\circ}, 210^{\circ}, 510^{\circ},\ …\) 
\(\therefore \theta\) \(=75^{\circ}, 105^{\circ}\ \ \ (0^{\circ} \leq \theta \leq 180^{\circ})\)

Advanced Trigonometry, 2ADV T2 SM-Bank 37

Solve the equation  \(\cos\left(\dfrac{3\theta}{2}\right) = \dfrac{1}{2}\)  for \(-90^{\circ} \leq \theta \leq 90^{\circ}\).   (2 marks)

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\(\theta = -40^{\circ}, 40^{\circ}\)

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\(\cos\left(\dfrac{3\theta}{2}\right) = \dfrac{1}{2}\)

\(\text{Reference angle:}\ \cos\,60^{\circ} = \dfrac{1}{2}\)

\(\text{Since cos is positive in 1st/4th quadrants:}\)

\(\dfrac{3\theta}{2}\) \(= -60^{\circ}, 60^{\circ}, 360^{\circ},\ …\)
\(\theta\) \(= -40^{\circ}, 40^{\circ}, 240^{\circ},\ …\)
  \(= -40^{\circ}, 40^{\circ}\ \ \ (-90^{\circ} \leq \theta\leq 90^{\circ})\)

Advanced Trigonometry, 2ADV T2 SM-Bank 33

Given  \(\cos\,\theta = -\dfrac{12}{37}\)  for  \(0^{\circ} \lt \theta \lt 180^{\circ}\),

find the exact value of \(\sin\,\theta\).   (2 marks)

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\(\dfrac{35}{37}\)

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\(\cos\,\theta = -\dfrac{12}{37}\)

\(\text{Graphically:}\)

\(x=\sqrt{37^2-12^2}=35\) 

\(\therefore\ \sin\,\theta=\dfrac{35}{37}\)

Advanced Trigonometry, 2ADV T2 SM-Bank 43v2

Find the exact value of

\(\cos(-240^{\circ})\).   (2 marks)

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

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\(\cos(-240^{\circ})= \cos\,120^{\circ}\)

\(\text{Reference angle:}\ 180-120=60^{\circ}\)

\(\text{Since cos is negative in 2nd quadrant:}\)

\(\cos(-240^{\circ})= -\cos\,60^{\circ}=-\dfrac{1}{2}\)

Advanced Trigonometry, 2ADV T2 2009 HSC 1e

Find the exact value of \(\theta\) such that  \(2\cos\,\theta = 1\), where  \(0^{\circ} \leq \theta \leq 90^{\circ}\).   (2 marks)

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 \(\theta = 60^{\circ}\)

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\(2 \cos\,\theta\) \(= 1\)
\(\cos\,\theta\) \(= \dfrac{1}{2}\)
\(\therefore \theta\) \(= 60^{\circ},\ \ \ \ 0^{\circ} \leq \theta \leq 90^{\circ}\)

Advanced Trigonometry, 2ADV T2 2005 HSC 2a

Solve  \(\cos\,\theta = \dfrac{1}{\sqrt{2}}\)  for  \(0^{\circ} ≤ \theta ≤ 360^{\circ}\).   (2 marks)

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\(45^{\circ}, 315^{\circ}`

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\(\cos\,\theta = \dfrac{1}{\sqrt{2}}\)  for  \(0^{\circ} ≤ \theta ≤ 360^{\circ}\)

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

\(\text{Since cos is positive in 1st/4th quadrants:}\)

\(\theta\) \(= 45^{\circ}, 360-45`
  \(= 45^{\circ}, 315^{\circ}\)

 

Advanced Trigonometry, SMB-011

  1. Determine the reference angle for \(240^{\circ}\).   (1 mark)

  2. Using part (a) or otherwise, give the exact value for \(\cos 240^{\circ}\)   (2 marks)

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a.   \(60^{\circ}\)

b.   \(-\dfrac{1}{2}\)

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a.   \(240^{\circ}\ \text{is in the 3rd quadrant.}\)

\(\text{Reference angle:}\ \ 180+\theta=240^{\circ}\ \ \Rightarrow \ \ \theta = 60^{\circ}\)
  

b.   \(\cos 60^{\circ} = \dfrac{1}{2}\)

\(\Rightarrow \ \cos \theta<0\ \ \text{in 3rd quadrant}\)

\(\therefore \cos 240^{\circ} = -\dfrac{1}{2}\)